english medium ûÁ ©’í∫’ ´÷ üµ¿u ´’ç · use euclid's division lemma to show...

´’ çí∫ ∞¡ ¢√®Ωç 7 °∂œ v • ´ J 2017

★ Mathematics is a scoring subject,

plan your study schedule wisely to

achieve top-most grade.

★ Focus on studying and practicing

in the areas where you are weak.

★ Prepare a list of all important

formulae and try to understand

what each term stands for in the

formula.

★ Try to read the formulae regularly.

English Medium ûÁ ©’í∫’ ÷ üµ¿u ’ç

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4 Marks Questions

Q: Prove that 3√2 − 5√

3 is Irrational.Sol: Let us assume that 3√

2 − 5√3 is rational

pLet 3√

2 − 5√3 =

qwhere p, q are integers and q ≠ 0.

p3√

2 = + 5√3

q Squaring on both the sides

p (3√

2 )2= ( + 5 √

3 ) 2

qp2 p

9 × 2 = + 25 × 3 + 2 × × 5 √3

q2 qp2 p

18 = + 75 + 10√3 q2 q

p p2−10√

3 = + 75 − 18q q2

p2 + 57q2=

q2

p2 + 57q2 q√

3 = ( ) ( )q2 −10pp2 + 57q2

= − ( )10pqp2 + 57q2

as p and q are integers − ( )10pqis rational which makes √

3 as rational.This contradicts the fact that √

3 is Irrational∴Our assumption is wrong.Hence 3√

2 − 5√3 is Irrational.

Q: Use Euclid's division lemma to show that thesquare of an odd positive integer can be ofthe form 6q + 1 or 6q + 3 for some integer q.

Sol: Let 'a' be an odd positive integer and let b = 6

By Euclid division lemma,a = 6m + r where 0 ≤ r < 6

i.e, 'a' can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 for some integer m.As 'a' is an odd integer it takes the form 6m + 1, 6m + 3 or 6m + 5If a = 6m + 1,a2 = (6m + 1)2 = 36m2 + 12m + 1

= 6(6m2 + 2m) + 1= 6q + 1 where q is an Integer

If a = 6m + 3, a2 = (6m + 3)2 = 36m2 + 36m + 9

= 6(6m2 + 6m + 1) + 3= 6q + 3 where q is an integer

If a = 6m + 5, a2 = (6m + 5)2 = 36m2 + 60m + 25

= 6(6m2 + 10m + 4) + 1= 6q + 1 where q is an integer

Thus the square of an odd integer can be ofthe form 6q + 1 or 6q + 3 for some integer q.

Q: A is the set of factors of 16 and B is the setof factors of 24. Find A ∪ B and A ∩ Bthrough venn diagram. Verify n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Sol: A is the set of factors of 16A = {1, 2, 4, 8, 16}B is the set of factors of 24B = {1, 2, 3, 4, 6, 8, 12, 24}

i) A ∪ B = {1, 2, 3, 4, 6, 8, 12, 16, 24}ii) A ∩ B = {1, 2, 4, 8}

n(A ∪ B) = 9n(A ∩ B) = 4n(A) = 5n(B) = 8L.H.S. = n(A ∪ B) = 9

R.H.S. = n(A) + n(B) − n(A ∩ B)= 5 + 8 − 4 = 9

L.H.S = R.H.S. Hence verifiedQ: If two zeroes of the polynomial

x4 + 3x3 − 5x2 − 17x − 6 are 1 ± √2 . Find

other zeroes.Sol: 1 + √

2 and 1 − √2 are the two zeroes of the

polynomial x4 + 3x3 − 5x2 − 17x − 6Consider, [x − (1 + √

2 )] [x − (1 − √2)]

= (x − 1 − √2) (x − 1 + √

2)= (x − 1)2 − (√

2 )2= x2 − 2x + 1 − 2= x2 − 2x − 1 divide the polynomial.

x2 − 2x − 1) x4 + 3x3 − 5x2 −17x − 6 (x2 + 5x + 6x4 − 2x3 − x2− + +

5x3 − 4x2 − 17x − 65x3 − 10x2 − 5x

− + +

6x2 − 12x − 66x2 − 12x − 6− + +

0∴ x4 + 3x3 − 5x2 −17x − 6

= (x2 − 2x − 1) (x2 + 5x + 6)= (x2 − 2x − 1) (x2 + 2x + 3x + 6)= (x2 − 2x − 1)(x + 2) (x + 3)

∴ The other zeroes of the polynomial are −2, and −3.

Q: A Quadrilateral ABCD and a circle are drawnon a graph sheet with scale 1 cm = 1 unit onboth the axes. The vertices of theQuadrilateral are A(−2, 2), B(2, 2), C(2, −2)and D(−2, −2). The circle with centre (0, 0)and radius 2 units is removed from the

above Quadrilateral ABCD. Find the area ofthe remaining region in ABCD.

Sol: Given, the vertices of the Quadrilateral areA(−2, 2), B(2, 2), C(2, −2) and D(−2, −2)AB = x2 − x1 = 2 + 2 = 4 unitsBC = y2 − y1 = −2 − 2 = 4 unitsCD = x2 − x1 = −2 − 2 = 4 unitsDA = y2 − y1 = 2 + 2 = 4 units

Also, the vertices are on the lines y = 2, x = 2, y = −2 and x = −2

⇒ Each angle of the Quadrilateral ABCD is 90°∴ ABCD is a square of side 4 units

Area of the square ABCD = 4 × 4= 16 sq. units

Area of the circle with radius 2 units and cen-tre at O(0, 0) = Π r2

22= × 2 × 2788= sq. units.7

Area of the remaining part of the squareABCD = Area of the square ABCD − Area of thecircle

88 112 − 88= 16 − =

7 7 = 3.43 sq.units (approximately)

1 3 16 2 4 6

8 12 24

A B

PAPER - I

Q: If x2 + y2 = 14xy then prove that 2 log (x + y) = 4 log 2 + log x + log y

Sol: x2 + y2 = 14xyx2 + y2 + 2xy = 16xy(x + y)2 = 16xylog (x + y)2 = log 16 xy

= log 24 xylog (x + y)2 = log 24 + log x + log y

[ ... loga xy = loga x + loga y]2 log (x + y) = 4 log 2 + log x + log y

[ ... loga xm = m loga x]1Q: Find x if 3 log 5 + 2 log 3 − log 16 = log x2

1Sol: 3 log 5 + 2 log 3 − log 162

= log 53 + log 32 − log 161/2

[... m logax = logaxm]log 53 × 32

=

161/2

[... logaxy = loga x + logay xand loga = loga x − logay]y

125 × 9= log

41125

= log = log x (given)4

1125⇒ x =

4Q: Solve 5x = 7x − 3

Sol: x log10 5 = (x − 3) log10 7= x log10 7 − 3 log10 7

x log10 5 − x log10 7 = − 3 log10 7ie., x log10 7 − x log10 5 = 3 log10 7x (log10 7 − log10 5) = 3 log10 7

3 log10 7x =

log10 7 − log10 5Q: Find the H.C.F. of 136 and 250 by using

Euclid Division Algorithm.Sol: By Euclid Division Algorithm

250 = 136 × 1 + 114136 = 114 × 1 + 22 114 = 22 × 5 + 422 = 4 × 5 + 24 = 2 × 2 + 0The H.C.F of 136 and 250 is 2

Q: A = {1, 2, 3, 4, 5}, B = {2, 3, 5}Find i) A∪B and ii) A∩B. What do you infer?

Sol: A = {1, 2, 3, 4, 5}, B = {2, 3, 5}A∪B = {1, 2, 3, 4, 5} ∪ {2, 3, 5}

= {1, 2, 3, 4, 5}A∩B = {1, 2, 3, 4, 5} ∩ {2, 3, 5}

= {2, 3, 5}As B A we havei) A ∪ B = A ii) A ∩ B = B

Q: P = {p, q, r}, Q = {a, b, r}, find P − Q and Q − P. What do you notice?

Sol: P = {p, q, r}, Q = {a, b, r}P − Q = {p, q, r} − {a, b, r}

= {p, q}Q − P = { a, b, r } − {p, q, r}

= {a, b}P − Q ≠ Q − P also P − Q and Q − P are dis-

joint sets.Q: If n(A) = 8, n(B) = 12, n(A ∩ B) = 3 find

n(A ∪ B).Sol: n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

= 8 + 12 − 3 = 17Q: A = {2, 3, 5, 7}, B = {1, 3, 5, 7, 9}

Find (A ∪ B) − (A ∩ B).Sol: A = {2, 3, 5, 7}, B = {1, 3, 5, 7, 9}

A ∪ B = {2, 3, 5, 7} ∪ {1, 3, 5, 7, 9}= {1, 2, 3, 5, 7, 9}

A ∩ B = {2, 3, 5, 7} ∩ {1, 3, 5, 7, 9}= {3, 5}

(A ∪ B) − (A ∩ B) = {1, 2, 3, 5, 7, 9} − {3, 5}= {1, 2, 7, 9}

Q: If n(A − B) = 5, n(B − A) = 7, n(A ∩ B) = 3Find n(A ∪ B).

Sol:

n (A ∪ B) = n (A − B) + n (A ∩ B) + n (B − A)= 5 + 3 + 7= 15

Q: Find the relation among the sets A, B and Cfrom the following Venn diagram.

Sol: i) A Bii) A and C are disjoint sets.iii) B and C are also disjoint sets.

Q: For any quadratic polynomial ax2 + bx + c,a≠ 0. How are the graphs when i) a > 0 ii) a < 0?

Sol: The graph of the quadratic polynomialax2 + bx + c, a ≠ 0 is a parabola.If a > 0 the parabola opens upwardsand if a < 0 the parabola opens downwards.

Q: Find the zeroes of the polynomial x2 − 5 andverify the relationship between the zeroesand the coefficients.

Sol: x2 − 5 = (x + √5 ) (x − √5 )

∴ x = √5 and x = − √5 are the two zeroes

of the polynomial x2 − 5Sum of the zeroes = √5 + (− √

5)= 0

− (Coefficient of x)=

Coefficient of x2

Product of the zeroes = (√5) (−√5) = −5

Constant term= Coefficient of x2

Q: Write a polynomial of degree 10 that has 7terms. The coefficient of its first term is thesum of its degree and the number of terms.

Sol: The number of terms in the required poly-nomial = 7

Degree of the polynomial = 10The coefficient of the first term = 10 + 7 = 17

One of the polynomial that satisfies the aboveconditions is

17x10 + 2x8 − 3x7 + 5x6 − 8x5 + 2x2 + 6Q: Write i) A quadratic polynomial that has one

zero ii) A quadratic polynomial that has twozeroes.

Sol: Consider (x + 5)2 = x2 + 10x + 25The polynomial x2 + 10x + 25 has one zero

x = −5Consider (x + 3) (x + 2) = x2 + 5x + 6The polynomial x2 + 5x + 6 has two zeroes

x = −2 and x = −3.

2 Marks Questions

µ1. Use Euclid's division lemma to show that

cube of any positive integer is of the form4m, 4m + 1 or 4m + 3 for some integer m.

2. Prove that √3 + 7√2 is irrational.

3. Divide the polynomial 5x4 + 3x3 − 4x2 + 2x+ 1 by the polynomial x2 + 2x − 5 and ver-ify the division algorithm.

4. Solve the pair of linear equation3x + 2y = 8; 5x − 3y = 7by Substitution method.

5. If the nth terms of the two A.P. 9, 7, 5 ...and 24, 21, 8 ... are the same. Find thevalue of n and that term.

Practice Questions (4 Marks)

n (A − B) = 5

n (B − A) = 7

µ

AB

C

µ

⊃

⊃

Q: Find the sum and product of the zeroes 3x3 − 2x2 + 5x + 12.

Sol: If α, β, γ are the zeroes of the polynomials−bThe sum of the roots, α + β + γ = a

−(−2)= 3

2= 3−d

The product of the roots, α, β, γ = a−12

= = −43

Q: Solve 2x + 3y = 5 and 5x − 2y = 3 by substi-tution method.

Sol: 2x + 3y = 5 ....... (1)5x − 2y = 3 ....... (2)

5 − 2xFrom equation (1) y =

3Substituting in equation (2)

5 − 2x5x − 2 ( ) = 33

10 4x5x − + = 33 3

15x − 10 + 4x = 919x = 19

⇒ x = 1Substituting x = 1 in equation (1)2(1) + 3y = 5

3y = 5 − 23

y = 3

y = 1∴ x = 1, y = 1 is the solution of the given

pair of linear equationQ: Solve 2x − 3y = 1, 5x + 2y = 12 by Elimination

method.

Sol: 2x − 3y = 1 ........ (1)5x + 2y = 12 ......... (2)The L.C.M. of the Co−efficients of x in the

above pair of linear equation is 10.∴ To eliminate the variable 'x' from the

above equations, multiply equation (1) with 5and equation (2) with 2.

10x − 15y = 510x/ + 4y = 24 − − −

−19y = −19y = 1

Substituting y = 1 in equation (1)2x − 3(1) = 1

2x = 4 ⇒ x = 2∴ x = 2, y = 1 is the solution of the pair of linear

equations.Q: For what values of p, will the pair of linear

equations 2x + 3y = 1 and (3p − 1)x + (1 − 2p)y = 2p + 3 have no solution.

Sol: 2x + 3y − 1 = 0(3p − 1)x + (1 − 2p) y − (2p + 3) = 0

a1 = 2, b1 = 3, c1 = −1a2 = 3p − 1, b2 = 1 − 2p, c2 = −(2p + 3)The pair of linear equations do not have

solution.a1 b1 c1

⇒ = ≠ a2 b2 c22 3 −1

= ≠ 3p − 1 1 − 2p −(2p + 3)

2 3 =

3p − 1 1 − 2p ⇒ 2(1 − 2p) = 3(3p − 1)⇒ 2 − 4p = 9p − 3 ⇒ 9p + 4p = 2 + 3⇒ 13p = 5

5⇒ p =

13

Q: The sum of the successors of two positiveintegers is 42 and the difference of their pre-decessors is 12. Find the numbers.

Sol: Let the two positive integers be x and ythen the successors of x and y are x + 1and y + 1, and the predecessors of x and yare x − 1 and y − 1.

As per the problem(x + 1) + (y + 1) = 42⇒ x + y = 40 ................ (1)and (x − 1) − (y − 1) = 12⇒ x − y = 12 .............. (2)

Adding (1) and (2)x + y = 40x − y = 12

2x = 52x = 26Substituting x = 26 in (1)26 + y = 40⇒ y = 40 − 26 = 14∴ The required numbers are 14 and 26.

Q: If α, β are the roots of the equation x2 − 6x + 8 = 0 then the value of α2 + β2.

Sol: α, β are the roots of the equationx2 − 6x + 8 = 0a = 1, b = −6, c = 8

−bα + β = = 6a

cαβ = = 8aα2 + β2 = (α + β)2 − 2αβ

= (6)2 − 2(8)= 36 − 16 = 20

Q: The roots of the quadratic equation ax2 + bx + c = 0, (a ≠ 0) are α and β. Findthe quadratic equation in x whose roots areαβ2 and βα2.

Sol: α, β are the roots of ax2 + bx + c = 0

−b⇒ α + β = a

cαβ = a

Consider αβ2 + βα2

= αβ (α + β)c −b= ( )a a−bc= a2

Also consider (αβ2 ) + (βα2)= α3β3

c 3= ( )a∴ The Quadratic equation whose roots are αβ2

and βα2 is given byx2 − (αβ2 + βα2) x + (αβ)3 = 0

−bc cx2 − ( ) x + ( )3

= 0a2 a

⇒ a3 x2 + abc x + c3 = 0.

2 Marks Questions

Q: Solve the following pair of equations.4 1 8 3

+ = 2; + = 5 x − 1 y − 2 x − 1 y − 2

x ≠ 1, y ≠ 2Sol: The given pair of equations

4 1 + = 2 ................ (1)x − 1 y − 28 3

+ = 5 ................ (2)x − 1 y − 2

1 1Let = a and = b

x − 1 y − 2Then the equations (1) and (2) become4a + b = 2 ................ (3)8a + 3b = 5 ................ (4)Multiplying equation (3) by 2 and subtracting

equation (4)from it, we get

8a + 2b = 48a/ + 3b = 5 − − −

−b = −1⇒ b = 1

Substituting the value of b in equation (3)4a + 1 = 2

4a = 11

a = 4

1 1 1∴ a = ⇒ = 4 x − 1 4⇒ x − 1 = 4⇒ x = 5

1b = 1 ⇒ = 1

y − 2⇒ y − 2 = 1⇒ y = 3 ∴ x = 5 and y = 3.

Q: Find the sum of all three digit numbers thatends with 5. How many such numbers arethere?

Sol: The three digit numbers that ends with 5are 105, 115, 125, 135 ... 995 which is anA.P.

a1 = 105 ; a2 = 115d = 115 − 105

= 10an = a1 + (n − 1)d995 = 105 + (n − 1)10995 − 105 = (n − 1) 10890 = (n − 1)10⇒ n − 1 = 89 ⇒ n = 90There are 90 three digit numbers that endswith 5Sn = n2 [a1 + an]

90S90 = [105 + 995]

2= 45 × 1100= 49,500

Q: Shriya and Ridhi together can complete a taskin 12 hours. If Shriya works for 4 hours andRidhi for 9 hours, only half of the task can becompleted. How long would it take for each ofthem to complete the task separately?

Sol: Let the time taken by Shriya alone to com-plete the task be 'x' hours and the timetaken by Ridhi alone be 'y' hours.

Shriya takes 'x' hours to complete the task⇒ in 1 hour she completes

1 () part of the task x

i.e., in 4 hours she completes 4() parts of the taskx

Similarly, In 9 hours Ridhi completes9( ) parts of the task y

Now, as per the problem4 9 1 + = ........................ (1)x y 2Both of them, together complete the task in

12 hours12 12

So + = 1 ................ (2)x y

1 1Let = p and = q

x yThen the equations (1) and

(2) reduces to1

4p + 9q = ......... (3)2

12p + 12q = 1 ......... (4)Multiplying equation (3) by 3 and subtracting equation (4) from it, we get

312p + 27q =

212p + 12q = 1− − −

315q = − 1

21

= 21

⇒ q = 30

Substituting the value of q in equation (4)

1212p + = 1

3012 18 3

12p = 1 − = = 30 30 5

3 1p = ×

5 121

= 201 1 1

p = ⇒ = 20 x 20

⇒ x = 201 1 1

and q = ⇒ = 30 y 30

⇒ y = 30∴ Shriya alone can complete the task in 20

hours and Ridhi alone can complete the task in30 hours.

4 Marks Questions

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Nñ‰ûª© ÷ ô ´’çí∫ ∞¡ ¢√®Ωç 7 °∂œ v• ´ J 2017 °æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπ III

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπIV

Q: Find the H.C.F. of 15, 18 and 21 by primefactorization method.

Sol: 15 = 3 × 518 = 2 × 32

21 = 3 × 7H.C.F. (15, 18, 21) = 3

Q: 'The sum of the two Irrational numbers isalways Irrational' − Do you agree?

Sol: Consider two irrational numbersa = √5 and b = − √

5a + b = √5 + (− √

5)= 0 which is a Rational number.

∴ The sum of two Irrational numbers is notalways Irrational.

81Q: Expand log .

3281

Sol: log = log 81 − log 32 32

x[... loga = loga x − loga y]y= log 34 − log 25

= 4 log 3 − 5 log 2 [... loga xn = n logax]32

Q: Determine the value of log ( )2 3125532 25

Sol: log ( ) = log 2 3125 2 55 5 5

2 5 2= log () = 5 log ()2 5 2 5 5 5

= 5 [... loga a = 1]Q: What is the relationship between a, b if r = 0

in a = bq + r where a, b are positive integersand where 0 ≤ r < b for some integers q & r .

Sol: If r = 0

a = bq + r= bq + 0

i.e., a = bq⇒ b is the factor of a.

Q: Write A = {x / x is a vowel in the word 'assessment'}B = {x / x is a vowel in the word 'assignment'}in Roster form. Are these two sets equal? Ifnot what is the relation between them?

Sol: A = {a, e}B = {a, i, e}A ≠ B, but A B

Q: Write P = {2, 9, 28, 65, 126} in Set−builderform.

Sol: P = {2, 9, 28, 65, 126}= {13 + 1, 23 + 1, 33 + 1, 43 + 1, 53 + 1}∴ P = {x / x = y3 + 1, y ∈ N, y < 6 }

Q: Draw the Venn diagram to denote the set ofBoys and the set of girls of your class.

Sol:

The set of boys and the set of girls are dis−joint sets.

Q: List the subsets of the set A = { p, q, r }Sol: As the number of element in the set A is 3.

The number of subsets of A = 23 = 8They are

A1 = φ A5 = { p, q } A2 = { p } A6 = { p, r } A3 = { q } A7 = { q, r } A4 = { r } A8 = { p, q, r }

Q: If A = { 2, 3, 4, 5 }, B = { 4, 5, 6, 7 }. Find A∩Band draw the Venn diagram.

Sol: A = { 2, 3, 4, 5 }B = { 4, 5, 6, 7 }A ∩ B = { 4, 5 }

Q: "Two sets A and B are said to be equal if theyhave same number of elements." Do youagree with the statement? Explain.

Sol: Consider A = { p, q, r }and B = { a, b, c }n(A) = 3 and n(B) = 3i.e., both the sets A and B have same num-

ber of elements. But they do not have the sameelements

∴ A ≠ BTwo sets A and B are said to be equal if they

have same number of elements and also thesame elements.

i.e., A = B if A B and B A.Q: Write a polynomial of degree 5 with three

terms. How many such polynomials can bewritten?

Sol: 3x5 + 2x − 1 is a polynomial of degree 5and it has 3 terms. We can write infinitepolynomials of such type.

Q: The polynomial of degree n is given by p(x) = a0 xn + a1 x n−1 + a2 x n−2 + ....... + an−1 x + an where a0, a1, a2, ....... an are realcoefficients and a0 ≠ 0. Why do you think thecondition a0 ≠ 0 is necessary while writingthe polynomial.

Sol: p(x) = a0 xn + a1 xn−1 + a2 x n−2 + ....... +an−1 x + an is a nth degree polynomial. Ifa0 = 0 then it becomes

p(x) = a1 xn−1 + a2 xn−2 + ....... + an−1 x +an which is a polynomial of degree n − 1, and notdegree n.

∴ It is necessary to write the condition a0 ≠ 0.2

Q: Check whether is a zero of the 3

polynomial p(x) = 2x2 + 3x + 5.Sol: p(x) = 2x2 + 3x + 5

2 2 2 2p() = 2 () + 3 () + 5 3 3 3

8 71= + 2 + 5 = ≠ 0

9 92

∴ x = is not the zero of the polynomial p(x)3

Q: If the zero of the polynomial p(x) = 5x + b is−3. Find the value of b.

Sol: p(x) = 5x + b−3 is the zero of p(x)⇒ p(−3) = 0p(−3) = 5x − 3 + b = 0−15 + b = 0⇒ b = 15

Q: Write an example for a finite and an examplefor an infinite Arithmetic progressions.

