jee main practice

8/17/2019 Jee Main Practice

http://slidepdf.com/reader/full/jee-main-practice 1/7

MATHEMATICS TODAY

| APRIL ‘15 31

1. Te number o common divisors o 10800 and

9000 is

(a) 49 (b) 36 (c) 25 (d) 64

2. A, B are two students in a group o n students. I

the number o ways o assigning the n students to a line

o n single rooms such that A and B are not in adjacent

rooms is 3600, then n =(a) 10 (b) 8 (c) 7 (d) 9

3. Te sides AB, BC, CA o a triangle ABC have n,

n + 1, n + 2 (n ∈ N , n ≥ 3) interior points respectively on

them. I the number o triangles ormed by any three o

these (3n + 3) points is 205, then n =

(a) 7 (b) 6 (c) 4 (d) 3

4. I the tangent and normal o a rectangular

hyperbola xy = 4 cut off intercepts a1, a2 on x -axis and

b1, b2 on y -axis respectively, then a1a2 + b1b2 =

(a) c2 (b) 2c2 (c) 0 (d) 4c2

5. Te length o common tangent o hyperbolas

x y 2 2

4 31− = and

x y 2 2

3 41 0− + = is

(a) 27

(b) 7 2 (c) 7 (d) 72

6. tan cot− −

+

∫ ∫

1

0

21

0

2

x dx x dx =

(where [.] represents the greatest integer unction )

(a) 1 – cot 2 (b) 1 + cot 2

(c) 2(1 + cot 2) (d) 2(1 – cot 2)

7. I4 6

9 4

9 42e e

e e

dx Ax B e C x x

x x x −

−= + − +

−

− ∫ ln( ) , then

A+ B =

(a)−9

35 (b)

35

36 (c)

9

36 (d) None

8. I f x x x dx b a

f x c( )sin cos( )

ln ( ) ,=−

+ ∫ 1

2 2 2 then

f (x ) can be

(a)1

2 2 2 2a x b x cos sin+ (b)

12 2 2 2b x a x cos sin+

(c) a2cos2x + b2 sin2x (d) b2 cos2x + a2 sin2x

9. I sin2x – 2sinx – 1 = 0 has exactly our different

solutions in x ∈ [0, np], then minimum value o ‘n’ can

be (n ∈ N )(a) 4 (b) 3 (c) 2 (d) 1

10. P (a, b) is a point in first quadrant. I two circles

which pass through ‘P ’ and touch both the coordinate

axes cut at right angles, then

(a) a2 – 6ab + b2 = 0 (b) a2 + 2ab – b2 = 0

(c) a2 – 4ab + b2 = 0 (d) a2 – 8ab + b2 = 0

11. AB is diameter o a semi circle K , C is an arbitrarypoint on the semicircle (other than A orB ) and ‘S’ is the

centre o the circle inscribed in D ABC , then measure o (a) angle ASB changes as C moves on K (b) angle ASB is the same or all positions o ‘C ’ but

it cannot be determined without knowing theradius.

(c) angle ASB = 135° or all positions o C (d) angle ASB = 150° or all positions o C

Practice Paper 2015

JEE Main

Exam on

4th April

* ALOK KUMAR, B.Tech, IIT Kanpur

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).

He trains IIT and Olympiad aspirants.



MATHEMATICS TODAY

| APRIL ‘1532

12.x

x dx

7

2 51( )−= ∫

(a)x

x C

8

2 41( )−+ (b)

x

x C

8

2 84 1( )−+

(c)x

x C

8

2 48 1( )−+ (d)

x

x C

8

2 44 1( )−+

13. I f x x x x x ( ) ( ) ( )( )= + + + + + +1 1 1 1 2 4 then

f x dx ( ) = ∫ 0

100

(a) 5010 (b) 5050(c) 5100 (d) 5000

14. Degree o differential equation o the curve

y a e

x

a= −

−1 , where ‘a’ being the parameter is

(a) 1 (b) 2(c) 3 (d) not defined

15. I solution o y d y

dx

dy

dx x

2

2

2

+

= is given by

y x

C x C 23

1 2= + +l

, then l =

(a) 2 (b) 3(c) 5 (d) 4

16. I x ∈ R, then the maximum value o

y a x x x b= − + +2 2 2( )( ) is

(a) a2 + b2 (b) a2 – b2

(c) a2 + 2b2 (d) none o these

17. Te number o three digit numbers o the orm xyz

such that x < y and z ≤ y is(a) 176 (b) 278(c) 276 (d) 240

18. I (1.5)30 = k, then the value o ( . ) ,1 52

29n

n=∑ is

(a) 2k – 3 (b) k + 1(c) 2k + 7 (d) 2k – 9/2

19. Te set o values o x or