Sol: If the Arithmetic progression has finite num-ber of terms, it is called as a finite A.P

Ex: 5, 8, 11, 14 ........ 32If the Arithmetic progression has infinite

terms it is called as an Infinite A.PEx: 3, 10, 17, 24, 38 ......Q: If the sum of the first ten terms of an A.P is

100 and the sum of the first eleven terms is121, find the eleventh term of the A.P

Sol: The sum of the first ten terms of an A.PS10 = 100

and the sum of the first eleven terms of the A.PS11 = 121the nth term of the A.Pan = Sn − Sn − 1∴ a11 = S11 − S10

= 121 − 100 = 21

1 Mark Questions

Boys Girls µ

1. The logarithmic form of 251/2 = 5 A) log 25 = 5 B) log 5 = 251 1

2 21 1C) log25 5 = D) log5 25 = 2 2

12. log10 = −3 in the exponential form

1000A) 10003 = 10 B) 10−3 = 1000

1 1C) 103 = D) 10−3 = 1000 1000

3. If logax = loga3 + loga2 . Then the value of xisA) 3 B) 2 C) 6 D) 5

4. The law of logarithm applied to expand 2log is5

A) logaxy = loga x + loga y xB) loga = logax − loga y y

C) loga xm = m loga x

D) loga a = 11

5. The value of log =12 161 1 1

A) B) C) 4 D) 2 16 4

6. The product of two Irrational numbers isA) Always a rational numberB) Always an Irrational numberC) Need not be an Irrational numberD) None of the above

7. In the Euclid Division Algorithm a = bq + r. Ifa = 15, b = 2 then there exists two positiveintegers q and r s.t. the values of r are

A) 0 B) 0, 1C) 0, 1, 2 D) 0, 1, 2, 3

8. (17 × 11 × 2) + (17 × 11 × 5) is aA) Prime number B) Composite numberC) Neither Prime nor CompositeD) Both Prime and Composite

9. The last digit of 625 isA) 6 B) 2 C) 5 D) 4

10. The denominator of a rational numberwhose decimal expansion terminates is ofthe form (n, m are non negative integers)A) 2n × 5m B) 2n × 3m

C) 3n × 5m D) 2n × 3m × 5n

11. logb b√b =

1 3A) B) 1 C) D) 22 212. If x = log2 5, y = log2 7 then log2 35 in terms

of x and yxA) x B) y C) xy D) y

13. The Irrational number from the following isA) 2.35 B) 2.3535.....C) 2.335335...... D) 2.353353335.......

14. A ∪ φ = A) A B) φ C) A or φ D) A and φ

15. If A B thenA) A ∪ B = A B) A ∩ B ≠ AC) A ∪ B = B D) A ∩ B = B

16. If A − B = A thenA) A B B) B C C) A and B are equal sets D) A and B are disjoint sets

17. The venn diagram to represent the set ofprimes (P) and the set of composite num-bers (C) less than 10

A) B)

C) D)

18. The venn diagram that represents A ∩ B is

A) B)

C) D)

19. Which of the following is not correct?A) A ∪ B = B ∪ A B) A ∩ B = B ∩ AC) A − B = B − A D) A − B ≠ B − A

20. Mutually Disjoint sets from the following:A) A ∪ B, A − B, AB) A ∩ B, B − A, BC) A − B, B − A, A ∩ BD) A − B, B − A, A ∪ B

21. If A = {a, b, c, d}, B = {a, e, i, o} thenA) A B B) B A C) A = B D) A ≠ B

22. P = {x/x2 = 4, 3x = 9} is a A) Infinite set B) Empty set

C) Universal set D) None of the above23. A − B =

A) {x/x ∈ A or x ∉ B}B) {x/x ∉ A and x ∈ B}C) {x/x ∈ A and x ∉ B}D) {x/x ∉ A or x ∈ B}

24. Null set is denoted byA) φ B) { φ } C) {0} D) 0

25. If A ∩ B = φ then n (A ∪ B) =A) n(A) B) n(B)C) n(A) + n(B) D) n(A) − n(B)

26. A polynomial from the following is1 1

A) x + B) x x − 1C) 3 √

x + 2 D) 2x + 3

27. If the zero of the polynomial 3x2 + k is −5then k =

A) 75 B) −75 C) 57 D) −5728. A cubic polynomial with two terms is

A) 2x3 B) 3x3 + 5 C) 3x3 + 2x + 1 D) 3x3 + 2x2 + 5x + 2

29. The degree of the polynomial 5x2 + 2x3 + 5x + 1 is

A) 5 B) 2 C) 3 D) 1 30. The product of the roots of the quadratic

equation px2 + qx + r = 0 (p ≠ 0) is−q q r q

A) B) C) D) p p p r

Objective Questions

A23

67

45

B µ

1-C2-D3-C 4-B 5-C 6-C

7-B 8-B 9-A10-A11-C 12-C

13-D 14-A15-C 16-D 17-D 18-C

19-C 20-C 21-D 22-B 23-C 24-A

25-C 26-D 27-B 28-B29-C30-C.

⊃

⊃ ⊃

⊃ ⊃

2 4 63 1 8 95 7 10

P C

I

µ

µ

µ

P C µ2

3 57

46 8 9

10

P

P

C µ2

3 57

46 8

9

4 6

10 89

23 5

7

A B

µA B µA B

µA B

I

I

P

C

⊃⊃ ⊃

Answers

Q: Find the distance between (5, 0) & (8, 0).Sol: Distance between (x1, 0) (x2, 0)

= x2 − x1Distance between (5, 0) (8, 0)

= 8 − 5 = 3 units.Q: Identify the coordinates (0, 5) & (6, 0) and

find the area of a triangle.Sol:

1 1Area of a triangle = bh = × 6 × 52 2 = 15 Sq. Units.

Q: 'C' is the mid point of A (1, 2), B (9, 10), thenfind the mid point of BC.

Sol: Mid point of A (1, 2), B (9, 10) is C

1+ 9 2 + 10 10 12= ( , ) = ( , ) = (5, 6)

2 2 2 2∴ Mid point of B (9, 10), C (5, 6) is

9 + 5 10 + 6 14 16= ( , ) = (, ) = (7, 8)

2 2 2 2Q: Show that points (a, b + c) (b, c + a) & (c,

a + b) are collinear.Sol: A (a, b + c) B (b, c + a) C (c, a + b)

x1 y1 x2 y2 x3 y3Area of ΔABC 1= x1 (y2 − y3) + x2 (y3 −− y1) + x3 (y1 − y2)2

1= a (c + a − a − b) + b (a + b − b − c) 2+ c (b + c − c − a)

1= ac − ab + ba − bc + cb − ac21

= × 0 = 0. Since, the area is zero2

Therefore, the points must be collinear.Q: From figure AP : PB = 2 : 1, find the

coordinates of A & B.

mx2 + nx1 my2 + ny1Sol: P (4, 3) = ( , )m + n m + n2(0) + 1(x) 2(y) + 1(0)

(4, 3) = ( , )2 + 1 2 + 1x 2y

(4, 3) = ( , )3 3x 2y

∴ = 4 = 33 3 9⇒ x = 12 ∴ y = = 4.52∴ A (12, 0), B (0, 4.5)

Q: For what value of x, the area of the triangleformed by the points (5, −1), (x, y), (6, 3)

11is sq.units.2Sol: Area of triangle

1= x1 (y2 − y3) + x2 (y3 − y1) + x3 (y1 − y2)211 1 = 5 (4 − 3) + x (3 + 1) + 6 (−1 − 4)2 2

11 = 5 + 4x − 30 11 = 4x − 25

4x = 11 + 254x = 36

36∴ x = = 9

4

Q:

In ΔABC, AD is the median then verify it istrue or false.

Sol: D is the mid point of B (3, −2), C(5, 2) 3 + 5 −2 + 2 8 ( , ) = ( , 0) = (4, 0)

2 2 2

∴ Area of ΔABD ie, A (4, −6), B (3, −2), D (4, 0) is1

= 4 (−2 −0) + 3 (0 + 6) + 4 (−6 + 2)21 1

= −8 +18 −16 = −6= 3 Sq.units2 2

Area of ΔADC i.e A (4, −6), D (4, 0), C (5, 2) is1

= 4 (0 − 2) + 4 (2 + 6) + 5 (−6 − 0)21 1

= − 8 + 32 − 30= −6 2 2= 3 Sq.units∴ Area of ΔABD = Area of ΔADC∴ AD is the median of ΔABC is true.

Q: D (1, 1), E (2, −3) & F (3, 4) are the midpoints of ABC, then find the A, B, C.

Sol:

D (1, 1) = (x4, y4)E (2, −3) = (x5, y5)F (3, 4) = (x6, y6)

∴ A (x1, y1) = (x6 + x5 − x4, y6 + y5 − y4) = (3 + 2 − 1, 4 − 3 − 1) = (4, 0)

B (x2, y2) = (x4 + x6 − x5, y4 + y6 − y5) = (1 + 3 − 2, 1 + 4 + 3) = (2, 8)

C (x3, y3) = (x4 + x5 − x6, y4 + y5 − y6) = (1 + 2 − 3, 1 − 3 − 4) = (0, −6)

y

o x← 6→

(6, 0)

(0, 5)

↑5

↓

A (4, −6)

B (3, −2) C (5, 2)D

A (x1, y1)

B (x2, y2) C (x3, y3)D (1, 1)

E (2, −3)F (3, 4)

Q:

In the figure ΔABC is right angled at B and Dis the mid point of BC, then find the AD.(Here AB = 6 cm, AC = 10 cm)

Sol: In ΔABC, ∠B = 90°

∴ AC2 = BC2 + AB2

102 = x2 + 62

⇒ x2 = 102 − 62 = 100 − 36 = 64 ∴ x = √

64 = 8 cm

8∴ BD = DC = = 4 cm

2In ΔABD, ∠B = 90°

AD2 = BD2 + AB2

= 42 + 62 = 16 + 36 = 52

∴ AD = √52 = √

4 × 13 = 2 √

13 cm

Q:

In the figure, AB = 8 cm, BC = 6 cm, CD = 2cm & AC = DE then BE = ?

Sol: In ΔABC, ∠B = 90°

∴ AC2 = BC2 + AB2 = 62 + 82

= 36 + 64 = 100

∴ AC = √100 = 10 cm

In ΔBED, ∠B = 90°

∴ DE2 = BD2 + BE2 (... DE = AC)AC2 = 82 + BE2

102 = 82 + BE2

∴ BE2 = 100 − 64 = 36

∴ BE = 6 cmQ: Determine the point on a line segment

7.5 cm, which divides it is a given ratio 2 : 3.Sol:

Sol: i) We draw a line segment AB = 7.5 cm, ACmaking any acute angle with AB.

ii) We mark points A1, A2, A3, A4, A5 at

equal distance i.e., AA1 = A1A2 = A2A3 = A3A4 = A4A5

iii) We join A5 to B.iv) From A2, we draw a line parallel to A5 B to

intersect AB in P. 'P' is the required pointon AB such that AP : PB = 2 : 3.

Q: A point 'O' in the interior of a rectangle ABCDis joined with each of the vertices A, B, C &D, prove that OB2 + OD2 = OC2 + OA2

Sol: In ΔODE, OD2 = OE2 + DE2

In ΔOEC, OC2 = OE2 + EC2

In ΔAOF, AO2 = AF2 + OF2

In ΔBOF, BO2 = FB2 + OF2

LHS OB2 + OD2 = OF2 + FB2 + OE2 + DE2

= OF2 + EC2 + OE2 + AF2

( ... FB = EC & DE = AF)= (OE2 + EC2) + (OF2 + AF2)= OC2 + OA2 RHS

Q: In the figure, DE I I AC & DC I I AP, prove that

BE BC = EC CP

Sol: In ΔBPA, we have DC I I APBC BD

∴ = .........(1) CP DA

(By Thales Theorem)In ΔBCA, we have DE I I AC

BE BD∴ = .........(2)

EC DABC BE

From (1) & (2) = CP EC

A

B D4x

4 C

106

8 cm

A

E

B ← 6→C ←2→D

AC

E D

1.

ABCD is trapezium with AB I I DC, if AC,BD intersect at E and ΔAED ∼ ΔBEC thenprove that AD = BC.

2. In an equilateral triangle with side ‘d’,prove that

a √3

(i) The altitude is of length = 2√

3 a2

(ii) The area of the triangle = 4

3. In ΔPQR, X, Y are points on sides PQ &PR respectively. Such that QX = RY, if ∠Q= ∠R show that XY I I QR.

Try these ...D C

A B

E

B

P

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PAPER - II

1, 2 and 4 Marks Questions

B (0, y)

A (x, 0)

P (4, 3)

1.

2

0 X

Y

A

C

BP 7.5 cm

A1

A2

A3

A4

A5A BF

E

O

CD

.

4 Marks Questions

´’ çí∫ ∞¡ ¢√®Ωç 7 °∂œ v • ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπVI

Q: A square of side 6 cm is inscribed in a circle.Find the area enclosed between the circleand the square?

Sol: In ΔABC, AC2 = AB2 + BC2

= 62 + 62

√

√

AC = 36 + 36 = 72

= √36 × 2

AC = 6√2 cm

Radius of the circle 1 1

= AO = AC = × 6√2

2 2= 3 √

2 22

Area of circle = Πr2 = (3 √2 )2

722 396

= × 9 × 2 = cm27 7

Area of the square = 6 × 6 = 36 cm2

∴ Required shaded area = Area of circle - Area of square

396= − 36

7396 − 252 144

= = cm27 7

Q: Ramesh bent a steel wire in the form ofsquare, encloses an area of 484 sq. cm. Thesame wire he bent in the form of a circle.Find the area of circle.

Sol: a2 = 484∴ a = √

484 = 22 cm

Perimeter of a square = 4 × 22 = 88 cmCircumference of the circle = 88 cm

222Πr = 88 = 2 × × r = 88

7

7 1∴ r = 88 × × = 14 cm

22 222

∴Area of circle = Πr2 = × 14 × 14 7

= 616 cm2

Q: Find the area of shaded region, where ABCDis a square of side 14 cm.

Sol: Area of square =14 × 14 = 196 sq.cm

Area of 4 circles = 4 × Πr2

22 7 7= 4 × × ×

7 2 2= 22 × 7 = 154 sq.cm

∴Area of shaded part = Area of square − Area of 4 circles

= 196 − 154 = 42 sq.cm

Q: An agricultural field in the form of a rectangleof length 20 m & width 14 m. A pit 6 m long,3 m wide and 2.5 m deep is dug in a cornerof the field and the earth taken out of the pitis spread uniformly over the remaining areaof the field. Find out the extent to which thelevel of the field has been raised.

Sol:

Volume of dig out = lbh= 6 × 3 × 2.5= 45 m3

Volume of Part I = lbh= 11 × 6 × h= 66 h m3

Volume of Part II= 14 × 14 × h m3

= 196 h m3

∴ The total volume of Part I & II= (66 h + 196 h) m3

= (262 h) m3

262 h m3 = Volume of the earth dig out= 45 m3

45∴ h = = 17.18 cm (Approximately)

262Q: Three cubes of volume 64 cm3 each are

joined end to end to form a solid. Find thesurface area of the cuboid so formed.

Sol:

l = 12 cm, b = 4 cm, h = 4 cm

∴Surface area of a cuboid= 2 (lb + bh + lh)= 2 (12 × 4 + 4 × 4 + 12 × 4)= 2 (48 + 16 + 48) = 2 × 112 = 224 cm2

Q: A cone of height 24 cm has a curved surfacearea 550 cm2. Find its volume.

Sol: Curved surface area = Πrl22

550 = × r × √r2 + 576 cm2

7550 × 7

= = r × √r2 + 57622

(25 × 7)2 = r2 (r2 + 576)r4 + 576 r2 = 625 × 49r4 + 576 r2 − 625 × 49 = 0(r2 − 49)(r2 + 625) = 0r2 + 625 ≠ 0r2 − 49 = 0 = r2 = 49 ∴ r = √

49 = 7 cm1

∴ Volume of a cone = Πr2h31 22= × × 7 × 7 × 243 7

= 1232 cm3

D

A B

O6

6

6 cm

C20 m

14 m11 m

3 m

6 m 14 m

2.5 m

III

24

r

l

4 Marks Questions

D C

7

7

A B

7

7

1. In the figure ABCD isa square with side =14 m.Find the area ofshaded region? (91 cm2)

2. Draw a circle with radius 5 cm. From a point11 cm away from its centre, construct thepair of tangents to the circle

3. Metalic spheres of radii 6 cm, 8 cm & 10 cmrespectively, we melted to form a single solidsphere. Find the radius of the resultingsphere.

4. Find the number of spherical bullets of radii is1 mm each that can be made out of cylindri-cal solid of radius 4 cm and height 6 cm.

4 4

4 4 4

Try these ...

O.

RA

P

B

C

Q

D

Q:

Cheque whether PS is bisector of ∠P ofΔPSR if PQ = 6 cm, PR = 8 cm, QS = 1.5 cm,SR = 2 cm.

PQ QS 6 1.5Sol: = ⇒ =

PR SR 8 2PS is bisector of ∠P is correct.

Q: In ΔABC such that AB BD = & ∠B = 60°, AC DC

∠C = 70°, find ∠CAD?

AB BDSol: = then AD is the bisector of ∠A.

AC DC∴ ∠A + ∠B + ∠C = 180°

∴ ∠A = 180° − 130° = 50°

∠A 50∴ ∠CAD = = = 25°

2 2Q:

In ΔABC, AB = 5.2 cm, CF = 2.4 cm, BE = 2.6cm then find AC?

1 1 Sol: × AB × FC = × AC × BE

2 21 1 × 5.2 × 2.4 = × AC × 2.62 2

5.2 × 2.4∴ AC = = 4.8 cm

2.6

Q:

In a figure find radius of the circle.Sol: OP2 = OA2 + AP2

52 = r2 + 42 ⇒ r2 = 25 − 16 = 9∴ r = √

9 = 3 cmQ:

Find the area of a shaded part.

xSol: Area of a sector = × Πr2

36090 22

= × × 7 × 7 = 38.5 cm2360 7

Q: If volumes of a two spheres ratio is 64 : 27,then find the ratios of radius.

4 4Sol: Πr1

3 : Πr23 = 64 : 27

3 3r1

3 : r23 = 43 : 33

∴ r1 : r2 = 4 : 3Q: Perimeter of a sector is 27.2 m whose

radius is 5.7 m, find the length of the sector.Sol: 2r + l = Perimeter of a sector

2(5.7) + l = 27.2l = 27.2 − 11.4 = 15.8 m

Q: In the figure ΔABCis circumscribing acircle. Find thelength of BC.

Sol: BP = 3 cm, AQ = 4 cm∴ QC = AC − AQ = 12 − 4 = 8 cm∴ BR = BP = 3 cm & QC = PC = 8 cm∴ BP + PC = BC = 3 + 8 = 11 cm

Q: The radii of bases of cylinder and cone arein the ratio 3 : 4 and their heights are in theratio 2 : 3. Can we find the ratio between thevolume of cylinder to that of cone or not?

1Sol: Volume of cylinder & cone is Πr2h : Πr2h

31Π (3)2 × 2 : × Π (4)2 × 3

318Π : 16Π ⇒ 9 : 8

Yes, we find the ratio between the volume ofcylinder to that cone is 9 : 8.

Q: Find the curved surface area of a right cir-cular cone of height 15 cm & base diameter16 cm.

Sol: l2 = r2 + h2 = 82 + 152

= 64 + 225 = 289∴ l = √

289 = 17 cm

∴ Curved surface area = Πrl= Π (8) (17) = 136Π cm2

Q: Find the maximum volume of a cone thatcan be covered out of a solid hemisphere ofradius 'r'. 1

Sol: Volume of a cone = Πr2h, 3but here h = r

1 Πr3= Πr2 × r = cubic units.

3 3

1 & 2 Marks Questions

P

Q S R

A

B D C60° 70°

A

F E

B<

>

C2.6 2.4

5.2

Q: If tan θ is not defined then θ =?Sol: θ = 90°

15Q: If tan C = ,

17is it true or not? Justify.

Opposite side 8Sol: tan C = =

Adjacent side 1515

∴ tan C = is wrong17

cos2 θQ: If = 3, then find θ.

cot2 θ − cos2 θ

Sol: cos2 θ = 3 cot2 θ − 3 cos2 θ

cos2 θ 3cos2 θ + 3 cos2 θ = 3 cot2 θ ⇒ =

cot2 θ 43

sin2 θ = 43 √

3sin θ = √

= 4 2

√3

sin 60° = ⇒ θ = 60°2

Q: log tan 1° + log tan 2° + log tan 3° + .........+ log (tan 89°) = ?

Sol: log [(tan 1° . tan 89°) (tan 2° . tan 88°)...... tan 45°]

log [(tan 1° × cot 1°) (tan 2° × cot 2°) ........tan 45°]log (1 × 1 × ........ × 1)

= log 1 = 0Q: If A + B = 90° then tan A × tan B = ?Sol: A + B = 90°

A = 90 − Bcot A = cot (90 − B)= tan Btan A . tan B = tan A × cot A = 1

Q: If α + β = 90°, α = 2β then cos2 α + sin2 β = ?Sol: 2β + β = 90°

90β = = 30°

3∴ α = 60°

∴ cos2 60° + sin2 30°

1 2 1 2 1 1= () + ( ) = +

2 2 4 42 1

= = 4 2

1 and 2 Marks Questions

A

8

90° θ

17

15B C

A4

P5Or

7 cm

7 cm

A

R Q 12 cm

B P C

←3→

←4

→ ←

→

14

l2 = r2 + 242

√

l = r2 + 576

rr

a ra

.

15

8

l = 17 cm

d = 16 cm16

r = − = 8 cm2

A C D

EB

30°60

°

1500 √ 3 m

D← q →← p →

B C

h

A

θ 90 − θ

Q: The shadow of a vertical tower on levelground increases by 15 m, when the altitudeof the sun changes from angle of elevation60° to 45°. Draw the figure.

Sol:

Q: A tower is 100 m high. Find the angle of ele-vation if its top from a point 100 √

3 m away

from its foot.Sol:

In ΔABCAB 100 1

tan θ = = = BC 100 √

3 √

3

1tan 30° =

√3

∴ θ = 30°

Q: Shiva said that if angle of elevation is 45°,then the height and distance is same, giveone example.

Sol: htan 45° = 10

h1 = ⇒ h = 10 m

10∴ Distance and heights are same if angle ofelevation is 45°.

Q: If the angle depression of an object from a 75m height tower is 30°. Then the distance ofobject from the base of tower is 25 √

3 m. Is

it correct? Sol: In ΔABC

75tan 30° =

x1 75 = ⇒ x = 75 √

3 m

√3 x

∴ Given answer 25 √3 is wrong.

Q: A tree casts a shadow 5 m long on theground, when the angle of elevation of thesun is 45°. Find the height of the tree.

Sol: In ΔABC, 5

tan 45° = x

51 =

x∴ x = 5 m

Q: A bag contains 17 cards, bearing numbers1, 2, 3, ..... 17. A card drawn at random fromthe bag. Find the probability that number onthe drawn card is prime.

Sol: Prime numbers: 2, 3, 5, 7, 11, 13, 17

Total possibilities = 7 No.of cards = 17

7∴ P (prime) =

17Q: A child has a die whose six faces show the

letters as given below E E N A D U

The die is thrown random once. What is theprobability of getting E?

2 1Sol: P(E) = =

6 3Q: Madhu said that probability of one thing is

7−2

. Is it correct? Justify your answer.

Sol: It is incorrect, because probability always inbetween 0 & 1. i.e., 0 ≤ P(E) ≤ 1.

Q: Two dice are thrown at the same time, findthe probability of getting different numberson both the dice.

30 Sol: P (Getting Different no.) =

36 5

= 6

Q: P(E) + P( E−

) = 1. Give one example.Sol: A coin is thrown, getting head P(E)

1=

2 1 1getting tail P( E

−) = 1 − =

2 2 1 1

∴ P(E) + P( E ) = + 2 2

= 1N

Q: Explain median = l + − cf × h2[ ]f

Sol: l = Lower limit of the median class

f = Frequency of the median classh = Size of the median class

cf = Cumulative frequency of the class pre-ceding of the median class

N = ΣƒiQ: If Σƒi = N = 80, Σƒiui = −26, A = 55, h = 10,

find x = ? 1

Sol: x = A + h [ Σƒiui]N

1= 55 + 10 [ × (−26)]80= 55 − 3.25 = 51.75

Q: l = 60, N = 100, F = 46, f = 20, h = 10, thenfind median.

N − cf 2

Sol: Me = l + [ ] × hf

50 − 46= 60 + ( ) × 10

204

= 60 + 2

= 60 + 2 = 62

Q: Write the formula of mode for frequency dis-tribution.

ƒ − ƒ1Sol: Mode = l + ( ) × h2ƒ − ƒ1 − ƒ2

Q: In the X class, 36 members are studying. 7members are interested in Telugu. 4 mem-bers are interested in Hindi. 15 students areinterested in Mathematics. 10 members areinterested in Social Studies. What is themode?

Sol: Mathematics.

1 & 2 Marks Questions

A

h

15 m xB C D60° 45°

100

100 √3

A

θ

B C

h

A

C10B45°

75 m

A

CxB30°

30°

x

A

C5 mB45°

Q: Show that

1 − sin A 1 + sin A 2√ + √

=

1 + sin A 1 − sin A cos A

1 − sin A 1 − sin ASol: √

× 1 + sin A 1 − sin A

(1 − sin A)2 (1 − sin A)2

√

√

= = 1 − sin2 A cos2 A1 − sin A .........(1)

cos A

1 + sin A 1 + sin A

√ × =1 − sin A 1 + sin A

(1 + sin A)2 (1 + sin A)2

√

√

= 1 − sin2 A cos2 A1 + sin A ..........(2)

cos ALHS (1) + (2)

1 − sin A 1 + sin A +

cos A cos A1 − sin A + 1 + sin A 2

= = cos A cos A

= RHSQ: Show that

cot θ cosec + 1 + = 2 sec θcosec θ + 1 cot θ

Sol: LHScot2 θ + cosec2 θ + 1 + 2 cosec θ

cot θ (cosec θ + 1)[cosec2 θ − cot2 θ = 1 cosec2 θ = 1 + cot2 θ]

2 cosec2 θ + 2 cosec θ

cot θ (cosec θ + 1)2 cosec θ [cosec θ +1]

= cot θ [cosec θ + 1]

12 ( )sin θ 2

= = cos θ cos θ

sin θ= 2 sec θ RHS

b a cos x + b sin xQ: tan x = then find = ?

a a cos x − b sin xsin x

a + b ( )a cos x + b sin x cos xSol: =

a cos x − b sin x sin xa − b ( )cos x

a + b tan x=

a − b tan x

ba + b ( )a=

ba − b ()a

a2 + b2a = a2 − b2

aa2 + b2

= a2 − b2

Finding the speed of jet...

Q: The angle of elevation of a jet plane from apoint 'A' on the ground is 60°. After a flight of30 seconds, the angle of elevation changesto 30°. If the jet plane is flying at a constantheight of 1500 √

3 m. Find the speed of jetplane. (√

3 = 1.732)Sol:

1500 √3 m

In ΔADE, tan 30° = AD1 1500 √

3 = √

3 AD∴AD = 1500 × (√3 )2 = 4500 m

1500 √3

In ΔABC, tan 60° = AC1500 √

3 √

3 = AC

⇒ AC = 1500 mDistance travelled in 30 sec = BE = AD − AC

= 4500 − 1500 = 3000 mDistance 3000

∴ Speed = = = 100 m/sectime 30

Q: The angle of elevation of the top of a tower,as seen from two points B & D situated in thesame line and at distances p & q respective-ly, from the foot of the tower, are comple-mentry. Prove that the height of the tower is√

pq .h

Sol: In ΔABC, tan θ = p∴ h = p × tan θ ...... (1)In ΔADC, htan (90 − θ) = q

hcot θ = q

⇒ h = q × cot θ ..... (2)Multiplying (1) & (2)h × h = p × tan θ × q × cot θh × h = pq (tan θ × cot θ)

h2 = pq × 1 = pq ∴ h = √pq units

Number of blue marbles...?

Q: A jar contains 36 marbles, some are greenand others are blue. If a marble is drawn atrandom from the jar, the probability that it isgreen is 2/3. Find the number of blue mar-bles in the jar.

Sol: Total number of out comes = 36Favourable out comes for green marbles= x

x 2∴ P (G) = But given P (G) =

36 3x 2 72

∴ = ⇒ x = = 2436 3 3

∴ Number of green marbles = 24∴ Number of blue marbles = 36 − 24

= 12Q: Cards marked with the numbers 3 to 102 are

placed in a box and mixed thoroughly. Onecard drawn from this box. Find the probabili-ty that the number on the card isi) even numberii) a number on the card is less than 15iii) a number which is perfect squareiv) a prime number less than 25

Sol: Cards marked with the numbers 3, 4, 5, ...102

i.e total number of cards = 100∴ Total number of out comes = 100i) Even numbers → 4, 6, 8, ...... 102

Total number of even numbers = 5050 1

∴ P(Even) = = 100 2

ii) A number on the card is less than 153, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

∴ Total number of favourable out comes = 12

12 3∴P (< 15) = = 100 25iii) Perfect squares

4, 9, 16, 25, 36, 49, 64, 81, 100∴ Total favourable out comes = 9

9∴ P (Perfect squares) =

100iv) Prime number less than 25 is 5, 7, 11, 13,

17, 19, 23Total favourable out comes = 7

7∴ P (Prime no. < 25) =

100

4 Marks Questions

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1. The quadratic equation among the following isA) x(x + 1) + 5 = (x + 3) (x − 3)B) (x + 1)3 = x3 − 5C) x2 − 3x + 2 = x(x + 5)D) 3x2 − 4 = (x + 5) (3x − 1)

2. To solve a quadratic equation ax2 + bx + c = 0where a≠ 0 the quadratic formula is given by

√

√

−b + b2 − 4ac −b − b2 − 4acA) B)

2a 2a

√

√

−b ± b2 − 4ac b ± b2 − 4acC) D)

2a 2a3. The constant to be added on both the sides of

the quadratic equation x2 − x − 6 = 0 to solveit by the method of completing the squares is

1 1 1 1A) B) C) D) 2 4 6 8

4. px2 + qx + r = 0, p ≠ 0 is a quadratic equationwhere p + q + r = 0 then the roots are

q −q r −rA) 1, B) 1, C) 1, D) 1, p p p p

5. ''The product of the two consecutive naturalnumbers is 20''. The quadratic equation thatrepresents the statement is

A) x(x + 1) = 20 B) (x − 1)x = −20C) x2 + (x + 1)2 = 20 D) (x − 1)2 + x2 = 20

6. The distance between the points (−5, 0) and(8, 0) is A) 5 Units B) 8 Units C) 13 Units D) 3 Units

7. If an = 6n + 5 then an + 1 =A) 6n + 6 B) 6n + 11 C) 6n + 15 D) n + 11

8. If an = 2n - 1 then a5 =A) 8 B) 12 C) 16 D) 10

9. The product of the terms a2 and a4 in the .1 1

progression , , 1, .... 4 2

A) a1 B) a3 C) a5 D) a710. The sum of n terms of an Arithmetic

Progression is given byn n

(i) Sn = [a1 + an] (ii) Sn = [2a1+ (n − 1)d]2 2

A) Only (i) B) Only (ii)C) Either (i) or (ii) D) Neither (i) nor (ii)

10 11. Σ (2i + 3) =

i = 1A) 100 B) 120 C) 140 D) 160

12. The next term of the A.P √12, √

27, √

48.. is

A) √52 B) √

65 C) √

70 D) √

75

13. Which of the following progression is not anA.PA) 5, 10, 15, 20.... B) 2, −3, −8, −13 ...

1 3 5 7C) −3, 3, −3, 3 ... D) , , , ...2 2 2 2

14. In a Geometric Progression an−1 is given byA) a1rn − 1 B) a1r n + 1

C) a1r n + 2 D) a1 r n − 2

SSC PUBLIC EXAMS 2017 - MODEL PAPERPAPER - I

English Version(Parts A and B)

Max. Marks: 40 Time: 2 hrs. 45 min.

Instructions:i) In the time duration of 2 hours 45 min-

utes, 15 minutes is allotted to read andunderstand the question paper.

ii) Answer the questions under Part-A on aseparate answer book.

iii) Write the answers to the questions underPart - B on the question paper itself andattach it to the answer book of Part - A.

PART - A

Max.Marks: 35 Time: 2 hrs. 15 min

NOTE:

i) Answer all the questions from the giventhree Sections I, II and III of Part - A.

ii) In Section - III, every question has inter-nal choice. Answer any one alternative.

SECTION - I

i) Answer all the questions

ii) Each question carries 1 mark.(7 ×× 1 = 7 M)

1. Find the H.C.F of 352 and 125 by prime fac-torization method.

2. Write A = {1, 5, 10, 17, 26} in Set-builderform.

3. The length of a rectangular plot is 3 m morethan twice its breadth. Express the perime-ter of the plot as a linear polynomial in x.

4. Find the Quadratic polynomial whosezeroes are a + b and a − b where a, b arereal numbers.

5. Check whether the pair of linear equationsin two variables 3x + 2y − 11 = 0 and −5y + 2x − 2 = 0 are consistent.

6. The sum n terms of an A.P. is given by Sn = n(n + 4). Find its nth term. How do yourelate the nth term 'an' and the sum of nterms 'Sn' of an A.P.?

7. If the vertices of a triangle are A(2, 3), B(−1, 5), C(−2, −3). Find its centriod.

SECTION - IIi) Answer all the questions.ii) Each question carries 2 marks.

(6 ×× 2 = 12 M)8. Verify the relationship between the zeroes of

the polynomial x2 − 16 and its coefficients byfinding its zeroes.

9. If loga6 = x and loga3 = y. Write logaa/2 interms of x and y.

10. Check whether the points (2, 3), (0, 6) and(6, −3) are collinear.

11. The sum of the reciprocals of the age ofSneha 3 years ago and 6 years from now is1/6. Find her present age.

12. If A B and B ∩ C = φ. Draw the Venndiagram showing the relationship amongthe sets A, B and C.

13. The ratio of the incomes of two persons is 5 : 4 and the ratio of their expenditures is 3 : 2. If each of them manages to saveRs.3,000 per month. Find their monthlyincome.

SECTION - IIIi) Answer all the following questions.ii) In this section, each question has internal

choice.iii) Each question carries 4 marks.iv) Answer any one alternative.

(4 ×× 4 = 16 M)14. Prove that 5√

3 − √7 is an Irrational

number.(OR)

Use Euclid's division lemma to show that thesquare of an odd positive integer is of theform 6q + 1 or 6q + 3 for some integer q.

15. Draw the graph of the polynomial p(x) = x2 − x − 6 and find its zeroes from thegraph.

(OR)Draw the graph of the following pair of linearequations in two variables and find their solu-tion from the graph3x − 2y = −14x + y = 17

16. The students of a class contribute for a pro-gramme. Each student contributed thesame amount. Had there been 15 more stu-dents in the class and each student hadcontributed Rs.40 less, the total amountcontributed would have increased fromRs.3,000 to Rs.3,200. Find the number ofstudents in the class.

(OR)

Solve the following pair of equations byreducing them to a pair of liner equations.

x + y x − y = 4 ; = −2xy xy

1 1 117. If , and are in A.P. q + r r + p p + q

Then show that p2, q2 and r2 are in A.P.(OR)

Find the co-ordinates of the points of trisec-tion of the line segment joining the points (2, 5) and (−7, 3)

PART - BInstructions:

i) Choose the correct answer and write thecorresponding alphabet (A, B, C or D) inthe given answer booklet.

ii) Answer all the questions.1

iii) Each question carries mark.2

1(10 ×× = 5 M)

2

118. The exponential form of log2 = −4 is ( )

161 2 1

A) () = −4 B) (−4)2 = 16 16

1 1C) 24 = D) 2−4 =

16 1619. The sum of two Irrational numbers ( )

A) Always an irrational numberB) Always a rational numberC) Need not be an Irrational numberD) None of the above

20. The nth term of a G.P. is given by.... ( )A) an = a1 rn B) an = a1 rn−1

C) an = a1 rn +1 D) an = a1 r2n

21. The number of zeroes of the polynomial −3Π 3Πp(x), x ∈ [ , ] from the graph 2 2 ( )

A) 2 B) 3 C) 4 D) 522. If the quadratic equation px2 + qx + r = 0

have two distinct real roots then ( )A) q2 = 4pr B) q2 > 4prC) q2 < 4pr D) p2 = 4qr

23. If n(A − B) = 57, n(A∪B) = 120, n(B − A) =48 then n (A∩B) = ( )

A) 75 B) 84 C) 51 D) 1524. The elements of the set A − B from the

following venn diagram ( )

A) A − B = {1, 2, 3, 4, 5, 12}B) A − B = {2, 4}C) A − B = {1, 5, 3, 12}D) A − B = {10, 6, 7}

25. The linear equation that represents astraight line parallel to 3x + 2y = 7 from thefollowing. ( )

A) 2x + 3y = 7 B) 6x + 4y = 14C) 6x + 4y + 14 = 0 D) 2x + 3y + 7 = 0

26. The dimensions of a rectangular plot aregiven as follows

The quadratic polynomial that represents itsarea is ( )A) 2x2 + 5 B) 2x2 + 5xC) 5x2 + 2 D) 5x2 + 2x

27. One end of the diameter of a circle whosecentre (−1, 4) is (3, 5) then its other end is

( )A) (−5, 3) B) (3, −5)C) (−5, − 3) D) (−3, −5)

⊃

1

.. . . . ..

. −1

Π−3Π

2

3Π2

−Π2

Π2

−Π

1 10 5 3

24 6

12 7

A B µ

A B

x

D 2x + 5 C

PART - B Answers

Mathematics

18-D19-C20-B

21-C22-B23-D

24-C25-B26-B

27-A.

Objective Questions

1-B2-C3-B

4-C 5-A6-C

7-B 8-C 9-B

10-C 11-C 12-D

13-C 14-D.

Answers

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SSC PUBLIC EXAMS 2017 - MODEL PAPERMathematics

Paper - II Parts A & B

Max. Marks: 40 Time: 2 Hrs. 45 Min

PART - AInstructions:

i) Read the question paper carefully andunderstand. Time allotted for 15 mnts.

ii) Write all the questions in all 4Sections.

iii) Internal choice in the Section - IV.

SECTION - I

Instructions: i) Write ALL the questions. Each question

carries 1 mark.(7 ×× 1 = 7 M)

1. In ΔABC, PQ I I BC & AP : PB = 1 : 2 thenwhat is the ratio of ΔAPQ & ΔABC areas?

2. Laxman tell that the tangent lines of the endpoints of a diameter of a circle are perpen-dicular. Are you accept this argument?

3.

In the figure AB = 6 cm, AP = 4 cm then findOP?

4. Find the value of cos2 16° − sin2 74°.5. Three cubes whose edges measure 3 cm,

4 cm & 5 cm respectively to form a singlecube. Find its edge.

6. If three coins are tossed simultaneously, thenfind the probability of getting at least onehead.

7. Raju said that the mean of a set of observa-tions is equal to their Sum divided by the totalnumber of observations. Write it in notation.

SECTION - II

Instructions: i) Write ALL the questions. Each ques-

tion carries 2 marks. (6 ×× 2 = 12 M)8. A man goes 5 km due to East and turn 12 km

due to North, then find the distance betweenstarting point to end point.

9. In the figure OA = 7−2 cm.

Find the area of a shaded part.

10. Show that tan2 θ − sin2 θ = tan2 θ × sin2 θ.

11.

What are the angles of depression from theobserving positions P & Q of the object at A?

12. A bag contains cards which are numberedfrom 5 to 90. A card is drawn at random fromthe bag. Find the probability(i) A two digit number (ii) A number which is a perfect cube.

13. A frequency distribution table, l = 30, h = 10,f = 15, f1 = 10 & f2 = 6 then find the mode.

SECTION - III

Instructions:

i) Write ALL the questions.

ii) Each question carries 4 marks.

iii) Each question having internal choice.

(4 × 4 = 16 M)

14. An aeroplane at an altitude of 1500 metersfinds that two ships are sailing towards it inthe same direction. The angles of depres-sion of the ships as observed from the aero-plane are 60° & 45° respectively. Find thedistance between the two ships.

(OR)

An urn contains 10 oranges, 7 apples and15 mangoes. A fruit is drawn at random.What is the probability of drawing(i) an orange (ii) a mango (iii) not orange (iv) not apple

15. A square water tank has its sides equal to50 m. There are four semicircular grassyplots all round it. Find the cost of turfing theplots at Rs. 2/- m2.

(OR)

A solid sphere of radius 6 cm is melted andthen cast into a small spherical balls each ofdiameter 0.6 cm. Find the number of ballsthus obtained.

16. If the median of the following distributiontable is 25, find the missing frequency.

(OR)

sin2 A − sin2 BS.T. tan2 A − tan2 B =

cos2 A . cos2 B17. Construct a triangle whose measures AB =

6.5 cm, ∠B = 50°, ∠A = 60°, and constructa triangle similar to a given triangle ABC which it sides equal to 2

−3

of corresponding sides of Δ ABC

(OR)

Draw a circle of radius 3 cm. From a point 7cm away from its centre, construct the pairof tangents to the circle.

PART - B

SECTION - IV

Instructions: i) Choose the correct answer and write

in the given brackets.

ii) Write all the questions.1

iii) Each question carries mark.2

1(10 ×× = 5 M)

2

18. If ΔABC ~ ΔDEF, ∠A = 80°, ∠F = 40° then∠C = ? ( )A) 50° B) 60° C) 70° D) 120°

19.

20. The lengths of the tangents drawn fromexternal point to a circle is ..... ( )A) equal B) unequalC) equal and unequal D) None

1 121. If tan θ + = 2 then tan2 θ + =

tan θ tan2 θ( )

A) 1 B) 2 C) 4 D) 3 22. cosec 30° + cot 45° = ? ( )

√3 + 1

A) 3 B) 2 C) D) 12√

2 xi − 25

23. If ui = , Σ fiui = 20, Σ fi = 10010

then x− = ? ( )A) 23 B) 24 C) 27 D) 25

24. If tan θ + cot θ = x = then x2 + 5 = ? ( )A) 1 B) 25 C) 0 D) 6

25. If 25 natural numbers are arranged in aAscending order. Then median of that is...

( )A) 12 B) 14 C) 13 D) 15

26. If the height of the vertical pole is 10 m. If a20 m ladder makes an angle is θ, then θ =

( )A) 0° B) 30° C) 45° D) 60°

27. A number is selected at random from thenumbers 3, 4, 6, 6, 6, 7, 7, 7, 7, 7. The prob-ability that the selected numbers is theiraverage is ... ( )

1 2 3 7A) B) C) D) 10 10 10 10

P

B

A

O

B

O A

Q P

A B C45°°

60°°

CI 0-10 10-20 20-30 30-40 40-50 total

f 5 15 20 x 2 60

O

P A Q

53

18-B19-B

20-A21-B

22-A23-C

24-D25-C

26-B27-C.

PART - B Answers

1. The points (−4, 0), (4, 0), (0, 3) are the ver-tices of a ....

A) Right triangle B) Isosceles triangleC) Equilateral triangle D) Scalene triangle

2. If the points A(1, 2), O(0, 0), C(a, b) arecollinear then

A) a = b B) a = 2b C) 2a = b D) a = −b3. The perimeter of triangle with vertices (0, 4)

(0, 0) & (3, 0) isA) 5 B) 12 C) 11 D) 7 + √5

4. The fourth vertex D of a parallelogram ABCDwhose three vertices are A(−2, 3), B(6, 7) &C(8, 3) is

A) (0, 1) B) (−1, 0) C) (0, −1) D) (1, 0)

5. The distance between the points (a cos 35°,0), (0, a cos 55°) is ...

A) a B) 2a C) 3a D) None of the above

6. The ratio in which the line segment joiningP(x1, y1), Q (x2, y2) is divided by X-axis

A) y1 : y2 B) −y1 : y2C) x1 : x2 D) −x1 : x2

7. Slope of the given coordinates (0, 0) (sin 45°,cos 45°)

2 √3 A) 1 B) 0 C) D)

√3 2

8. The coordinates of the circumcentre of thetriangle formed by the points O(0, 0), A(0, b),B(a, 0) are ...

b aA) (a, b) B) ( , )2 2a bC) ( , ) D) (b, a)2 2

9. The distance of the points (4, 8) from the X-axis is

A) 4 B) 8 C) 12 D) 4√2

10. If points (a, 0) (0, b) & (1, 1) are collinear, 1 1then + =a b

A) −1 B) 0 C) 2 D) 111. Which of the following statement is wrong?

i) All circles are similar ii) All squares are similariii) All squares are congruentiv) Two triangles are similar, if their

corresponding sides are proportionalA) i B) ii C) iii D) iv

12. Which of the following statement is correct?A) All isosceles triangles are similarB) Any two similar figures are congruentC) Any two triangles are similar, if their cor-

responding sides are equalD) Any two congruent figures are similar

13. AD is the bisector of∠A, then x =?

A) 2 B) 8 C) 6 D) 1014. In ΔABC, X, Y, Z are the midpoints of AB,

BC, CA then ratio of areas ΔABC & ΔXYZ isA) 1 : 4 B) 2 : 3 C) 3 : 2 D) 4 : 1

15. Which of the following cannot be the sidesof a right traingle?A) 6, 8, 10 B) 2, 1, √

5C) 9, 5, 7 D) 5, 12, 13

16.

Δ ABC ~ Δ DEF then perimeter of Δ DEF = A) 14 B) 18 C) 15 D) 26

AB BC17. In the Δ ABC & ΔPQR, = then they PQ QR

will be similar whenA) ∠B = ∠Q B) ∠A = ∠PC) ∠B = ∠P D) ∠A = ∠R

18. Sum of the medians in a triangle is alwaysA) Less than the perimeter of a triangleB) Greater than the perimeter of a triangleC) Equal to the perimeter of a triangleD) None of the above

19. A vertical stick 12 cm long casts a shadow8 cm long on the ground. At the same timea tower cast the shadow 40 m long on theground. Height of the tower in meters is?A) 30 B) 60 C) 45 D) 90

20. PQ II RS then

A) ∠P = ∠R B) ∠Q = ∠S C) ΔPOQ ~ ΔSOR D) PQ = SR

Objective Questions

B

5 4

A

CDx + 2 x

A

B C

D

E F

2.53

2 4

1-B 2-C 3-B 4-C

5-A6-B 7-A8-C

9-B 10-D 11-C 12-D

13-B 14-D 15-C 16-C

17-A18-A19-B 20-C

Answers

.

PR

Q

O

S

In the figure PQ = ? ( )A) 5 cm B) 8 cm C) 10 cm D) √

34 cm

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017 °æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπ IX

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπVIII ´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017 ´’ çí∫ ∞¡ ¢√®Ωç 7 °∂œ v • ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπX

v°æ: 3√2 − 5√

3 ÅØËC äéπ éπ®Ω ùÃßª ’ Ææçêu ÅEE®Ω ÷°œç îªçúÕ.

≤ƒüµ¿†: 3√2 − 5√

3 ÅØËC äéπ Åéπ ®Ω ùÃßª ’ Ææçêu ÅEÜ£œ«ç îªçúÕ.

p3√

2 − 5√3 = , q

Éçü¿’™ p, q © ’ °æ‹®Ωg Ææç êu© ’, q ≠ 0.p

3√2 = + 5√

3 q

É®Ω ’ ¢Áj °æ¤™« ´®Ω_ç îËßª ’í¬,p

(3√2 )2

= ( + 5 √3 ) 2

qp2 p

9 × 2 = + 25 × 3 + 2 × × 5 √3

q2 qp2 p

18 = + 75 + 10√3

q2 qp p2

−10√3 = + 75 − 18

q q2

p2 p2 + 57q2 + 57 = q2 q2

p2 + 57q2 q√

3 = ( ) ( )q2 −10pp2 + 57q2

= − ( )10pqp2+ 57q2

p, q © ’ °æ‹®Ωg Ææçêu© ’, é¬•öÀd, − ( )10pqÅØËC äéπ Åéπ ®Ω ùÃßª ’ Ææçêu Åçõ‰ √

3 èπÿú≈ äéπÅéπ ®Ω ùÃßª ’ Ææ çêu Å´¤ ûª ’çC.

ÉC ÅÆæûªuç. áçü¿’ éπçõ‰ √3 ÅØËC ä éπ éπ®Ω ùÃßª ’

Ææçêu ÅØË Ææû√u EéÀ N®Ω ’ü¿l¥ ¶µ«´†, Åçü¿’Íé ÉC N®Ó üµΔ ¶«Ææç, é¬•öÀd 3√

2 − 5√3 ÅØËC éπ®Ω ùÃßª ’Ææçêu

Å´¤ûª ’çC.v°æ: q àüÁjØ√ äéπ °æ‹®Ωg Ææçêu Å®· † °æ¤púø’ v°æA üµ¿† ¶‰Æ œ

°æ‹®Ωg ÆæçêuéÀ îÁç C † ´®Ω_ç 6q + 1 ™‰üΔ 6q + 3®Ω ÷°æç™ Öçô ’ç ü¿E îª ÷°æçúÕ.

≤ƒüµ¿†: a àüÁjØ√ üµ¿† ¶‰Æ œ °æ‹®Ωg Ææçêu, b = 6 Å† ’éÓçúÕ.ßª‚éÀxú˛ ¶µ«í∫ £æ…®Ω ¨Ï≠æ NCµE Å† ’ Ææ Jç*a = 6m + r, 0 ≤ r < 6, m äéπ üµ¿† °æ‹®Ωg ÆæçêuÅçõ‰ a ÅØËC 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 ™‰üΔ 6m + 5 ®Ω ÷°æç™ Öçô ’çC.é¬F a E äéπ üµ¿† ¶‰Æ œ °æ‹®Ωg Ææçêuí¬ BÆæ ’éÓ´úøç´©x a Íé´©ç 6m + 1, 6m + 3 ™‰üΔ 6m + 5®Ω ÷°æç™ Öçô ’çC.a = 6m + 1 Å®·ûËa2 = (6m + 1)2 = 36m2 + 12m + 1

= 6(6m2 + 2m) + 1= 6q + 1 (q äéπ üµ¿† °æ‹®Ωg Ææçêu)

a = 6m + 3 Å®·ûËa2 = (6m + 3)2 = 36m2 + 36m + 9

= 6(6m2 + 6m + 1) + 3= 6q + 3 (q äéπ üµ¿† °æ‹®Ωg Ææçêu)

a = 6m + 5 Å®·ûËa2 = (6m + 5)2 = 36m2 + 60m + 25

= 6(6m2 + 10m + 4) + 1= 6q + 1 (q äéπ üµ¿† °æ‹®Ωg Ææçêu)

∴ v°æA üµ¿† ¶‰Æ œ °æ‹®Ωg Ææçêu 6q + 1 ™‰üΔ 6q +3 ®Ω ÷°æç™ Öçô ’çC.

v°æ: A ÅØËC 16 éÀ îÁçC† é¬®Ω ù«ç é¬© ÆæN’A, B ÅØËC24 é À î Á ç C † é ¬ ® Ω ù« ç é ¬ © Æ æ N ’ A Å ® · û Ë¢ÁØ˛*vû√Eo UÆ œ A ∪ B, A ∩ B ÆæN’ ûª ’ ©† ’®√ßª ’çúÕ.n(A ∪ B) = n(A) + n(B) − n(A ∩ B) E ÆæJ îª ÷ úøçúÕ.

≤ƒüµ¿†: A ÅØËC 16 éÀ îÁçC† é¬®Ω ù«ç é¬© ÆæN’A.A = {1, 2, 4, 8, 16}B ÅØËC 24 éÀ îÁçC† é¬®Ω ù«ç é¬© ÆæN’A.B = {1, 2, 3, 4, 6, 8, 12, 24}

i) A ∪ B = {1, 2, 3, 4, 6, 8, 12, 16, 24}ii) A ∩ B = {1, 2, 4, 8}n(A ∪ B) = 9

n(A ∩ B) = 4n(A) = 5n(B) = 8

L.H.S. = n(A ∪ B) = 9R.H.S. = n(A) + n(B) − n(A ∩ B)

= 5 + 8 − 4 = 9L.H.S. = R.H.S.

v°æ: äéπ îªûª ’ ®Ω ’s ¥ïç ABCD, äéπ ´%ûªhç 1 ÂÆç.O’. = 1ßª‚Eö¸ ÊÆ\™ ¸í¬ Ö†o ví¬°∂ é¬T ûªçÂ°j U¨»®Ω ’. îªûª ’ ®Ω ’s ¥ïç Q®√{© ’ A (−2, 2), B (2, 2), C (2, −2), D (−2, −2). Éçü¿’™ † ’ç* Íéçvü¿ç (0, 0), ¢√u≤ƒ®Ωl¥ç2 ßª‚Eô ’xí¬ Ö†o ´%ûªh ¶µ«í¬Eo BÆ œ¢Ë »®Ω ’. N’T L† îªûª ’ ®Ω ’s ¥ï ¢Áj » ™«uEo éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: É*a †N: îªûª ’ ®Ω ’s ¥ï Q®√{© ’ A(−2, 2), B(2, 2),C(2, −2), D(−2, −2)

AB = x2 − x1 = 2 + 2 = 4 ßª‚Eô ’x BC = y2 − y1 = −2, −2 = 4 ßª‚Eô ’xCD = x2 − x1 = −2, −2 = 4 ßª‚Eô ’xDA = y2 − y1 = 2 + 2 = 4 ßª‚Eô ’x

îªûª ’ ®Ω ’s ¥ï Q®√{© ’ y = 2, x = 2, y = −2, x = −2 Í®ê ©Â°j Öçúøôç ´©x üΔE v°æA éÓùç 90° Å´¤ ûª ’çC.∴ ABCD 4 ßª‚Eô ’x ¶µº’ïçí¬ Ö†o ä éπ îªûª ’ ®ΩvÆæç.ABCD îªûª ’ ®ΩvÆæ ¢Áj¨»©uç = 4 × 4 = 16 îª.ßª‚ Eô ’x

´%ûªh Íéçvü¿ç (0, 0), ¢√u≤ƒ®Ωl¥ç 2 ßª‚Eô ’x ´%ûªh ¢Áj¨»©uç = Πr2

22= × 2 × 2788= îª.ßª‚ Eô ’x7

îªûª ’ ®Ω ’s ¥ïç™ E N’T L† ¶µ«í∫ç ¢Áj¨»©uç = îªûª ’ ®Ω ’s ¥ïç ¢Áj¨»©uç − ´%ûªh ¢Áj¨»©uç

88 112 − 88= 16 − =

7 7= 3.43 îª. ßª‚Eô ’x (Ææ ’´ ÷ ®Ω ’í¬)

v°æ: x4 + 3x3 − 5x2 − 17x − 6 ÅØË •£æ› °æ CéÀ È®çúø’¨¡⁄Ø√u© ’ 1 ± √2 Å®·ûË, N’T L† ¨¡⁄Ø√u ©† ’éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: 1 + √2, 1 − √

2 © ’ x4 + 3x3 − 5x2 − 17x − 6 Å ØË •£æ› °æCéÀîÁçC† È®çúø’ ¨¡⁄Ø√u© ’.

Åçõ‰ [x − (1 + √2 )] [x − (1 − √

2)]= (x − 1 − √

2) (x − 1 + √2)

= (x − 1)2 − (√2 )2 = x2 − 2x + 1 − 2

= x2 − 2x − 1 ûÓ •£æ› °æ CE ¶µ«Tç îª ´îª ’a.x2 − 2x − 1) x4 + 3x3 − 5x2 −17x − 6 (x2 + 5x + 6

x4 − 2x3 − x2− + +

5x3 − 4x2 − 17x − 65x3 − 10x2 − 5x

− + +

6x2 − 12x − 66x2 − 12x − 6− + +

0

∴ x4 + 3x3 − 5x2 −17x − 6 = (x2 − 2x − 1) (x2 + 5x + 6)

= (x2 − 2x − 1) (x2 + 2x + 3x + 6)= (x2 − 2x − 1)(x + 2) (x + 3)

∴ x4 + 3x − 5x2 − 17 − 6 éÀ îÁçC† N’T L†¨¡⁄Ø√u© ’ −2, −3

v°æ: x2 + y2 = 14xy Å®·ûË 2 log (x + y) = 4 log 2 + log x + log y Å E îª ÷°æç úÕ.

≤ƒ üµ¿ †: x2 + y2 = 14xy x2 + y2 + 2xy = 16xy(x + y)2 = 16xylog (x + y)2 = log 16 xy

= log 24 xylog (x + y)2 = log 24 + log x + log y

[ ... loga xy = loga x + loga y]2 log (x + y) = 4 log 2 + log x + log y

[ ... loga xm = m loga x]1 v°æ: 3 log 5 + 2 log 3 − log 16 = log x2

Å®·ûË x N© ’´ éπ† ’éÓ\çúÕ. 1≤ƒ üµ¿ †: 3 log 5 + 2 log 3 − log 16 2

= log 53 + log 32 − log 161/2

[... m logax = logaxm]log 53 × 32

=

161/2

[... logaxy = loga x + logay xand loga = loga x − logay]y

125 × 9= log

41125

= log = log x ( ü¿ û√hç ¨¡ç)4

1125⇒ x =

4 v°æ: 5x = 7x − 3 Å®·ûË x N© ’ ´† ’ éπ† ’éÓ\çúÕ.≤ƒ üµ¿ †: x log10 5 = (x − 3) log10 7

= x log10 7 − 3 log10 7x log10 5 − x log10 7 = − 3 log10 7ie., x log10 7 − x log10 5 = 3 log10 7

x (log10 7 − log10 5) = 3 log10 73 log10 7

x = log10 7 − log10 5

v°æ: ßª‚éÀxú˛ ¶µ«í∫ £æ…®Ω Ø√ußª ÷Eo Ö°æ ßÁ÷ Tç* 136,250 © í∫.≤ƒ.é¬. éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: 250 = 136 × 1 + 114136 = 114 × 1 + 22114 = 22 × 5 + 422 = 4 × 5 + 24 = 2 × 2 + 0

136, 250 © í∫.≤ƒ.é¬. = 2 v°æ: A = {1, 2, 3, 4, 5}, B = {2, 3, 5} Å®· ûË

i) A∪B, ii) A∩B © † ’ éπ † ’éÓ\ç úÕ. O’®Ω ’ ví∫£œ«ç *ç C ®√ßª ’ç úÕ.

≤ƒ üµ¿ †: A = {1, 2, 3, 4, 5}, B = {2, 3, 5}A∪B = {1, 2, 3, 4, 5} ∪ {2, 3, 5}

= {1, 2, 3, 4, 5}A∩B = {1, 2, 3, 4, 5} ∩ {2, 3, 5}

= {2, 3, 5}B A

i) A ∪ B = A ii) A ∩ B = B v°æ: P = {p, q, r}, Q = {a, b, r} Å®·ûË, P − Q,

Q − P © † ’ éπ † ’èπ◊\ E, O’Í®ç ví∫£œ«ç î √®Ó ®√ ßª ’ç úÕ.≤ƒ üµ¿ †: P = {p, q, r}, Q = {a, b, r}

P − Q = {p, q, r} − {a, b, r}= {p, q}

Q − P = { a, b, r } − {p, q, r}= {a, b}

P − Q ≠ Q − P, P − Q, Q − P © ’ Nßª·éπh ÆæN’ ûª ’© ’.

v°æ: n(A) = 8, n(B) = 12, n(A ∩ B) = 3 Å®·ûË n(A ∪ B) E éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: n(A ∪ B) = n(A) + n(B) − n(A ∩ B)= 8 + 12 − 3 = 17

v°æ: A = {2, 3, 5, 7}, B = {1, 3, 5, 7, 9} Å®·ûË (A ∪ B) − (A ∩ B) © † ’ éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: A = {2, 3, 5, 7}, B = {1, 3, 5, 7, 9}A ∪ B = {2, 3, 5, 7} ∪ {1, 3, 5, 7, 9}

= {1, 2, 3, 5, 7, 9}A ∩ B = {2, 3, 5, 7} ∩ {1, 3, 5, 7, 9}

= {3, 5}(A ∪ B) − (A ∩ B)

= {1, 2, 3, 5, 7, 9} − {3, 5}= {1, 2, 7, 9}

v°æ: n(A − B) = 5, n(B − A) = 7, n(A ∩ B) = 3 Å®·ûËn(A ∪ B) E éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †:

n (A ∪ B) = n (A − B) + n (A ∩ B) + n (B − A)= 5 + 3 + 7= 15

v°æ: éÀçC ¢ÁØ˛ *vûªç † ’ç* ÆæN’ ûª ’© ’ A, B, C © ´ ’üµ¿uÆæç•ç üµΔEo ûÁ©°æçúÕ.

≤ƒ üµ¿ †: i) A Bii) A ´ ’Jßª· C © ’ Nßª·éπh ÆæN’ ûª ’© ’.iii) B ´ ’Jßª· C © ’ èπÿú≈ Nßª·éπh ÆæN’ ûª ’© ’.

v°æ: i) a > 0, ii) a < 0 Å®· † °æ¤púø’ ´®Ω_ •£æ› °æC ax2 + bx + c, a ≠ 0 © Í®ë« *vû√© ’ á™«Öçö«ßÁ÷ ûÁ© °æçúÕ.

≤ƒ üµ¿ †: ax2 + bx + c, a ≠ 0 ´®Ω_ •£æ› °æC Í®ë« *vûªçäéπ °æ®√ ´ ©ßª ’ç

a > 0 Å®· † °æ¤púø’ Â°j¢Áj°æ¤ '∪' ´îËa °æ®√ ´ ©ßª ’çÅ´¤ ûª ’çC. a < 0 Å®· † °æ¤púø’ éÀçC ¢Áj°æ¤ '∩' ´îËa°æ®√ ´ ©ßª ’ç Å´¤ ûª ’çC.

v°æ: x2 − 5 ´®Ω_ •£æ› °æC ¨¡⁄Ø√u ©† ’ éπ† ’éÓ\çúÕ. üΔE¨¡⁄Ø√u ©èπ◊, í∫ ’ù é¬ ©èπ◊ ’üµ¿u Ææç•ç üµΔEo ÆæJ îª ÷ úøçúÕ.

≤ƒ üµ¿ †: x2 − 5 = (x + √5 ) (x − √5 )

∴ x = √5 , x = − √5 © ’ ´®Ω_ •£æ› °æC x2 − 5

ßÁ·éπ\ ¨¡⁄ Ø√u © ’ ¨¡⁄Ø√u© ¢Á·ûªhç = √5 + (− √

5)= 0

− ( x ßÁ·éπ\ í∫ ’ùéπç)=

x2 ßÁ·éπ\ í∫ ’ùéπç¨¡⁄Ø√u© ©•l¥ç = (√5) (−√

5) = −5Æ œn®Ω °æü¿ç

= x2 ßÁ·éπ\ í∫ ’ùéπç

v°æ: °æJ ÷ùç 10 Å®·u, 7 °æüΔ©’ Ö†o •£æ› °æ CE®√ßª ’çúÕ. ¢Á·ü¿öÀ °æü¿ç ßÁ·éπ\ í∫’ùéπç •£æ› °æC °æJ ´ ÷ùç, °æüΔ© Ææçêu ¢Á·û√h EéÀ Ææ ÷ †¢Á ’i Öçú≈L.

≤ƒ üµ¿ †: •£æ› °æC °æüΔ© Ææçêu = 7•£æ« °æC °æJ ´ ÷ùç = 10

∴ ¢Á·ü¿öÀ °æü¿ç í∫ ’ùéπç = 10 + 7 = 17 17x10 + 2x8 − 3x7 + 5x6 − 8x5 + 2x2 + 6

Å™«çöÀ äéπ •£æ› °æC. v°æ: i) äéπ ¨¡⁄†uç Ö†o ´®Ω_ •£æ› °æ CE ®√ßª ’çúÕ.

ii) È®çúø’ ¨¡⁄Ø√u© ’ Ö†o ´®Ω_ •£æ› °æ CE ®√ßª ’çúÕ.≤ƒ üµ¿ †: (x + 5)2 = x2 + 10x + 25

x2 + 10x + 25 ´®Ω_ •£æ› °æC x = −5 ÅØË äÍé¨¡⁄Ø√uEo éπLT ÖçC

(x + 3) (x + 2) = x2 + 5x + 6x2 + 5x + 6 ´®Ω_ •£æ› °æC È®çúø’ ¨¡⁄Ø√u © † ’ éπLTÖçC. ÅN x = −3, x = −2.

í∫ ùÀ ûªçÊ°°æ®˝ - I

4 ´÷®Ω’\© v°æ ¡o©’

2 ´÷®Ω’\© v°æ ¡o©’

⊃

n (A − B) = 5

n (B − A) = 7

µ

AB

C

µ

⊃

1 3 16 2 4 6

8 12 24

A B µ

v°æ: 15, 18, 21 © í∫.≤ƒ. é¬.† ’ v°æüµΔ† é¬®Ω ù«ç é¬© ©•l¥°æü¿l¥ A™ éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: 15 = 3 × 518 = 2 × 32

21 = 3 × 715, 18, 21 © í∫.≤ƒ.é¬. = 3

v°æ: ' È ® ç ú ø ’ é π ® Ω ù Ã ß ª ’ Æ æ ç ê u © ¢ Á · û ª h ç á © x ° æ ¤ p ú ø ÷Åéπ®ΩùÃßª ’ Ææçêu Å´¤ûª ’çC— – O’ Ææ´ ÷ üµΔ Ø√EoÆæ´ ’ Jnç îªçúÕ.

≤ƒ üµ¿ †: È®çúø’ éπ®Ω ùÃßª ’ Ææçêu© ’ a = √5 , b = − √5

Å† ’èπ◊çõ‰ a + b = √5 + (− √

5) = 0 äéπ Åéπ ®Ω ùÃßª ’ Ææçêu∴ È®çúø’ éπ®Ω ùÃßª ’ Ææçêu© ¢Á·ûªhç á©x °æpúø÷ éπ®Ω ùÃßª ’

Ææçêu Å†úøç ÆæJé¬ü¿’.Åçõ‰ È®çúø’ éπ®Ω ùÃßª ’ Ææçêu© ¢Á·ûªhç á©x °æ¤púø÷

éπ®Ω ùÃßª ’ Ææçêu é¬éπ §Ú ´îª ’a.81 v°æ: log † ’ NÆæh Jç îªçúÕ.32

81≤ƒ üµ¿ †: log = log 81 − log 32 32

x[... loga = loga x − loga y]y= log 34 − log 25

= 4 log 3 − 5 log 2 [... loga xn = n logax]32 v°æ: log ( ) N© ’ ´† ’ í∫ùÀç îªçúÕ.

2 3125532 25

≤ƒ üµ¿ †: log ( ) = log 2 3125 2 55

5 52 5 2= log () = 5 log ()2 5 2 5

5 5= 5 [... loga a = 1]

v°æ: a, b © ’ ü µ¿† °æ‹ ®Ωg Ææç êu© ’ Å®·ûË ßª‚éÀxú˛¶µ«í∫£æ…®Ω Ø√ußª ’ç a = bq + r, 0 ≤ r < b ™ r =0 Å®·ûË a, b © ´ ’üµ¿u Ææç•çüµ¿ç àN’öÀ?

≤ƒ üµ¿ †: a = bq + r r = 0 Å®·ûËa = bq

Åçõ‰ b, a ßÁ·éπ\ é¬®Ω ù«çéπç Å´¤ ûª ’çC v°æ: P = {2, 9, 28, 65, 126} Ææ N’ A E Ææ N’ A E®√t ù

®Ω ÷°æç ™ ®√ ßª ’ç úÕ.≤ƒ üµ¿ †: P = {2, 9, 28, 65, 126}

= {13 + 1, 23 + 1, 33 + 1, 43 + 1, 53 + 1}∴ P = {x / x = y3 + 1, y ∈ N, y < 6 }

v°æ: O’ §ƒ®∏Ω ¨»© ¶«© ’®Ω ’, ¶«L éπ© ÆæN’ ûª ’ ©† ’ ¢ÁØ˛*vûªç üΔy®√ îª ÷°æçúÕ.

≤ƒ üµ¿ †:

v°æ: ´‚úø’ °æüΔ© ’ Ö†o 5 ´ °æJ ´ ÷ù •£æ› °æ CE®√ßª ’çúÕ. É™« áEo •£æ› °æ ü¿’ ©† ’ ®√ßÁ·îª ’a?

≤ƒ üµ¿ †: 3x5 + 2x − 1 •£æ› °æC °æJ ´ ÷ùç 5. •£æ› °æ C™ ´‚úø’ °æüΔ© ’ ÖØ√o®·. ´‚úø’ °æüΔ© ’ÖçúÕ, 5 ´ °æJ ´ ÷ùç í∫© •£æ› °æ ü¿’ ©† ’ î √™«(Å†çûªç) ®√ßÁ·îª ’a.

v°æ: p(x) = 5x + b •£æ› °æC ßÁ·éπ\ ¨¡⁄†uç −3 Å®·ûË'b' N© ’ ´† ’ éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: p(x) = 5x + bx = −3 p(x) ßÁ·éπ\ ¨¡⁄†uçÅçõ‰ p(−3) = 0p (−3) = 5x − 3 + b = 0

− 15 + b = 0⇒ b = 15

v°æ: äéπ éÓùç üΔE Ææç°æ‹ ®Ωéπ éÓùç È®çúÕç ûª© éπçõ‰ 3°áèπ◊\´ ÖçC. Ñ N≠æßª ÷Eo Í®&ßª ’ ÆæO’ éπ ®Ω ù«©

ïûªí¬ ûÁL ßª ’ ñ‰ ßª ’çúÕ. ≤ƒ üµ¿ †: x + y = 180° ...... (1)

x = 2y + 3i.e., x − 2y − 3 = 0 ..... (2)

v°æ: Åçéπ v¨Ï ú µÕ ™ E 'n' °æüΔ© ¢Á·ûªhç éπ† ’éÓ\´ ú≈ EéÀÆæ ÷vûªç ®√ßª ’çúÕ.

≤ƒüµ¿†: Åçéπ v¨Ï ú µÕ ™ E 'n' °æüΔ© ¢Á·ûªhçnSn = [2 a1 + (n − 1)d]2

a1: Åçéπ v¨Ï ú µÕ ™ E ¢Á·ü¿öÀ °æü¿çd: ≤ƒ´ ÷ †u ¶µ‰ü¿ç

v°æ: °æJ N’ûª Åçéπ v¨Ï ú µÕéÀ äéπöÀ, Å†çûª Åçéπv¨Ïú µÕéÀ äéπöÀÖüΔ £æ« ®Ω ù© ’ ®√ßª ’çúÕ.

≤ƒüµ¿†: °æJ N’ûª Ææçêu™ °æüΔ ©† ’ éπL T† Ö†o v¨Ïú µÕE°æJ N’ûª v¨Ïú µÕ Åçö«®Ω ’.

ÖüΔ: 5, 8, 11, 14.....32 ★ °æüΔ© Ææçêu Å°æ J N’ ûª ¢Á ’i† v¨Ïú µÕE Å†çûª v¨Ïú µÕ

Åçö«®Ω ’. ÖüΔ: 3, 10, 17, 24, 31, 38..... v°æ: ´‚© Gçü¿’´¤ Íéçvü¿çí¬ Ö†o ´%ûªh ¢√u≤ƒ EéÀ

îÁçC† *´J Gçü¿’´¤ (−5, 3) Å®·ûË È®çúÓ *´JGçü¿’´¤† ’ éπ† ’éÓ\çúÕ.

≤ƒ üµ¿ †: ¢√uÆæç™ E È®çúÓ *´J Gçü¿’´¤ (x2, y2)Å† ’èπ◊çõ‰ ´%ûªh Íéçvü¿ç Åçõ‰ ´‚©Gçü¿’´¤ (−5, 3), (x2, y2) © ´ ’ üµ¿u Gç ü¿’ ´¤ Å ´¤ ûª ’ç C.

x1 + x2 y1 + y2( , ) = (0, 0)2 2

−5 + x2 3 + y2( , ) = (0, 0)2 2 −5 + x2⇒ = 0 ⇒ x2 = 5

23 + y2 = 0 ⇒ y2 = −3

2 ∴ ¢√uÆæç™ E È®çúÓ *´J Gçü¿’´¤ (5, −3)

1 ´÷®Ω’\ v°æ ¡o©’

11. 252 = 5 † ’ Ææç´ ®Ω_ ´ ÷ † ®Ω ÷°æç™ ®√ÊÆh...

A) log 25 = 5 B) log 5 = 251 1 2 2

1 1C) log25 5 = D) log5 25 = 2 21

2. log10 = −3 °∂æ ÷û√çéπ ®Ω ÷°æç™ ®√ÊÆh...1000

A) 10003 = 10 B) 10−3 = 10001 1C) 103 = D) 10−3 =

1000 10003. logax = loga3 + loga2 Å®·ûË x =

A) 3 B) 2 C) 6 D) 52

4. log NÆæh ®Ωù éÀçü¿ É*a† à Ææç ®Ω_ ÷ † Ø√ußª ’ç5üΔy®√ ï®Ω ’ í∫ ’ ûª ’çC?

A) logaxy = loga x + loga y xB) loga = logax − loga y y

C) loga xm = m loga x

D) loga a = 11

5. log =1216

1 1 1A) B) C) 4 D) 2 16 4

6. È®çúø’ éπ®Ω ùÃßª ’ Ææçêu© ©•l¥ç ...A) á©x °æ¤púø÷ Åéπ ®Ω ùÃßª ’ Ææçêu Å´¤ûª ’çC.B) á©x °æ¤púø÷ éπ®Ω ùÃßª ’ Ææçêu Å´¤ûª ’çC.C) á©x °æ¤púø÷ éπ®Ω ùÃßª ’ Ææçêu é¬† ´ Ææ®Ωç ™‰ü¿’D) àD é¬ü¿’

7. ßª‚éÀxú˛ ¶µ«í∫ £æ…®Ω ¨Ï≠æ NCµ a = bq + r ™ a =15, b = 2 Å®·ûË 'r' N© ’ ´© ’A) 0 B) 0, 1C) 0, 1, 2, D) 0, 1, 2, 3

8. (17 × 11 × 2) + (17 × 11 × 5)

A) v°æüµΔ† Ææçêu B) Ææçßª·éπh ÆæçêuC) v°æüµΔ† Ææçêu ™‰üΔ Ææçßª·éπh ÆæçêuD) v°æüµΔ† Ææçêu, Ææçßª·éπh Ææçêu

9. 625 ™ *´J ÅçÈé?A) 6 B) 2 C) 5 D) 4

10. Åçûª ´ ’ßË’u ü¿¨»ç¨¡ ®Ω ÷°æç Ö†o Åéπ ®Ω ùÃßª ’ Ææçêu£æ…®√ EéÀ îÁçC† v°æüµΔ† é¬®Ω ù«ç é¬© ©•l¥ ®Ω ÷°æç(m, n© ’ ®Ω ’ù‰ ûª®Ω °æ‹®Ωg Ææçêu© ’)

A) 2n × 5m B) 2n × 3m

C) 3n × 5m D) 2n × 3m × 5n

11. logb b√b =

1 3A) B) 1 C) D) 22 212. x = log2 5, y = log2 7 Å®·ûË log2 35 éÀ îÁçC†

N© ’´ x, y ©™ A) x B) y C) xy D) xy

13. éÀçC ¢√ öÀ™ éπ®Ω ùÃßª ’ Ææ çêu ?A) 2.35 B) 2.3535.....C) 2.335335...... D) 2.353353335.......

14. A ∪ φ = A) A B) φ C) A ™‰üΔφ D) A, φ

15. A B Å®·ûËA) A ∪ B = A B) A ∩ B ≠ AC) A ∪ B = B D) A ∩ B = B

16. A − B = A Å®·ûËA) A B B) B CC) A, B Ææ´ ÷ † ÆæN’ûª ’© ’ D) A, B Nßª·éπh ÆæN’ ûª ’© ’

17. P 10 éπçõ‰ ûªèπ◊\ ¢Áj† v°æ üµΔ † Ææçêu© ÆæN’A, Q 10éπçõ‰ ûªèπ◊\ ¢Áj† Ææçßª·éπh Ææçêu© ÆæN’A. P, QÆæN’ ûª ’© ¢ÁØ˛ *vûªç

A) B)

C) D)

18. A ∩ B E Ææ ÷*çîË ¢Á Ø˛ *vûªç

A) B)

C) D)

19. éÀçC ¢√ öÀ™ Ææûªuç é¬EC?A) A ∪ B = B ∪ A B) A ∩ B = B ∩ AC) A − B = B − A D) A − B ≠ B − A

20. éÀçC ¢√ öÀ™ °æ®Ω Ææp®Ω Nßª·éπh ÆæN’ ûª ’© ’?A) A ∪ B, A − B, AB) A ∩ B, B − A, BC) A − B, B − A, A ∩ BD) A − B, B − A, A ∪ B

21. A = {a, b, c, d}, B = {a, e, i, o} Å®·ûËA) A B B) B A C) A = B D) A ≠ B

22. P = {x/x2 = 4, 3x = 9} äéπ

A) Å†çûª ÆæN’A B) ¨¡⁄†u ÆæN’AC) N¨¡y ÆæN’A D) àD é¬ü¿’

23. A − B =A) {x/x ∈ A ™‰üΔ x ∉ B}B) {x/x ∉ A ´ ’Jßª· x ∈ B}C) x/x ∈ A ´ ’Jßª· x ∉ B}D) {x/x ∉ A ™‰üΔ x ∈ B}

24. ¨¡⁄†u ÆæN’AA) φ B) { φ } C) {0} D) 0

25. A ∩ B = φ Å®·ûË n (A ∪ B) =A) n(A) B) n(B)C) n(A) + n(B) D) n(A) − n(B)

26. éÀçC ¢√ öÀ™ àC •£æ›°æC?1 1

A) x + B) x x − 1C) 3 √

x + 2 D) 2x + 3

27. 3x2 + k •£æ›°æC ¨¡⁄†uç r Å®·ûË k = A) 75 B) −75 C) 57 D) −57

28. È®çúø’ °æüΔ© ’ Ö†o äéπ °∂æ ’† •£æ›°æC?A) 2x3 B) 3x3 + 5 C) 3x3 + 2x + 1 D) 3x3 + 2x2 + 5x + 2

29. •£æ› °æC 5x2 + 2x3 + 5x + 1 ßÁ·éπ\ °æJ ÷ùç?A) 5 B) 2 C) 3 D) 1

•£æ› ∞Îj *a¥éπ v°æ ¨¡o ©’

1-C 2-D 3-C 4-B 5-C 6-C

7-B 8-B 9-A10-A11-C 12-C

13-D 14-A15-C 16-D 17-D 18-C

19-C 20-C21-D22-B23-C24-A

25-C26-D27-B28-B29-C.

Ææ ´ ÷ üµΔ Ø √ © ’

¶« ©’®Ω ’ ¶ « L é π © ’ µ

v ° æ ß ª’ Ao ç î ªç ú Õ .. .4 ´÷®Ω’\© v°æ ¡o©’

1. x2 + 7x + 12 •£æ› °æ C Í®ë« * vû√Eo UÆ œ,¨¡⁄Ø√u©† ’ éπ† ’éÓ\çúÕ.

2. éÀçC ÆæO’ éπ ®Ω ù«© ïûª† ’ Í®&ßª ’ ÆæO’ éπ ®Ω ù«© ïûª © ’í¬ ´ ÷Ja, ≤ƒüµ¿† † ’ éπ† ’éÓ\çúÕ.

4 1 8 3 + = 2; + = 5 x − 1 y − 2 x − 1 y − 2 x ≠ 1, y ≠ 2

3. äéπ °æEE vPßª ’, JCµ éπLÆ œ 12 í∫çô™ x °æ‹Jh îËßª ’ í∫ ©®Ω ’. vPßª ’ Ø√© ’í∫ ’ í∫çô© ’, JCµ ûÌN’tCí∫çô© ’ °æE îËÊÆh Ææí∫ç °æEE °æ‹Jh îËßª ’ í∫ ©®Ω ’.vPßª ’ ™‰üΔ JCµ äéπ\Í® Ç °æEE °æ‹JhîËßª ’ ú≈ EéÀ°æõ‰d Ææ´ ’ßª ’ç áçûª?

4. 3x + 2y = 12, 4x − 3y = −1 ÆæO’ éπ ®Ω ù«© ’ Ææçí∫ûªÆæO’ éπ ®Ω ù« ™‰¢Á÷ ÆæJ îª ÷ úøçúÕ. ¢√öÀéÀ ví¬°∂ Ußª ’çúÕ.

5. 2x2 − x − 10 = 0 † ’ ´®√_Eo °æ‹Jh îËßª ’úøç üΔy®√≤ƒCµç îªçúÕ.

6. äéπ võ„°‘ > ßª ’ç ™ E Ææ´ ÷ç ûª®Ω ¶µº’ñ«© ’ (x + 8)ÂÆç.O’., (2x + 3) ÂÆç.O’ Å®·u, ¢√öÀ ´ ’üµ¿uü¿÷®Ωç (x + 4) ÂÆç.O’ Å®·ûË üΔE ¢Áj¨» ™«uEoéπ† ’éÓ\çúÕ.

7. Gçü¿’ ´¤© ’ A (3, 2), B (−2, −3) ©ûÓ à®ΩpúË Í®ë«êçúøç vAü∑Δéπ®Ωù Gçü¿’´¤©† ’ éπ† ’éÓ\çúÕ.

8. 5 *´J ÅçÈéí¬ Ö†o ´‚úø’ ÅçÈé© Ææçêu©¢Á·û√hEo éπ† ’éÓ\çúÕ. Å™«çöÀ Ææçêu© ’ áEoÖØ√o®·?

9. äéπ Åçéπ v¨Ï ú µÕ ™ E ¢Á·ü¿öÀ °æC °æüΔ© ¢Á·ûªhçÆæ ’Ø√o, 6 ´ °æü¿ç 12 Å®·ûË ¢Á·ü¿öÀ É®Ω¢Áj °æüΔ©¢Á·û√hEo éπ† ’éÓ\çúÕ.

5 5 5æ10. 640, 320, 160 ... í∫ ’ù v¨Ïú µÕ, , , ...

128 64 32í∫ ’ù v¨Ï ú µø’© n ´ °æüΔ© ’ Ææ´ ÷ † ¢Á ’iûË n N© ’ ´† ’éπ† ’èπ◊\E, Ææ´ ÷ † ¢Á ’i† °æüΔEo ®√ßª ’çúÕ.

⊃

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P C

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µ

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µA B µA B

µA B

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017 °æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπ XI

⊃⊃

• öÃd °æ ô d úø ç ´÷ ØË ¨» †’. ..í∫ûªç™ °æKéπ~ © † í¬ØË ´·êu ¢Á ’i† v°æ¨¡o ©èπ◊ Ææç•ç Cµç *† v°æA Ææ´ ÷ üµΔ Ø√Eo •öÃd °æ õ‰d üΔEo. ´ ÷J† °æKéπ~NüµΔ †ç™ •öÃd °æ ôdúøç ´©x Ö°æ ßÁ÷í∫ç Öçúø ü¿E Ö§ƒ üµΔu ßª·© ’ îÁ°æpúøç, Æ ‘E ßª ’®Ω ’x Ææ© £æ…© ’ É´y úøçûÓ

îªC¢Ë °æü¿l¥A ´ ÷®Ω ’a èπ◊ Ø√o† ’. v°æA§ƒ®∏√uç ¨»Eo ûª®Ω í∫Aí∫C™ v¨¡ü¿l¥í¬ N†úøç, Å®Ωn ´ ’ ßË’u ´ ®Ωèπ◊ îªü¿ ´úøç Å© ¢√ô ’ îËÆæ ’ èπ◊ Ø√o† ’. °æKéπ~© ’ Ææ ’© ¶µºçí¬ ®√Æ œ´ ’ç* vÍíú˛ûÓ ´ ÷ §ƒ®∏Ω ¨» ©™ Åví∫ ≤ƒn †ç™ EL î √† ’. – ®√´¤© Ææ ’≠œtûª, í∫ú˛ î √çü¿,

ÇC ™« ¶«ü˛ >™«x

Nñ‰ûª© ÷ ô

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπXII

v°æ:

Â°j °æôç™ ΔABC ©ç• éÓù vA¶µº’ïç, ∠B = 90°, ADÅØËC BC éÀ îÁçC† ´ ’üµ¿u í∫ûª Í®ê. Å®·ûË AD = ?

(Ééπ\úø AB = 6 cm, AC = 10 cm)≤ƒüµ¿†: ΔABC ™ , ∠B = 90°

∴ AC2 = AB2 + BC2

102 = 62 + x2

∴ x2 = 100 − 36 = 64 ⇒ x = √64 = 8 cm

8∴ BD = DC = = 4 cm

2ΔABD ™ , ∠B = 90°

∴ AD2 = BD2 + AB2

= 42 + 62 = 16 + 36 = 52

∴ AD = √52 = √

4 × 13 = 2 √

13 cm

v°æ:

Â°j °æôç™ , AB = 8 cm, BC = 6 cm, CD = 2 cm& AC = DE Å®·ûË BE = ?

≤ƒüµ¿†: ΔABC ™ , ∠B = 90°

∴ AC2 = BC2 + AB2 = 62 + 82 = 36 + 64 = 100

∴ AC = √100 = 10 cm

ΔBED ™ , ∠B = 90°

∴ DE2 = BD2 + BE2 (... DE = AC)AC2 = 82 + BE2

102 = 82 + BE2

∴ BE2 = 100 − 64 = 36∴ BE = 6 cm

v°æ: 7.5 cm Í®ë« êçú≈Eo UÆ œ, üΔEo 2 : 3 E≠æp Ah™ N¶µº>çîªçúÕ.

≤ƒüµ¿†: i) 7.5 cm ûÓ Í®ë« êçúøç AB Ußª ÷L. ii) AB ûÓ Å©p éÓùç îËÊÆ™« AC Í®ê Ußª ÷L.iii) A1, A2, A3, A4, A5 Ææ´ ÷ † ü¿÷®√ ©ûÓ

AC Â°j U ßª ÷ L Åçõ‰AA1 = A1A2 = A2A3 = A3A4 = A4A5

iv) A5, B † ’ éπ©§ƒL.v) A2 P E éπ© §ƒL. A2P II A5B ÖçúË™« Ußª ÷L.

'P' é¬¢√ Lq† Gçü¿’´¤. AP : PB = 2 : 3.

v°æ: éÀç C °æ ôç™ ABCD D®Ω` îª ûª ’ ®ΩvÆæç. 'O' D®Ω` îª ûª ’ ®Ω vÆæç™ äéπ Gçü¿’´¤ 'O' † ’ç* A, B, C & D ©† ’éπL°œûË, OB2 + OD2 = OC2 + OA2 ÅE îª ÷°æçúÕ.

≤ƒüµ¿†: ΔODE ™ , OD2 = OE2 + DE2

ΔOEC ™ , OC2 = OE2 + EC2

ΔAOF ™ , AO2 = AF2 + OF2

ΔBOF ™ , BO2 = FB2 + OF2

LHS OB2 + OD2 = OF2 + FB2 + OE2 + DE2

= OF2 + EC2 + OE2 + AF2

( ... FB = EC & DE = AF)= (OE2 + EC2) + (OF2 + AF2)= OC2 + OA2 RHS

v°æ: °æéπ\°æ ôç™ DE II AC & DC II AP

BE BCÅ®· ûË =

EC CPÅE îª ÷°æçúÕ.

≤ƒüµ¿†: ΔBPA ™ , DC II APBC BD

∴ = .........(1) ( ü∑Ë™ ¸q Æ œüΔl¥çûªç v°æé¬®Ωç)CP DAΔBCA ™ , DE II AC

BE BD∴ = .........(2)

EC DA(1) & (2) © † ’ç*BC BE = CP EC

Ê°°æ®˝ – II

1, 2, 4 ´÷®Ω’\ © v°æ ¨¡o ©’v°æ: (5, 0), (8, 0) © ´ ’üµ¿u ü¿÷®Ωç áçûª?≤ƒüµ¿†: (x1, 0) (x2, 0) © ´ ’üµ¿u ü¿÷®Ωç = x2 - x1

(5, 0) (8, 0) © ´ ’üµ¿u ü¿÷®Ωç = 8 − 5 = 3 ßª‚Eô ’x

v°æ: (0, 5), (6, 0) Gçü¿’ ¤ ©† ’ E®Ω÷ °æéπ ûª©ç™ í∫ ’Jhç*,üΔE ÇüµΔ ®Ωçí¬ vA¶µº’ï ¢Áj »©uç éπ† ’éÓ\ î √a?

≤ƒüµ¿†:

1Δ ¢Áj¨»©uç = bh2

1= × 6 × 5 = 15 îª.ßª‚ Eô ’x.2v°æ: A(1, 2), B(9, 10) © ´ ’üµ¿u Gçü¿’´¤ C Å®·ûË BC

´ ’üµ¿u Gçü¿’´¤† ’ éπ† ’ éÓ\çúÕ.≤ƒüµ¿†: A(1, 2), B(9, 10) ´ ’ üµ¿u Gç ü¿’ ´¤ C

1 + 9 2 + 10 10 12= ( , )= ( , ) = (5, 6)

2 2 2 2∴ B(9, 10), C(5, 6) ´ ’üµ¿u Gç ü¿’´¤

9 + 5 10 + 6 14 16= (, ) = ( , ) = (7, 8)

2 2 2 2v°æ: A(a, b + c), B(b, c + a), C(c, a + b) © ’

ÆæÍ®&ßª ÷©E îª ÷°æçúÕ.≤ƒüµ¿†: Δ ABC ¢Áj¨»©uç =

1= x1 (y2 − y3) +x2 (y3 −− y1) + x3 (y1 − y2)2

1= a (c + a − a − b) + b (a + b − b − c) 2+ c (b + c − c − a)

1= ac − ab + ba − bc + bc − ac 21= × 0 = 0 2

∴ Δ ¢Áj¨»©uç Ææ ’Ø√o é¬•öÀd A, B, C © ’ ÆæÍ® & ßª ÷© ’.

v°æ: éÀçC °æôç™ AP : BP = 2 : 1 Å®·ûË A, BE®Ω ÷°æé¬©† ’ éπ† ’ éÓ\çúÕ.

mx2 + nx1 my2 + ny1≤ƒüµ¿†: P (4, 3) = ( , )m + n m + n2(0) + 1(x) 2(y) + 1(0)

(4, 3) = ( , )2 + 1 2 + 1 x 2y

(4, 3) = ( , )3 3x 2y

∴ = 4 = 33 3

9⇒ x = 12 ∴ y = = 4.52∴ A (12, 0), B (0, 4.5)

v°æ:

Â°j °æôç ΔABC ™ AD ´ ’üµ¿u í∫ûª Í®ê Å´¤ ûª ’çüΔ?é¬üΔ?

≤ƒ üµ¿ †: BC ´ ’üµ¿u Gçü¿’´¤ D = 3 + 5 −2 + 2 8 0( , ) = ( , ) = (4, 0)2 2 2 2

∴ Δ ABD ¢Áj¨»©uç Åçõ‰ A (4, −6), B (3, −2), D (4, 0)

1= 4 (−2 −0) + 3 (0 + 6) + 4(−6 + 2)2

1 1 = −8 + 18 −16 = −6= 3 îª. ßª‚.2 2

Δ ADC ¢Áj¨»©uç Åçõ‰ A(4, −6), D(4, 0), C(5, 2)1

= 4 (0 − 2) + 4 (2 + 6) + 5 (−6 − 0)21 1

= −8 + 32 − 30= −6 2 2= 3 îª. ßª‚.∴ ΔABD ¢Áj¨»©uç = ΔADC ¢Áj¨»©uç Åçõ‰ AD ÅØËC ΔABC ´ ’üµ¿u í∫ûª Í®ê Å´¤ ûª ’çC.

v°æ: x éÀ îÁçC† à N© ’ ´èπ◊ (5, −1) (x, 4) & (6, 3) 11vA¶µº’ï ¢Áj¨»©uç îªü¿®Ω°æ¤ ßª‚Eô ’x Å´¤ ûª ’çC?2

≤ƒüµ¿†: Δ ¢Áj¨»©uç = 1x1 (y2 − y3) + x2 (y3 − y1) + x3 (y1 − y2)2

11 1 = 5 (4 − 3) + x (3 + 1) + 6 (−1 − 4)2 2

11 = 5 + 4x − 30 11 = 4x − 25

4x − 25 = 114x = 11 + 254x = 36

36x = = 94v°æ: vA¶µº’ïç ABC ¶µº’ñ«© ´ ’üµ¿u Gçü¿’ ´¤© ’ D(1, 1),

E(2, −3) & F(3, 4) Å®·ûË vA¶µº’ï Q®√{ ©† ’éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: D(1, 1) = (x4, y4)E(2, −3) = (x5, y5)

F(3, 4) = (x6, y6)∴ A(x1 y1) = (x6 + x5 − x4, y6 + y5 − y4)

= (3 + 2 − 1, 4 − 3 − 1) = (4, 0)

B(x2 y2) = (x4 + x6 − x5, y4 + y6 − y5) = (1 + 3 − 2, 1 + 4 + 3) = (2, 8)

C(x3 y3) = (x4 + x5 − x6, y4 + y5 − y6) = (1 + 2 − 3, 1 − 3 − 4) = (0, −6)

y

x0 ←← 6 →→ (6, 0)

↑5↓

(0, 5)

B (0, y)

A (x, 0)

P (4, 3)

1

2

0 X

Y

A (4, −6)

B (3, −2) C (5, 2)

A (x1, y1)

B (x2, y2) C (x3, y3)

E (2, −3)F (3, 4)

D (1, 1)

4 ´÷®Ω’\© v°æ ¡o©’

A

B D C

C

106

8 cm

A

E

B ← 6 →C ←2 →D

A BP 7.5 cm

A1

A2

A3

A4

A5

A BF

E CD

A

B

P

C

E D

v°æßª’ Aoç îªçúÕ...D C

A B

E

O

1. ABCD v õ „ ° ‘ > ß ª ’ ç ™ DC II AB, AC, BD © ’ E´ü¿l êçúÕçîª ’èπ◊çö«®·.ΔAED ∼ ΔBEC Å®·ûËAD = BC ÅE îª ÷°æçúÕ.

2. Ææ´ ’ ¶«£æ› vA¶µº’ï ¶µº’ïç 'a' Å®·ûËa√

3i) Ææ´ ’ ¶«£æ› vA¶µº’ïç áûª ’h =

2√

3 a2

ii) Ææ´ ’ ¶«£æ› vA¶µº’ïç ¢Áj¨»©uç = 4

ÅE îª ÷°æçúÕ.3. ΔPQR ™ X, Y ©’ PQ & PR ©Â°j Gçü¿’ ¤©’ &

QX = RY & ∠Q = ∠R Å®·ûË XY II QR ÅEîª ÷°æçúÕ.

4. 5 cm ¢√u≤ƒ ®Ωl¥çûÓ ´%û√hEo EJtç îªçúÕ. ´%ûªhÍéçvü¿ç † ’ç* 11 cm ü¿÷®Ωç™ Ö†o Gçü¿’´¤† ’ç* Ç ´%û√h EéÀ Ææp®Ωz Í® ê© ’ Ußª ’çúÕ.

5. 6 cm, 8 cm, 10 cm ¢√u≤ƒ®√l¥© ’ Ö†o íÓ∞ « ©† ’éπJ Tç* äéπ Â°ü¿l íÓ∞¡çí¬ ûªßª ÷®Ω ’ îËÊÆh, üΔE¢√u≤ƒ®Ωl¥ç áçûª?

6. Ææ ÷h°æç ¢√u≤ƒ®Ωl¥ç 4 cm, áûª ’h 6 cm éπJ Tç* 1mm ¢√u≤ƒ®Ωl¥ç Ö†o íÓ∞ « © ’í¬ ûªßª ÷®Ω ’ îËÊÆh áEoíÓ∞ «© ’ ûªßª ÷®Ω´¤û√®·?

7. sec θ + tan θ = 5 Å®·ûË sin θ = ?8. sin x + sin2 x = 1 Å®·ûË cos2 x + cos4 x = ?9. a cos θ + b sin θ = p, a sin θ − b cos θ = q

Å®·ûË p2 + q2 = a2 + b2 ÅE îª ÷°æçúÕ.2 cos2 θ − 1

10. cot θ − tan θ = ÅE îª ÷°æçúÕ.sin θ. cos θ

1 + sin θ cos θ11. + = 2 sec θ

cos θ 1 + sin θ ÅE îª ÷°æçúÕ.

12. äéπ ´uéÀh ô´®˝ Åúø’í∫ ’ ¶µ «í∫ç † ’ç* í∫ ’ôd Â°j¶µ« í¬Eo 60° éÓùç™ îª ÷¨»úø’. í∫ ’ôd Åúø’í∫ ’ ¶µ«í∫ç† ’ç* ô´ ®˝Â°j ¶µ «í¬Eo 30° Ü®Ωn y éÓùç™ îª ÷¨»úø’. äéπ ¢Ë∞¡ ô´®˝ áûª ’h 50 O’. Å®·ûËí∫ ’ôd áûª ’h áçûª?

13. äéπ áûªh ®·† îÁô ’d 15 O’. ÖçC. í¬LéÀ éÌçûª¶µ«í∫ç NJT ØË©Â°j 60° éÓùç îËÊÆ™« °æúÕçC.Å®·ûË ¶µº÷N’ † ’ç* áçûª áûª ’h™ îÁô ’d NJ TçüÓ éπ† ’éÓ\çúÕ.

D

x

v°æ: PQ = 6 ÂÆç.O’., PR = 8Â Æ ç . O ’ . , QS = 1.5ÂÆç.O’., SR = 2 ÂÆç.O’.Å®·ûË ΔPSR ™ PSÅØËC ∠P éÀ Ææ´ ’ Cyêçúø†Í®ê Å´¤ ûª ’çüΔ?

PQ QS 6 1.5≤ƒüµ¿†: = ⇒ =

PR SR 8 21.5

⇒ 2

∴ PS ÅØËC ∠P éÀ îÁçC† éÓù Ææ´ ’Cy êçúø † Í®ê Å´¤ ûª ’çC.

v°æ: ΔABC ™

AB BD = AC DC

& ∠B = 60°, ∠C = 70°

Å®·ûË ∠CAD = ?

AB BD≤ƒüµ¿†: = Åçõ‰ AD ÅØËC ∠A éÀ îÁçC†

AC DCéÓù Ææ´ ’ Cy êç úø† Í®ê∴∠A + ∠B + ∠C = 180°

∠A = 180° − 130° = 50°

∠A 50∴∠CAD = = = 25°

2 2v°æ:

ΔABC ™ AB = 5.2 ÂÆç.O’., CF = 2.4, BE =2.6 Å®·ûË AC = ?

1 1≤ƒüµ¿†: × AB × CF = × AC × BE

2 21 1 × 5.2 × 2.4 = × AC × 2.62 2

5.2 × 2.4∴ AC = = 2 × 2.4 = 4.8 ÂÆç.O’.

2.6

v°æ: °æéπ\°æ ôç™ r =?

≤ƒüµ¿†: OP2 = OA2 + AP2

52 = r2 + 42 ⇒ r2 = 25 − 16 = 9 ⇒ r = √

9 = 3 ÂÆç.O’.

v°æ: Ê≠ú˛ îËÆ œ† v§ƒçûª ¢Áj¨» ™«uEoéπ† ’éÓ\çúÕ.

x≤ƒüµ¿†: ÂÆéπd®˝ ¢Áj¨»©uç = × Πr2360

90 22 77= × × 7 × 7 =

360 7 2= 38.5 ÂÆç.O’.2

v°æ: È®çúø’ íÓ∞ «© °∂æ ’† °æ J ´ ÷ ù«© E≠æpAh 64 : 27Å®·ûË ¢√u≤ƒ®√l¥© E≠æpAh?

4 4≤ƒüµ¿†: Π r1

3 : Π r23 = 64 : 273 3

r13 : r2

3 = 43 : 33

∴ r1 : r2 = 4 : 3v°æ: ÂÆéπd®˝ îª ’ô ’d éÌ ©ûª 27.2 O’., ¢√u≤ƒ®Ωl¥ç 5.7 O’.

Å®·ûË ÂÆéπd®˝ §Òúø´¤ = ?≤ƒüµ¿†: ÂÆéπd®˝ îª ’ô ’d éÌ ©ûª = 2r + l

27.2 = 2(5.7) + l∴ l = 27.2 − 11.4 = 15.8 O’.

v°æ: ° æ é π \ ° æ ô ç ™ E v A ¶ µ º’ ï ç ™ ´ % û √ h E o Å ç û ª J x " ç î √ ® Ω ’ .Å®·ûË BC = ?

≤ƒüµ¿†: BP = 3 ÂÆç.O’., AQ = 4 ÂÆç.O’.∴ QC = 12 − 4 = 8 ÂÆç.O’.∴ BR = BP = 3 ÂÆç.O’., QC = PC = 8 ÂÆç.O’.∴ BC = BP + PC

= 3 + 8 = 11 ÂÆç.O’.

v°æ: Å®Ωl¥ íÓ ∞¡ç™ Â°ü¿l °∂æ ’† °æ J ´ ÷ùç Ö†o ¨¡çèπ◊ ´¤† ’ûªßª ÷®Ω ’ îËÊÆh üΔE °∂æ ’† °æ J ´ ÷ùç áçûª?

(¢√u≤ƒ®Ωl¥ç = r)

≤ƒüµ¿†:1¨¡çèπ◊´¤ °∂æ ’.°æ = Πr2h3

(é¬F h = r) 1 Π r3= Π r2 × r = °∂æ ’. ßª‚Eô ’x3 3

v°æ: ¨¡çèπ◊´ ¤ ¶µ º÷¢√uÆæç 16 ÂÆç.O’., áûª ’h 15 ÂÆç.O’. Å®·ûË üΔE ´véπ ûª© ¢Áj¨»©uç áçûª?

≤ƒüµ¿†: l2 = r2 + h2 = 82 + 152

= 64 + 225 = 289∴ l = √

289 = 17 ÂÆç.O’.´véπ ûª© ¢Áj¨»©uç = Πrl= Π (8) (17) = 136 Π ÂÆç.O’.2

v°æ: Ææ ÷h°æç, ¨¡çèπ◊´¤© ¢√u≤ƒ ®√l¥© E≠æpAh 3 : 4 áûª ’h©E≠æpAh 2 : 3 Å®·ûË ¢√öÀ °∂æ ’† °æ J ´ ÷ ù«© E≠æpAhéπ† ’éÓ\´î √a?

≤ƒüµ¿†: Ææ ÷h°æç, ¨¡çèπ◊ ´¤© °∂æ ’† °æ J ´ ÷ ù«© E≠æpAh 1

= Πr2 h : Πr2 h3

1Π (3)2 × 2 : × Π (4)2 × 3

318Π : 16 Π

⇒ 9.8∴ éπ† ’ éÓ\ ´îª ’a. ¢√öÀ °∂æ ’.°æ. E≠æpAh = 9 : 8

1, 2 ´÷®Ω’\© v°æ ¡o©’

v°æ: äéπ ¢Ë ∞¡ tan θ E®Ωy *ûªç é¬èπ◊çõ‰ θ =?≤ƒüµ¿†: θ = 90°

áçü¿’ éπçõ‰ tan 90° = ∞

15v°æ: tan c = ÉC Ææûªu´ ÷?

17ÅÆæûªu´ ÷?Ææ´ ’JnçîªçúÕ.

á.¶µº’ 8≤ƒüµ¿†: tan c = = Ç.¶µº’ 15

15é¬•öÀd tan c = ÅØËC ÅÆæûªuç.17

cos2 θv°æ: = 3 Å®·ûË θ = ?

cot2 θ − cos2 θ

≤ƒüµ¿†: cos2 θ = 3 cot2 θ − 3 cos2 θ

cos2 θ + 3 cos2 θ = 3 cot2 θ

4 cos2 θ = 3 cot2 θ

cos2 θ 3⇒ =

cot2 θ 43sin2 θ = 4

3 √3sin θ = √ = 4 2

√3sin 60° = 2

⇒ θ = 60°v°æ: log tan 1° + log tan 2° + log tan 3° + .........

+ log (tan 89°) = ?≤ƒüµ¿†: log [tan 1° × tan 89°] (tan 2° × tan 88°)

× ...... tan 45°

log [tan 1° × cot 1°] [tan 2° × cot 2°) × .....tan 45°

log 1 × 1 × ..... × 1 = log 1 = 0v°æ: A + B = 90° Å®·ûË tan A × tan B = ?≤ƒüµ¿†: A + B = 90°

⇒ A = 90° − Bcot A = cot (90 − B) = tan B

∴ tan A × tan B = tan A × cot A = 1v°æ: α + β = 90°, α = 2β Å®·ûË cos2 α + sin2 β = ?≤ƒüµ¿†: 2β + β = 90°

903β = 90° ⇒ β = = 30°3∴ α = 60°

1 2 1 2∴ cos2 60° + sin2 30° = () + ()2 21 1 2 1 + = = 4 4 4 2

É C Ææ ûªu ´÷... Ææ ´’ Jnç îªç úÕ!

A

B Cθ

8 17

15

r

o P

A4

5

A

B D C60° 70°

P

Q RS

5.2

2.6 2.4

B C

A

F

E

A

B C

4

3

P

R Q 12

r

r

l = 17 cm

15

8

v°æ: ®Ω¢Ë’≠ äéπ Æ ‘d™ ¸¢Áj®Ω ’ûÓ îªûª ’®Ωv≤ƒEo ûªßª ÷ ®Ω ’ îË »úø’.üΔE ¢Áj¨»©uç 484 sq.cm. ÅüË Bí∫† ’ ´%ûªhçí¬´ ’L î √úø’. Å®·ûË üΔE ¢Áj¨»©uç áçûª?

≤ƒüµ¿†: a2 = 484∴ a = √

484 = 22 cm

îªûª ’®ΩvÆæç îª ’ô ’d éÌ ©ûª

= 4 × 22 = 88 cm´%ûªhç îª ’ô ’d éÌ ©ûª = 88 cm

2Πr = 8822

∴ 2 × × r = 887 7 1

∴ r = 88 × × = 14 cm22 2

22∴ ´%ûªh ¢Áj¨»©uç = Πr2 = × 14 × 14 = 616 cm2

7v°æ: 6 cm ¶µº’ïç Ö†o îªûª ’ ®ΩvÆæç ´%ûªhç™ ÉN’úÕ ÖçC.

Å®·ûË îªûª ’ ®Ω v≤ƒ EéÀ, ´ %û√h EéÀ ´ ’üµ¿u ¢Áj¨» ™«uEoéπ† ’éÓ\çúÕ.

≤ƒüµ¿†: ΔABC ™ , AC2 = AB2 + BC2

= 62 + 62

= 36 + 36 = 72

AC = √72

= √36 × 2

= 6√2 cm

´%ûªh ¢√u≤ƒ®Ωl¥ç =AC 6√

2 AO = = = 3√

2 cm2 2

22´%ûªh ¢Áj¨»©uç = Πr2 = [3√

2 ]27

22 396= × 9 × 2 = cm2

7 7îªûª ’ ®Ω vÆæ®Ω ¢Áj¨»©uç = 6 × 6 = 36 cm2

∴ Ê≠ú˛ îËÆ œ† v§ƒçûª ¢Áj¨»©uç =´%ûªh ¢Áj¨»©uç − îªûª ’ ®ΩvÆæ ¢Áj¨»©uç

396= − 36 = 3

7

áç û ª á ûª’h Â°®Ω’í∫ ’ ûª ’ç C..?v°æ: D®Ω`îªûª ’®Ωv≤ƒé¬ ®Ωç™ Ö†o §Ò©ç éÌ© ûª© ’ 20 m,

14 m. äéπ ´‚© † ’ç* 6 O’. §Òúø´¤, 3 O’.¢Áúø© ’p, 2.5 O’. ™ ûÁj† í∫ ’çûª† ’ ûªNy Ç ´ ’öÀdEN’í∫û√ §Ò©çÂ°j Ææ´ ÷ †çí¬ îªLxûË ÅC áçûª áûª ’hÂ°®Ω ’ í∫ ’ ûª ’çüÓ éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: í∫ ’çûª °∂æ ’.°æ. = lbh= 6 × 3 × 2.5 = 45 m3

I §ƒ®Ω ’d °∂æ ’.°æ. = lbh= 11 × 6 × h = 66 h m3

II §ƒ®Ω ’d °∂æ ’.°æ. = 14 × 14 × h m3 = 196 hm3

I & II §ƒ®Ω ’d© ¢Á·ûªhç °∂æ ’.°æ. = (66 h + 196 h)m3

= (262 h)m3

262 hm3 = ûªNy† ´ ’öÀd °∂æ ’.°æ. = 45 m3

45∴ h = = 17.18 cm ( Ææ ’ ´ ÷®Ω ’í¬)

262v°æ: äéÌ\éπ\ °∂æ ’†ç °∂æ ’† °æ J ´ ÷ùç 64 cm3 Ö†o 3

°∂æ ’Ø√ ©† ’ äéπ üΔE *´®Ω ´ ’®Ì éπöÀ éπL °œûË à®ΩpúË D®Ω` °∂æ ’†ç Ææç°æ‹ ®Ωg ûª© ¢Áj¨»©uç éπ† ’éÓ\çúÕ.

≤ƒüµ¿†:

°∂æ ’†ç °∂æ ’.°æ. V = a3 = 64 ⇒ a3 = 43

⇒ a = 4 cm∴ l = 4 + 4 + 4 = 12 ÂÆç.O’.b = 4 cm, h = 4 cm

Ææç°æ‹ ®Ωg ûª© ¢Áj¨»©uç (D®Ω` °∂æ ’†ç) = 2 (lb + bh + lh)= 2 (12 × 4 + 4 × 4 + 12 × 4)

= 2 (48 + 16 + 48) = 2 × 112 = 224 cm2

v°æ: äéπ ¨¡çèπ◊´¤ áûª ’h 24 cm, üΔE´ v é π û ª © ¢ Á j ¨ » © u ç 550ÂÆç.O’.2. Å®·ûË üΔE °∂æ ’† °æ J ´ ÷ùç éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: ¨¡çèπ◊´¤ ´véπ ûª© ¢Áj¨»©uç = Πrl

22 √

550 = × r × r2 + 576 cm2

7550 × 7

√

= r × r2 + 57622

(25 × 7)2 = r2 (r2 + 576)r4 + 576 r2 = 625 × 49r4 + 576 r2 − 625 × 49 = 0(r2 − 49)(r2 + 625) = 0r2 + 625 ≠ 0

∴ r2 − 49 = 0r2 = 49 ⇒ r = √

49 = 7 cm1

∴ ¨¡çèπ◊´¤ °∂æ ’.°æ. = Πr2h31 22

= × × 7 × 7 × 243 7= 1232 cm3

4 ´÷®Ω’\ © v°æ ¨¡o ©’

D

A B

O

6

6

6

C

a r

a

20 m

14 m

11 m

3 m

6 m 14 m

2.5 m

III

24 l

r

44

4 4 4

l2 = r2 + 242

√

l = r2 + 576

d = 16 cm16

r = = 8 cm2

7 cm

7 cm

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017 °æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπ XIII

ûÓ öÀ N üΔu®Ω ’n ©èπ◊ Å®Ωnç é¬ E Åç ¨» © ’ îÁ°æp úøç üΔy®√ Ææ ¶„bèπ◊d ™ Ø√èπ◊°æ ô ’d © Gµç *ç C. ûª Lx ü¿ç vúø’ © ’, Ö §ƒ üµΔu ßª· © v§Ú û√q£æ«ç ´ ©x Ç ûªt N ¨»yÆæç ûÓ °æ Kéπ~ © ’ ®√Æ œ ´ ’ç * vÍí ú˛ ≤ƒ Cµ ç î √ † ’.

– §ƒ E, Â£j« ü¿®√ ¶« ü˛

Nñ‰ûª© ÷ ô

90°

´’çí∫ ∞¡ ¢√ ®Ωç 7 °∂œ v• ´ J 2017°æ üÓ ûª®Ωí∫ A °æ Kéπ~ © v°æ ûËuéπ Ææç *éπXIV

1 − sin A 1 + sin A 2v°æ: √ + √

=

1 + sin A 1 − sin A cos AÅE îª ÷°æçúÕ.

1 − sin A 1 − sin A≤ƒüµ¿†: √

× 1 + sin A 1 − sin A

(1 − sin A)2 (1 − sin A)2√

√

= = 1 − sin2 A cos2 A

1 − sin A= .......(1)

cos A

1 + sin A 1 + sin A

√ × 1 − sin A 1 + sin A

(1 + sin A)2 (1 + sin A)2

√

√

= = 1 − sin2 A cos2 A

1 + sin A= ...............(2)

cos ALHS (1) + (2)

1 − sin A 1 + sin A +

cos A cos A1 − sin A + 1 + sin A 2

= = cos A cos A

RHScot θ cosec θ + 1v°æ: + = 2 sec θ

cosec θ +1 cot θÅE îª ÷°æçúÕ.

cot2 θ + cosec2 θ + 1 + 2 cosec θ≤ƒüµ¿†: LHS

cot θ (cosec θ + 1)[cosec2 θ − cot2 θ = 1cosec2 θ = 1 + cot2 θ]

(cot2 θ +1) + cosec2 θ + 2 cosec θ

cot θ (cosec θ + 1)

cosec2 θ + cosec2 θ + 2 cosec θ


2 cosec2 θ + 2 cosec θ=


2 cosec θ [cosec θ +1] =


12 ( )sin θ 2= =

cos θ cos θ

sin θ= 2 sec θ

RHS

N ´÷ †ç ¢Ë í∫ç éπ †’ éÓ\ ´ ú øç .. ?v°æ: äéπ ñ„ö¸ N´ ÷ †ç A † ’ç* 60° éÓùç™ Â°jéÀ

™‰*çC. 30 ÂÆéπ†x ûª®√yûª Ç Ü®Ωn y éÓùç 30° í¬´ ÷ J †ô ’x éπE °œç *çC. Ç N´ ÷ †ç äÍé áûª ’h™ Åçõ‰, 1500 √

3 m á ûª ’h ™ áí∫ ’ ®Ω ’ ûª ’çõ‰ üΔE¢Ëí¬Eo éπ† ’éÓ\çúÕ. (√3 = 1.732)

1500 √3

≤ƒüµ¿†: ΔADE ™ tan 30° = AD

1 1500 √3

= √

3 AD∴ AD = 1500 (√

3)2

= 4500 m

1500 √3

Δ ABC ™ tan 60° = AC1500 √

3√

3 = AC

∴ AC = 1500 m30 ÂÆéπ †x™ v°æßª ÷ ùÀç *† ü¿÷®Ωç = 4500 − 1500

= 3000 m.ü¿÷®Ωç 3000

∴ ¢Ëí∫ç = = = 100 O’./ÂÆ. é¬©ç 30

v°æ: äéπ ô´®˝† ’ B, D © † ’ç* îª ÷Ææ ’h Ø√o®Ω ’. ÉN äÍé´®Ω ’ Ææ™ ÖØ√o®·. ô´®˝ † ’ç* B, D éÀ Ö†oü¿÷®√© ’ P, Q. ÅN È®çúø’ °æ‹®Ωéπ éÓù«© ’. Å®·ûËüΔE áûª ’h √

PQ ÅE îª ÷°æçúÕ.h

≤ƒüµ¿†: ΔABC ™ tan θ = ph = p × tan θ ....... (1)ΔADC ™

htan (90 − θ) = q

hcot θ = q ⇒ h = q × cot θ ..... (2)(1) × (2)h × h = pq tan θ × cot θh2 = pq (1)

h = √pq ßª‚ E ô ’x

v°æ: éÌEo é¬®Ω ’f ©Â°j 3 † ’ç* 102 ´®Ωèπ◊ ®√Æ œ, ¢√öÀE äéπÂ°õ„d™ ¢ËÆ œ éπL §ƒ®Ω ’. Åçü¿’™ † ’ç* ßª ÷ ü¿% *a ¥ éπçí¬ äéπ é¬®Ω ’f BÆ œ † °æ¤púø’ ÅC(i) ÆæJ Ææçêu ÅßË’u Ææç¶µ« úûª (ii) 15 éπçõ‰ ûªèπ◊\´ Ææçêu Ö†o é¬®Ω ’f ÅßË’u

Ææç¶µ«úûª(iii) °æJ °æ‹®Ωg ´®Ω_ Ææçêu ÅßË’u Ææç¶µ« úûª(iv) 25 éπçõ‰ ûªèπ◊\´ ÅßË’u Ææç¶µ« u ûª† ’ éπ† ’éÓ\çúÕ.

≤ƒüµ¿†: ¢Á·ûªhç é¬®Ω ’f© ’ 3, 4, 5, ... 102 = 100

¢Á·ûªhç Å´ é¬ ¨»© ’ = 100

(i) ÆæJ Ææçêu© ’ → 4, 6, 8 ... 102Å´ é¬ ¨»© Ææçêu = 50

50 1∴ P (ÆæJ Ææçêu) = =

100 2

(ii) 15 éπçõ‰ ûªèπ◊\´ é¬®Ω ’f© ’ → 3, 4, 5, 6, 7, 8,9, 10, 11, 12, 13, 14

Å´ é¬ ¨»© Ææçêu = 1212 3

∴ P (< 15) = = 100 25

(iii) °æJ °æ‹®Ωg ´®Ω_ Ææçêu© ’ → 4, 9, 16, 25, 36,49, 64, 81, 100

Å´ é¬ ¨»© Ææçêu = 9 9

∴ P (°æJ °æ‹®Ωg ´®Ω_ Ææçêu) = 100

(iv) 25 éπçõ‰ ûªèπ◊\ ¢Áj† v°æüµΔ† Ææçêu© ’ → 5, 7,11, 13, 17, 19, 23

Å´ é¬ ¨»© Ææçêu = 77

∴ P (25 éπçõ‰ ûªèπ◊\´ v°æüµΔ† Ææçêu© ’) = 100

v°æ:

Ææí∫ô ’ = 9 Å®·ûË P = ?≤ƒüµ¿†:

∑fx∴ x− =

N290 + 10P

9 = 33 + P

290 + 10P = 297 + 9P10P − 9P = 297 − 290

P = 7

4 ´÷®Ω’\© v°æ ¡o©’

v°æ: äéπ áûªh®·† ô´ ®˝ Fúø 60° † ’ç* 45° ©èπ◊ ´ ÷J † °æ¤púø’ Fúø ô´®˝ Åúø’í∫ ’ ¶µ«í∫ç † ’ç* 15 O’.ü¿÷®Ωç † ’ç* ÉçéÌçûª ü ¿÷®Ωç ïJ TçC. DEéÀÆæç•ç Cµç *† °æôç Ußª ’çúÕ.

≤ƒüµ¿†:

v°æ: 100 O’. áûª ’h†o ô´®˝ † ’ç* 100√3 O’.

ü¿÷®Ωç™ Ö†o Gçü¿’´¤ † ’ç* 'θ' éÓùç™ îª ÷ÊÆh,Ç éÓùç áçûª?

≤ƒüµ¿†: ΔABC ™ 100

tan θ = 100√

3 1

= √

3 1

∴ tan 30° = √

3 ∴ θ = 30°

v°æ: äéπ ¢Ë∞¡ áûª ’hèπ◊, ü¿÷®√ EéÀ ´ ’üµ¿u éÓùç 45° Å®·ûËÅN È®çúø÷ Ææ´ ÷ † ´ ’E P´ Åçô ’ Ø√oúø’. äéπ ÖüΔ £æ« ®Ωù ÉýçúÕ.

≤ƒüµ¿†: ΔABC ™ htan 45° = 10

hl = ⇒ h = 10

10∴ éÓùç 45° Ö†o °æ¤púø’ áûª ’h, ü¿÷®√© ’ Ææ´ ÷ †ç.

v°æ: 75 O’. áûª ’h† Ö†o ¶µ«í∫ç † ’ç* éÀçü¿ Ö†o¶µ«í¬Eo 30° E´ ’o éÓ ùç™ îª ÷Æ œ † °æ¤púø’ Ç ´Ææ ’h ´¤èπ◊, Åúø’í∫ ’ ¶µ«í¬ EéÀ Ö†o ü¿÷®Ωç 25√

3 O’. ÉCÆæûªu´ ÷?

≤ƒüµ¿†: ΔABC ™ 75tan 30° = x

1 75 = √

3 x∴ x = 75 √3 O’.

∴ ´ ’†èπ◊ É*a† ï¢√• ’ 25 √3 O’. Å†úøç ÅÆæûªuç.

v°æ: äéπ ¶«uí∫ ’™ 17 é¬®Ω ’f© ’ ÖØ√o®·. ¢√öÀÂ°j 1, 2, 3.... 17 Ææçêu ©† ’ ´·vCç î √®Ω ’. Åçü¿’™ † ’ç* äéπé¬®Ω ’f† ’ ßª ÷ ü¿%*a ¥ éπçí¬ BÊÆh, ÅC v°æüµΔ† ÆæçêuÅßË’u Ææç¶µ« úûª áçûª?

ï: v°æüµΔ† Ææçêu© ’: 2, 3, 5, 7, 11, 13, 17Å´ é¬ ¨»© Ææçêu = 7

7∴ P (Prime) =

17v°æ: äéπ §ƒ*éπ ´·ë« ©Â°j äéπ ¶«© ’úø’ éÀçC Nüµ¿çí¬

®√¨»úø’.

E E N A D U

äéπ ¢Ë∞¡ üΔEo NÆ œ J † ôx ®·ûË E °æúøö« EéÀ Ææç¶µ« úûªáçûª?

2 1ï: P(E) = = 6 3

v°æ: ´ ’üµ¿’ É™« Åçô ’ Ø√oúø’ – 'äéπ üΔE Ææç¶µ« úûª 72 —– ÉC Ææûªu´ ÷? ÅÆæ ûªu´ ÷? Ææ´ ’ Jnç îªçúÕ.

ï: ÉC ÅÆæûªuç. áçü¿’ éπçõ‰ Ææç¶µ« úûª 0 , 1 ´ ’üµ¿u™ Öçô ’çC. Åçõ‰ 0 ≤ P(E) ≤ 1

v°æ: È®çúø’ §ƒ* éπ ©† ’ äÍé ≤ƒJ NÆ œ J † °æ¤púø’, ¢ËÍ®y®Ω ’Ææçêu© ’ ´îËa Ææç¶µ« úûª áçûª?

ï: ¢Á·ûªhç Å´ é¬ ¨»© Ææçêu = 6 × 6 = 36Åçü¿’™ ¢ËÍ®y®Ω ’ Ææçêu© ’ ´îËa Å´ é¬ »© ’ = 30

30 5∴ P (¢ËÍ®y®Ω ’ Ææçêu© ’) = = 36 6

n− − Cf2

v°æ: ´ ’üµ¿u í∫ûªç = l + ( ) × h ™ E °æüΔ ©† ’ f

N´ Jç îªçúÕ.ï: l = ´ ’üµ¿u í∫ûª ûª®Ω í∫A Cí∫ ’´ £æ«ü¿’l

n = ü¿û√hç ¨¡ç ™ E ®√¨¡Ÿ© ÆæçêuCf = ´ ’üµ¿u í∫ûª ûª®Ω í∫ AéÀ ´·çü¿’ ûª®Ω í∫A Ææç*ûª

§˘†” °æ¤†uçf = ´ ’üµ¿u í∫ûª ûª®Ω í∫A §˘†” °æ¤†uçh = ´ ’üµ¿u í∫ûª ûª®Ω í∫A §Òúø´¤

v°æ: l = 60, n = 100, CF = 46, f = 20, h = 10 Å®·ûË´ ’üµ¿u í∫ûªç áçûª?

n − Cf2ï: ´ ’üµ¿u í∫ûªç = l + [ ] × hf

50 − 46 = 60 + ( ) × 10204= 60 + 2

= 60 + 2 = 62

v°æ: äéπ ûª®Ω í∫ A ™ E NüΔu ®Ω ’n© Ææçêu 36. Åçü¿’™ àúø’ í∫ ’®Ω ’ ûÁ© ’í∫ ’ É≠æd °æ úø û√®Ω ’. †© ’í∫ ’JéÀ £œ«çD Åçõ‰É≠ædç. 15 ´ ’çC í∫ùÀ ûª ¨» ≤ƒYEo É≠æd °æ úø û√®Ω ’. 10´ ’çCéÀ ≤ƒç°∂œ’ éπ ¨»ÆæYç Åçõ‰ É≠ædç. ¶«£æ› ∞¡éπçàN’öÀ?

ï: í∫ùÀ ûª ¨»ÆæYç

A C D

EB

30°60

° 1500

√ 3

m

D← q →

← p →

B C

h

A

θ 90 − θ

x 4 6 9 10 15

f 5 10 10 P 8

x f fx

4 5 206 10 609 10 9010 P 10P15 8 120

1, 2 ´÷®Ω’\© v°æ ¡o©’

60° 45°

↑h↑

←15 m → ←x →

A

A

B

100

100√3 C

B DC

θ

A

B

h

10 C

45°

75 m

30°

30°A

B Cx

1.

Â°j °æôç™ Δ ABC ©ç• éÓù vA¶µº’ïç ∠B = 90°.AD ÅØËC BC ´ ’üµ¿u í∫ûª Í®ê Å®·ûË AD = ?

2. ΔABC ™ BC, AC© ´ ’üµ¿u Gçü¿’ ´¤© ’, P, Q,CQ : QA = 1 : 3, CP = 4 ÂÆç.O’. Å®·ûË BC= ?

3. È® ç úø ’ Æ æ ® Ω ÷ ° æ v A ¶µ º’ ñ « © î ª ’ ô ’ d é Ì © û ª © ’ 30ÂÆ ç.O’. , 20 ÂÆç.O’. ¢Á·ü¿öÀ vA¶µº’ïç™ E äéπ¶µº’ïç 12 ÂÆç.O’. È®çúÓ vA¶µº’ïç Ææü¿%¨¡ ¶µº’ïçéÌ©ûª áçûª?

4. PQ, PR ¶µº’ñ« ©Â°j M, N Gçü¿’ ´¤© ’, ΔPQRÅ®·ûË éÀçC Ææçü¿®√s ¥™ x MN II PQ Å´¤ûª ’çüΔ?éπ† ’ éÓ\çúÕ.(i) PM = 4 ÂÆç.O’., QM = 4.5 ÂÆç.O’., PN =

4 ÂÆç.O’., NR = 4.5 ÂÆç.O’. (ii) PQ = 1.28 ÂÆç.O’., PR = 2.56 ÂÆç.O’.,

PM = 0.16 ÂÆç.O’., PN = 0.32 ÂÆç.O’.

5.

AD 3Δ ABC ™ = & AC

DB 5

= 10 ÂÆç.O’. Å®·ûË AE = ?

6. äéπ îÁô ’d Fúø 5 O’. ü¿÷®Ωç™ °æúø’ ûª ’çC. ÇéÓùç 45° Å®·ûË îÁô ’d áûª ’h áçûª?

7. P(E) + P(E) = 1 é¬´ ú≈ EéÀ äéπ ÖüΔ £æ« ®ΩùÉýçúÕ.

8. ∑ fi = N = 80, ∑fiui = −26, A = 55, h = 10Å®·ûË x = ?

9. ¶«£æ› ∞¡éπç Ææ ÷vûªç ®√ßª ’çúÕ.

v°æßª’ Aoç îªçúÕ...A

B D C

4 5

A

BB

D E

A

C

N = 33 + P

∑fx = 290 + 10P

1. éÀçC ¢√ öÀ™ ´®Ω_ ÆæO’ éπ ®Ωùç?A) x(x + 1) + 5 = (x + 3) (x − 3)B) (x + 1)3 = x3 − 5C) x2 − 3x + 2 = x(x + 5)D) 3x2 − 4 = (x + 5) (3x − 1)

2. ´®Ω_ ÆæO’ éπ ®Ωùç ax2 + bx + c = 0, a ≠ 0 ´‚™«© ’éπ† ’éÓ\´ ú≈ EéÀ Ææ ÷vûªç

√

√

−b + b2 − 4ac −b − b2 − 4acA) B)

2a 2a

√

√

−b ± b2 − 4ac b ± b2 − 4acC) D)

2a 2a3. ´®√_Eo °æ‹Jh îËßª ’úøç üΔy®√ x2 − x − 6 = 0 ®Ω_ ÆæO’

éπ ®Ω ù«Eo ≤ƒCµçîªú≈ EéÀ, É®Ω ’ ¢Áj °æ¤™« éπ© §ƒ Lq† Ææçêu1 1 1 1A) B) C) D) 2 4 6 8

4. ´®Ω_ ÆæO’ éπ ®Ωùç px2 + qx + r = 0, p ≠ 0 ™ p + q + r = 0 Å®·ûË üΔE ´‚™«© ’

q −q r −rA) 1, B) 1, C) 1, D) 1, p p p p

5. ''È®çúø’ ´®Ω ’Ææ Ææ£æ«ï Ææçêu© ©•l¥ç 20—— EûÁLßª ’ñ‰ÊÆ ´®Ω_ ÆæO’ éπ ®Ωùç?A) x(x + 1) = 20 B) (x − 1)x = −20C) x2 + (x + 1)2 = 20 D) (x − 1)2 + x2 = 20

6. (−5, 0), (8,0) Gçü¿’ ´¤© ´ ’üµ¿u ü¿÷®ΩçA) 5 ßª‚Eô ’x B) 8 ßª‚Eô ’xC) 13 ßª‚Eô ’x D) 3 ßª‚Eô ’x

7. an = 6n + 5 Å®·ûË an + 1 =A) 6n + 6 B) 6n + 11 C) 6n + 15 D) n + 11

8. an = 2n - 1 Å®·ûË a5 =A) 8 B) 12 C) 16 D) 10

1 19. , , 1.... v¨Ïú µÕ™ a2, a4© ©•l¥ç.... éÀ Ææ´ ÷ †ç.

4 2A) a1 B) a3 C) a5 D) a7

10. Åçéπ v Ï ú µÕ ™ E ¢Á·ü¿öÀ n °æüΔ© ¢Á·û√h EéÀ Ææ ÷ vûªç?n n

(i) Sn = [a1 + an] (ii) Sn = [2a1+ (n − 1)d]2 2A) (i) ´ ÷vûª¢Ë’ B) (ii) ´ ÷vûª¢Ë’C) (i) ™‰üΔ (ii) D) àD é¬ü¿’10

11. Σ (2i + 3) =i = 1

A) 100 B) 120 C) 140 D) 160

12. √12, √

27, √

48 .... v¨Ïú µÕ ™ E ûª®√yA °æü¿ç

A) √52 B) √

65 C) √

70 D) √

75

13. éÀçC¢√ öÀ™ Åçéπ v Ïú µÕ é¬EC?A) 5, 10, 15, 20.... B) 2, −3, −8, −13 ...

1 3 5 7C) −3, 3, −3, 3 ... D) , , , ...2 2 2 2

14. í∫ ’ù v¨Ï ú µÕ ™ E an − 1 °æü¿çA) a1rn − 1 B) a1r n + 1

C) a1r n + 2 D) a1 r n − 2

15. äéπ v¨Ïú µÕ™ an = 3 + 2n2 Å®·ûË, ¢Á·ü¿öÀ ´‚úø’°æüΔ© ¢Á·ûªhç ?

A) 32 B) 35 C) 37 D) 3916. D®Ω` îª ûª ’ ®ΩvÆæç ABCD ™ A(3, 4), C (7, 7) Å®·ûË

B Q®Ω{ç E®Ω ÷ °æ é¬© ’?A) (3, 7) B) (4, 7) C) (7, 7) D) (3, 3)

17. (−2, −1), (k, 0), (4, l), (1, 2) © ’ Ææ´ ÷ç ûª®Ω îªûª ’ ®Ω ’s ¥ ñ« EéÀ Q®√{© ’ Å®·ûËA) k = 3, l = 1 B) k = 1, l = 3C) k = −3, l = 1 D) k = 3, l = −1

18. X- Åé ~¬ EéÀ Ææ´ ÷ç ûª®Ω Í®ê ¢√© ’?A) E®Ωy*ûªç é¬ü¿’ B) 0 C) 1 D) 2

SSC PUBLIC EXAMS 2017 - MODEL PAPERPAPER - I

Telugu Medium(Parts A and B)

Max. Marks: 40 Time: 2hrs. 45 min

Ææ÷ îª † ©’:i) Ææ´÷ üµΔ Ø√©’ ®√ßª’úøç v§ƒ®Ωç Gµç îª ú≈ EéÀ ´·çü¿’

v°æ¨¡o °æ vû√Eo èπ~◊ùoçí¬ îªCN Å´ í¬ £æ«† îËÆæ’ éÓçúÕ. v°æ¨¡o °æ vûªç ™E ÅEo v°æ¨¡o©†’ îªü¿´ ú≈ EéÀ15 EN’ ≥ƒ© Ææ´’ßª’ç Íéö« ®·ç î√®Ω’.

ii) Part - A ™ É*a† v°æ¨¡o ©èπ◊ Ææ´÷ üµΔ Ø√©’ NúÕí¬ï¢√ •’ °æ vûªç™ ®√ßª’çúÕ.

iii) Part - B ™ É*a† ÅEo v°æ¨¡o©èπ◊ Ææ´÷ üµΔ Ø√ ©†’ v°æ¨¡o °æ vûªç ™ØË ®√Æœ Part - A ï¢√•’ °ævû√ EéÀ ïûª îË ßª÷L.

PART - AMax.Marks: 35 Time: 2 hrs. 15Ææ÷ îª † ©’:

i) Part - A ™ É*a† I, II, III Sections ™ ÅEov°æ¨¡o ©èπ◊ Ææ´ ÷ üµΔ Ø√© ’ ï¢√ • ’ °æ vûªç ™ ØË ®√ßª ’çúÕ.

ii) Section - III ™ E v°æA v°æ¨¡oèπ◊ Åçûª ®Ω_ûª áç°œéπÖçô ’çC. é¬•öÀd 14 † ’ç* 17 ´®Ωèπ◊ Ö†o v°æ¨¡o ©èπ◊ Ææ´ ÷ üµΔ†ç ®√ÊÆ ô °æ¤púø’ v°æA v°æ¨¡o™ àüÁjØ√äéπ üΔEo á† ’o èπ◊E Ææ´ ÷ üµΔ†ç ®√ßª ’çúÕ.

SECTION - I

Ææ÷îª †©’: i) éÀçC ÅEo v°æ¨¡o ©èπ◊ ï¢√• ’ ®√ßª ’çúÕ.ii) v°æA v°æ¨¡oèπ◊ 1 ´ ÷®Ω ’\.

(7 ×× 1 = 7 M)1. 352, 125 © í∫.≤ƒ.é¬ † ’ v°æüµΔ† é¬®Ω ù«ç é¬© ©•l¥

°æü¿l¥ A™ éπ† ’éÓ\çúÕ.2. A = {1, 5, 10, 17, 26} Å®·ûË, A † ’ ÆæN’A

E®√tù ®Ω ÷°æç™ ®√ßª ’çúÕ.3. äéπ D®Ω` îª ûª ’ ®Ω v≤ƒ é¬®Ω Ææn©ç §Òúø´¤, ¢Áúø© ’p È®çúÕç

ûª© éπçõ‰ ´‚úø’ O’ô®Ω ’x áèπ◊\´ Å®·ûË üΔEîª ’ô ’d éÌ © ûª† ’ •£æ› °æC ®Ω ÷°æç™ ´uéπh°æ®ΩîªçúÕ.

4. a + b, a − b © ’ ¨¡⁄Ø√u © ’í¬ Ö†o ´®Ω_ •£æ› °æ CE®√ßª ’çúÕ.

5. éÀçC ÆæO’ éπ ®Ω ù«© ïûª Ææçí∫ûª ïûª Å´¤ ûª ’ç üË¢Á÷îª ÷úøçúÕ.3x + 2y − 11 = 0, −5y + 2x − 2 = 0

6. äéπ Åçéπ v¨Ïú µÕéÀ îÁçC† n °æ üΔ © ¢Á· ûªhç Sn = n (n + 4) Å®· ûË n ´ °æüΔEo éπ† ’éÓ\çúÕ. Åçéπ v¨Ï ú µÕ ™ E n °æüΔ© ¢Á·û√h EéÀ, n ´ °æüΔ EéÀ Ö†oÆæç•ç üµΔEo ®√ßª ’çúÕ.

7. äéπ vA¶µº’ï Q®√{© ’ A(2, 3), B(−1, 5), C(−2, −3)Å®·ûË üΔE í∫ ’®Ω ’ûªy ÍéçvüΔEo éπ† ’éÓ\çúÕ.

SECTION - II

Ææ÷îª †©’: i) ÅEo v°æ¨¡o ©èπ◊ Ææ´ ÷ üµΔ Ø√© ’ ®√ßª ’çúÕ.ii) v°æA v°æ¨¡oèπ◊ 2 ´ ÷®Ω ’\© ’.

(6 ×× 2 = 12 M)

8. x2 − 16 ÅØË •£æ› °æC ¨¡⁄Ø√u©† ’ éπ† ’èπ◊\E,¨¡⁄Ø√u ©èπ◊, •£æ› °æC í∫ ’ù é¬ ©èπ◊ ´ ’üµ¿u Ææç•ç üµΔEoÆæJ îª ÷ úøçúÕ.

9. loga6 = x, loga3 = y Å®·ûË logaa2 N© ’´† ’

x, y ©™ éπ† ’éÓ\çúÕ.10. A(2, 3), B(0, 6), C(6, −3) Gçü¿’ ´¤© ’ ÆæÍ® &

ßª ÷© ’ Å´¤û√ßË’¢Á÷ îª ÷úøçúÕ.11. 3 Ææç´ûªq ®√© éÀçü¿ô ÊÆo£æ« ´ßª ’Ææ ’ ´¤uvûª\´ ’ç,

6 Ææç´ûªq ®√© ûª®√yûª Ç¢Á ’ ´ßª ’Ææ ’ ´¤uvûª\ ´ ÷©¢Á·ûªhç 1/6 Å®·ûË Ç¢Á ’ v°æÆæ ’hûª ´ßª ’Ææ ’ áçûª?

12. A B, B ∩ C = φ Å®· † ÆæN’ ûª ’© ’ A, B, C©† ’ ¢ÁØ˛ *vûªç üΔy®√ îª ÷°æçúÕ.

13. Éü¿l®Ω ’ uèπ◊h© ÇüΔ ßª ÷© E≠æpAh 5 : 4, ¢√J ê®Ω ’a©E≠æpAh 3 : 2, ¢√®Ω ’ v°æA äéπ\®Ω ÷ ØÁ©èπ◊ ®Ω ÷.3,000 ≤Ò´·t ÇüΔ îËÊÆh, ¢√J ØÁ© ¢√K ÇüΔ ßª ÷ ©† ’ éπ† ’éÓ\çúÕ.

SECTION - III

Ææ÷îª†©’: i) ÅEo v°æ¨¡o ©èπ◊ Ææ´ ÷ üµΔ Ø√© ’ ®√ßª ’çúÕ.ii) Ñ Section ™ v°æA v°æ¨¡oèπ◊ Åçûª ®Ω_ûª áç°œéπ

Öçô ’çC.iii) v°æA v°æ¨¡o™ É*a† È®çúø’ Ææ´ ’Ææu™ x àüÁjØ√

äéπüΔEo á† ’o èπ◊E Ææ´ ÷ üµΔ†ç ®√ßª ’çúÕ.iv) v°æA v°æ¨¡oèπ◊ 4 ´ ÷ ®Ω ’\© ’.

(4 ×× 4 = 16 M)

14. 5√3 − √

7† ’ éπ®Ω ùÃßª ’ Ææçêu ÅE E®Ω ÷°œçîªçúÕ.(™‰üΔ)

ßª‚éÀxú˛ ¶µ«í∫ £æ…®Ω Ø√ußª ÷Eo Ö°æ ßÁ÷ Tç* à üÁjØ√äéπ ¶‰Æ œ üµ¿† °æ‹®Ωg Ææçêu ´®Ω_ç 6q + 1 ™‰üΔ 6q + 3 ®Ω ÷°æç™ Öçô ’ç ü¿E îª ÷°æçúÕ.

15. p(x) = x2 − x − 6 •£æ› °æC Í®ë« * vû√Eo UÆ œ ÇÍ®ë« *vûªç † ’ç* •£æ› °æC ¨¡⁄Ø√u ©† ’ éπ† ’éÓ\çúÕ.

(™‰üΔ)éÀçü¿ É*a† È®çúø’ îª®Ω ®√ ¨¡Ÿ© Í®&ßª ’ ÆæO’ éπ ®Ω ù«©ïûªèπ◊ Í®ë « * vû√Eo Ußª ’çúÕ. Ç Í®ë« *vûªç† ’ ç* Ç ÆæO’ éπ ®Ω ù« ©èπ◊ ≤ƒüµ¿ †† ’ éπ† ’éÓ\çúÕ.3x − 2y = −14x + y = 17

16. äéπ ûª®Ω í∫ A ™ E v°æA NüΔuJn äéπ é¬®Ωu véπ´ ’ E ®Ωy £æ« ùèπ◊ Ææ´ ÷ †çí¬ îªçüΔ Éî √a®Ω ’. äéπ¢Ë∞¡ Ç ûª®Ω í∫ A™ 15 ´ ’çC NüΔu ®Ω ’n© ’ áèπ◊\ ´í¬ ÖçúÕ Öçõ‰v°æA NüΔuJn ®Ω ÷.40 ûªèπ◊\ ´í¬ É´y í∫ L Íí ¢√®Ω ’,îªçüΔ ¢Á·ûªhç ®Ω ÷.3,000 † ’ç* ®Ω ÷.3,200 ©èπ◊Â°JT ÖçúËC. Ç ûª®Ω í∫ A ™ E NüΔu ®Ω ’n© Ææçêu† ’éπ† ’éÓ\çúÕ.

(™‰üΔ)éÀçC ÆæO’ éπ ®Ω ù«© ïûª† ’ Í®&ßª ’ ÆæO’ éπ ®Ω ù«©ïûªí¬ ´ ÷Ja, ¢√öÀ ≤ƒüµ¿† éπ† ’éÓ\çúÕ.x + y x − y = 4 ; = −2xy xy

1 1 117. , , © ’ Åçéπ v¨Ï ú µÕ™ ÖØ√o®·.

q + r r + p p + qÅ®·ûË p2, q2, r2 © ’ Åçéπ v¨Ï ú µÕ™ Öçö«ßª ’Eîª ÷°æçúÕ.

(™‰üΔ)Gçü¿’ ¤© ’ (2, 5), (−7, 3) ©ûÓ à®ΩpúË Í®ë« êçúøvAü∑Δ éπ ®Ωù Gçü¿’ ´¤ ©† ’ éπ† ’éÓ\çúÕ.

PART - BÆæ÷ îª † ©’:

i) ÅEo v°æ¨¡o©èπ◊ Ææ´ ÷ üµΔ Ø√© ’ ®√ßª ’çúÕ.1

ii) v°æA v°æ¨¡oèπ◊ ´ ÷®Ω ’\.2

1(10 ×× = 5 M)

2

118. log2 = −4 °∂æ ÷û√çéπ ®Ω ÷°æç? ( )

161 2 1

A) () = −4 B) (−4)2 = 16 16

1 1C) 24 = D) 2−4 =

16 1619. È®çúø’ éπ®Ω ùÃßª ’ Ææçêu© ¢Á·ûªhç? ( )

A) á©x °æ¤púø÷ éπ®Ω ùÃßª ’ Ææçêu Å´¤ ûª ’çCB) á©x °æ¤púø÷ Åéπ ®Ω ùÃßª ’ Ææçêu Å´¤ ûª ’çCC) äéπ éπ®Ω ùÃßª ’ Ææçêu é¬ † ´Ææ®Ωç ™‰ ü¿’D) àD é¬ü¿’

20. í∫ ’ù v¨Ïú µÕ™ n ´ °æü¿ç? ( )A) an = a1 rn B) an = a1 rn−1

C) an = a1 rn +1 D) an = a1 r2n

21. éÀçC Í®ë« *vûªç † ’ç* p(x) •£æ› °æ CéÀ Ö†o

−3Π 3Π¨¡⁄Ø√u© Ææçêu, x ∈ [ , ] ( )

2 2

A) 2 B) 3 C) 4 D) 522. ´®Ω_ ÆæO’ éπ ®Ωùç px2 + qx + r = 0 È®çúø’ ¢ËÍ®y®Ω ’

¢√Ææh´ ´‚™«© ’ éπLT Öçõ‰ ..... ( )A) q2 = 4pr B) q2 > 4pr

C) q2 < 4pr D) p2 = 4qr23. n(A − B) = 57, n(A∪B) = 120, n(B − A) = 48

Å®·ûË n (A∩B) ( )A) 75 B) 84 C) 51 D) 15

24. éÀçC ¢ÁØ˛ *vûªç † ’ç* A − B ÆæN’ A ™ E ´‚© é¬© ’.... ( )

A) A − B = {1, 2, 3, 4, 5, 12}B) A − B = {2, 4}C) A − B = {1, 5, 3, 12}D) A − B = {10, 6, 7}

25. éÀçC ¢√ öÀ™ 3x + 2y = 7 ÆæO’ éπ ®Ω ù« EéÀ Ææ´ ÷ç ûª®ΩÍ®ê† ’ Ææ ÷*çîË Í®&ßª ’ ÆæO’ éπ ®Ωùç? ( )

A) 2x + 3y = 7 B) 6x + 4y = 14C) 6x + 4y + 14 = 0 D) 2x + 3y + 7 = 0

26. éÀçü¿ Ê°®Ì\†o D®Ω` îª ûª ’ ®Ω v≤ƒ é¬®Ω Ææn©ç ¢Áj¨» ™«uEoûÁL ßª ’ñ‰ÊÆ ´®Ω_ •£æ› °æC? ( )

A) 2x2 + 5 B) 2x2 + 5xC) 5x2 + 2 D) 5x2 + 2x

27. (−1, 4) Gçü¿’´¤ ´ü¿l Íéçvü¿ç Ö†o ´%ûªh ¢√u≤ƒ EéÀ(3, 5) Gçü¿’´¤ äéπ *´J Gçü¿’´¤ Å®·ûËÈ®çúÓC....? ( )A) (−5, 3) B) (3, −5)C) (−5, 3) D) (−3, −5)

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MATHEMATICS

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MathematicsPaper-II

Telugu MediumParts A & B

Max. Marks: 40 Time: 2 Hrs 45 Min

PART - AÆæ÷îª† ©’:

i) °æKéπ~ ®√ßª’ ú≈ EéÀ ´·çü¿’ v°æ¨¡o °æ vû√Eo π~◊ùoçí¬ îªCN Å´ í¬ £æ«† îËÆæ’éÓçúÕ. DEéÓÆæç 15 EN’ ≥ƒ© Ææ´’ßª’ç Íéö« ®·ç î√®Ω’.

ii) É*a† 4 N¶µ« í¬© †’ç* ÅEo v°æ¨¡o © π◊Ææ´÷ üµΔ Ø√©’ ®√ßª’çúÕ.

iii) Section - IV ™E ©é~¬u ûªtéπ v°æ¨¡o © π◊Ææ´÷üµΔ†ç v¶«Èéö¸™ ®√ßª÷L.

iv) Section - III ™E v°æA v°æ ¨¡o π◊ Åçûª ®Ω_ûªáç°œéπ Öçô’çC.

SECTION - I

Ææ÷îª† ©’: ÅEo v°æ¨¡o © π◊ Ææ´÷ üµΔ Ø√©’ ®√ßª’çúÕ.v°æA v°æ¨¡o π◊ 1 ´÷®Ω’\.

(7 ×× 1 = 7 M)

1. Δ ABC ™ PQ II BC & AP : PB = 1 : 2 Å®·ûËΔ APQ & Δ ABC ¢Áj¨» ™«u© E≠æpAh áçûª?

2. äéπ ´%û√h EéÀ Ææp®Ωz Í® ê©’ UÊÆh, Ç Ææp®Ωz Gçü¿’ ¤©’,´%ûªh ¢√u≤ƒ EéÀ * ®Ω© ´ü¿l ©ç•çí¬ Öçö« ßª ’E©éπ~ tù˝ Åçô ’ Ø√oúø’. O’ Ææ ÷ üµΔ Ø√Eo Ææ ’JnçîªçúÕ.

3. °æéπ\ °æôç™ AB = 6ÂÆç.O’. AP = 4 ÂÆç.O’.Å®·ûË OP = ?

4. cos2 16° − sin2 74° N© ’´ áçûª?5. 3 °∂æ ’Ø√© ¶µº’ñ«© ’ 3 ÂÆç.O’., 4 ÂÆç.O’., 5 ÂÆç.O’.

¢√öÀE éπL°œ äÍé °∂æ ’†çí¬ ûªßª ÷®Ω ’ îËÊÆh üΔE ¶µº’ïçáçûª?

6. 3 Ø√ù‰ ©† ’ áí∫ ’ ®Ω ¢Ë Æ œ † °æ¤púø’, éπFÆæç äéπ ¶Ô´ ’t°æúø ö« EéÀ Ææç¶µ« úûª áçûª?

7. ®√¨¡Ÿ© ¢Á·û√hEo ®√¨¡Ÿ© ÆæçêuûÓ ¶µ«Tç îªí¬ Ææí∫ô ’´Ææ ’hçü¿E ®√V ÅØ√oúø’. DEo ≤ƒçÍé Aéπ ®Ω ÷°æç™ ®√ßª ’çúÕ.

SECTION - II

Ææ÷îª †©’: i) ÅEo v°æ¨¡o © π◊ Ææ´÷ üµΔ Ø√©’ ®√ßª’çúÕ.ii) v°æA v°æ¨¡o π◊ 2 ´÷®Ω’\©’.

(6 ×× 2 = 12 M)

8. äéπ úéÀh ûª ÷®Ω ’p ¢Áj°æ¤ 5 éÀ.O’. †úÕ* Åéπ\úÕ † ’ç*Öûªh®Ωç ¢Áj°æ¤ 12 éÀ.O’. ¢Á∞ «xúø’. Å®·ûË ¢Á·ü¿öÀ≤ƒn†ç † ’ç* *´J ≤ƒn†ç ´®Ωèπ◊ Ö†o ü¿÷®Ωçáçûª?

9. ° æ ô ç ™ OA = 72 Â Æ ç . O ’ .Å ® · û Ë Ê ≠ ú ˛ î Ë Æ œ † v § ƒ ç û ª¢Áj¨»©uç áçûª?

10. tan2 θ − sin2 θ = tan2 θ × sin2 θ ÅE îª ÷°æçúÕ.

11.

Â°j °æôç™ A † ’ îª ÷úøö« EéÀ P, Q © ´ü¿l E´ ’o éÓù«©† ’ éπ† ’éÓ\çúÕ.

12. äéπ ¶«uí∫ ’™ 5 † ’ç* 90 ´®Ωèπ◊ Ææçêu© ’ ®√Æ œ†é¬®Ω ’f© ’ ÖØ√o®·. üΔE † ’ç* äéπ é¬®Ω ’f† ’ßª ÷ü¿%*a ¥ éπçí¬ BÊÆh, ÅN éÀçC Nüµ¿çí¬ ®√´ ú≈ EéÀÆæç¶µ« ú ûª† ’ éπ† ’ éÓ\çúÕ.i) È®çúøç Èé© Ææçêu ii) °æJ °æ‹®Ωg °∂æ ’†ç

13. äéπ §˘†” °æ¤†u N¶µ« ï† °æöÀd éπ™ l = 30, h = 10,f = 15, f1 = 10 & f2 = 6 Å®·ûË ¶« £æ› ∞¡éπçáçûª?

SECTION - III

Ææ÷îª †©’: i) ÅEo v°æ¨¡o © π◊ Ææ´÷ üµΔ Ø√©’ ®√ßª’çúÕ.ii) v°æA v°æ¨¡o π◊ 4 ´÷®Ω’\©’.

(4 ×× 4 = 16 M)

14. 1500 O’. áûª ’h™ áí∫ ’ ®Ω ’ ûª ’†o N´ ÷ †ç † ’ç*†C ™ E È®çúø’ °æúø ´ ©† ’ í∫´ ’EÊÆh, ÅN 60°, 45°E´ ’o éÓ ùç™ éπE°œç î √®·. ÅN È®çúø÷ äÍé ¢Áj°æ¤

¢Á∞¡Ÿhçõ‰ ¢√öÀ ´ ’üµ¿u ü¿÷®Ωç áçûª?(™‰üΔ)

äéπ í¬V èπ◊çúŒ™ 10 Ø√Jçï, 7 ÊÆ°æ¤© ’, 15´ ÷ N’úÕ °æçúø’x ÖØ√o®·. üΔE † ’ç* ßª ÷ü¿% *a ¥ éπçí¬ äéπ °æçúø’† ’ BÊÆh ÅC éÀçC ¢√ öÀ™ äéπöÀé¬´ ú≈ EéÀ Ææç¶µ« úûª éπ† ’éÓ\çúÕ.

i) Ø√Jçï é¬´ ú≈ EéÀ ii) ´ ÷N’úÕ °æçúø’ é¬´ú≈EéÀ iii) Ø√Jçï é¬éπ§Ú´ ú≈ EéÀ iv) ÊÆ°æ¤ é¬éπ§Ú´ú≈ EéÀ

15. äéπ îªûª ’ ®Ω v≤ƒ é¬®Ω FöÀ ö«uçèπ◊ ¶µº’ïç 50 O’. üΔEv°æA ¶µº’ñ« EéÀ äéπ Å®Ωl¥ ´% û√h é¬®Ω í∫úÕfE Â°çî √ ©† ’èπ◊Ø√o®Ω ’. v°æA îª.O’.èπ◊ ÅßË’u ê®Ω ’a ®Ω ÷.2Å®·ûË, ¢Á·ûªhç áçûª ê®Ωa ´¤ ûª ’çC?

(™‰üΔ)6 ÂÆç.O’. ¢√u≤ƒ®Ωl¥ç Ö†o íÓ∞ «Eo éπJ Tç* 0.6ÂÆç.O’. ¢√uÆæç Ö†o *†o íÓ∞ « © ’í¬ ûªßª ÷®Ω ’îËÊÆh, ¢√öÀ Ææçêu† ’ éπ† ’éÓ\çúÕ.

16. éÀçC §˘†” N ¶µ« ï† °æöÀdéπ ´ ’üµ¿u í∫ûªç 25 Å®·ûËx = ?

(™‰üΔ)sin2 A − sin2 B

tan2 A − tan2 B = cos2 A . cos2 B

ÅE îª ÷°æçúÕ.17. AB = 6.5 ÂÆç.O’., ∠B = 50°, ∠A = 60° éÌ© ûª

©ûÓ ΔABC E EJtçîªçúÕ. ΔABC éÀ Ææy®Ω ÷ °æçí¬Öç ô ÷, üΔE ¶µº’ñ«™ x 23 ´çûª ’ ÖçúË™« Å† ’ ®Ω ÷°æ¶µº’ñ«© ’ éπL T† vA¶µº’ ñ«Eo EJtçîªçúÕ.

(™‰üΔ)3 ÂÆç.O’. ¢√u≤ƒ®Ωl¥ç ûÓ äéπ ´%û√hEo Ußª ’çúÕ. ´%ûªhÍéçvü¿ç † ’ç* 7 ÂÆç.O’. ü¿÷®Ωç™ Ö†o Gçü¿’´¤† ’ç* ´%û√h EéÀ Ææp®Ωz Í® ê© ’ Ußª ’çúÕ.

PART - B

SECTION - IVÆæ÷îª †©’: i) ÆæÈ®j† Ææ´÷ üµΔ Ø√Eo í∫’Jhç* üΔE Èé ü¿’ ®Ω’í¬ Ö†ov¶«Èéöx ®√ßª’çúÕ.

ii) ÅEo v°æ¨¡o © π◊ Ææ´÷üµΔ Ø√©’ ®√ßª’çúÕ.

1iii) v°æA v°æ¨¡o π◊ ´÷®Ω’\.

2 1(10 ×× = 5 M)

2

18. ΔABC ~ Δ DEF, ∠A = 80°, ∠F = 40° Å®·ûË∠C = ? ( )A) 50° B) 60° C) 70° D) 120°

19. °æôç™ PQ = ( ) A) 5 cm B) 8 cmC) 10 cm D) √

34 cm20. ¶«£æ«u Gçü¿’´¤ † ’ç* ´%û√h EéÀ UÆ œ† Ææp®Ωz Í® ê©

§Òúø ´¤© ’? ( )A) Ææ´ ÷ †ç B) ÅÆæ ´ ÷ †çC) A & B D) àD é¬ü¿’

1 121. tan θ + = 2 Å®·ûË tan2 θ + =

tan θ tan2 θ

( )A) 1 B) 2 C) 4 D) 3

22. cosec 30° + cot 45° = ? ( ) √

3 + 1A) 3 B) 2 C) D) 1

2√2

xi − 2523. ui = , Σ fiui = 20, Σ fi = 100 Å®·ûË 10

Ææí∫ô ’ ( x− ) = ? ( ) A) 23 B) 24 C) 27 D) 25

24. tan θ × cot θ = x Å®·ûË x2 + 5 = ? ( ) A) 1 B) 25 C) 0 D) 6

25. 25 Ææ£æ«ï Ææçêu ©† ’ Ç®Ó£æ«ùvéπ ´ ’ç™ Å´ ’JÊÆh¢√öÀ ´ ’üµ¿u í∫ûªç áçûª? ( ) A) 12 B) 14 C) 13 D) 15

26. 10 O’. áûª ’h† Ö†o Ææhç¶µ«EéÀ 20 O’. §Òúø¢Áj†EîÁa† θ éÓùç îËÆæ ’hçõ‰, θ = ( ) A) 0° B) 30° C) 45° D) 60°

27. 3, 4, 6, 6, 6, 7, 7, 7, 7, 7 © † ’ç* ßª ÷ü¿% *a ¥ éπçí¬ äéπ Ææçêu† ’ BÊÆh ÅC Ç Ææçêu© Ææí∫ô ’ÅßË’u Ææç¶µ« úûª áç ûª? ( )

1 2 3 7A) B) C) D)

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1. (−4, 0), (4, 0), (0,3) © ’ à vA¶µº’ï Q®√{© ’?A) ©ç•éÓù vA¶µº’ïç B) Ææ´ ’ Cy ¶« £æ› vA¶µº’ïçC) Ææ´ ’ ¶«£æ› vA¶µº’ïç D) N≠æ ´ ’ ¶«£æ› vA¶µº’ïç

2. A(1, 2), O(0, 0), C(a, b) © ’ Æ æ Í ® & ß ª ÷ © ’Å®·ûË....

A) a = b B) a = 2b C) 2a = b D) a = −b3. (0, 4) (0, 0), (3, 0) vA¶µº’ï Q®√{ ™„jûË ¢√öÀ

îª ’ô ’déÌ©ûªA) 5 B) 12 C) 11 D) 7 + √5

4. ABCD Ææ´ ÷ç ûª®Ω îªûª ’ ®Ω ’s ¥ï 3 Q®√{© ’ A(−2, 3),B(6, 7) & C(8, 3) Å®·ûË Ø√© ’íÓ Q®Ω{ç D N© ’´?

A) (0, 1) B) (−1, 0) C) (0, −1) D) (1, 0)

5. (a cos 35°, 0), (0, a cos 55°) © ´ ’üµ¿u ü¿÷®ΩçA) a B) 2a C) 3a D) àD é¬ü¿’

6. P(x1, y1), Q (x2, y2) E®Ω ÷ °æ é¬ ©† ’ X-Åéπ~ç

N¶µº> çîË E≠æpAh?A) y1 : y2 B) −y1 : y2C) x1 : x2 D) −x1 : x2

7. (0, 0) (sin 45°, cos 45°) Gçü¿’ ´¤ ©† ’ éπL T†Ææ®Ω∞¡ Í®ê ¢√© ’?

2 √3 A) 1 B) 0 C) D)

√3 2

8. O(0, 0), A(0, b), B(a, 0) Gçü¿’ ´¤© ’ Q®√{ © ’í¬ éπL T† vA¶µº’ï í∫ ’®Ω ’ûªy Íéç vü¿ç

b aA) (a, b) B) ( , )2 2a bC) ( , ) D) (b, a)2 2

9. (4, 8) Gçü¿’ ´¤èπ◊ X - Åé ~¬ EéÀ Ö†o ü¿÷®Ωç?A) 4 B) 8 C) 12 D) 4√

2 10. (a, 0), (0, b) & (1, 1) E®Ω ÷ °æ é¬© ’ ÆæÍ® & ßª ÷© ’

1 1Å®·ûË + =a bA) −1 B) 0 C) 2 D) 1

11. éÀçC ¢√é¬u™ x àC ÅÆæûªuç?i) ÅEo ´%û√h© ’ Ææ®Ω ÷ §ƒ© ’ii) ÅEo îªûª ’ ®Ω v≤ƒ© ’ Ææ®Ω ÷ §ƒ© ’

iii) ÅEo îªûª ’ ®Ω v≤ƒ© ’ Ææ®Ωy Ææ ´ ÷ †çiv) È®çúø’ vA¶µº’ ñ«© ¶µº’ñ«© E≠æpAh Å† ’ ®Ω ÷ °æ ¢Á ’iûË

ÅN Ææ®Ω ÷ §ƒ© ’.A) i B) ii C) iii D) iv

12. éÀçC ¢√é¬u™ x àC Ææûªuç?i) ÅEo Ææ´ ’ Cy ¶«£æ› vA¶µº’ ñ«© ’ Ææ®Ω ÷ §ƒ© ’ii) È®çúø’ Ææ®Ω ÷°æ °æö«© ’ Ææ®Ωy Ææ ÷ †çiii) È®çúø’ Ææ®Ω ÷°æ vA¶µº’ï ¶µº’ñ«© ’ Ææ´ ÷ †çiv) È®çúø’ Ææ®Ωy Ææ ´ ÷ † vA¶µº’ ñ«© ’ Ææ®Ω ÷ §ƒ© ’A) i B) ii C) iii D) iv

13.

AD ÅØËC ∠A éÀ îÁçC† éÓù Ææ ’ Cy êç úø† Í®ê Å®·ûË x = ?

A) 2 B) 8 C) 6 D) 10

14. ΔABC ™ AB, BC, CA © ´ ’üµ¿u Gçü¿’ ¤© ’ X,Y, Z Å®·ûË ΔABC & ΔXYZ © ¢Áj » ™«u© E≠æpAh?

A) 1 : 4 B) 2 : 3 C) 3 : 2 D) 4 : 1 15. éÀçC¢√ öÀ™ àN ©ç• éÓù vA¶µº’ï ¶µº’ñ«© éÌ©

ûª© ’ é¬´¤?A) 6, 8, 10 B) 2, 1, √

5

C) 9, 5, 7 D) 5, 12, 13

16. vA¶µ º’ïç ™ E ´ ’üµ¿u í∫û√© ¢Á·ûªhç? A) vA¶µº’ïç îª ’ô ’d éÌ ©ûª éπçõ‰ ûªèπ◊\´B) vA¶µº’ïç îª ’ô ’d éÌ ©ûª éπçõ‰ áèπ◊\Ć) vA¶µº’ïç îª ’ô ’d éÌ © ûªèπ◊ Ææ´ ÷ †ç D) àD é¬ü¿’

17. 12 ÂÆç.O’. áûª ’h†o éπv®Ω Fúø 8 ÂÆç.O’. ü¿÷®Ωç™ °æúø’ ûª ’çC. ÉüË Ææ´ ’ ßª ’ç™ äéπ ô´®˝ Fúø 40ÂÆç.O’. ü¿÷®Ωç™ °æúø’ ûª ’çC. Å®·ûË ô´®˝ áûª ’hO’ô®Ωx™ ...A) 30 B) 60 C) 45 D) 90

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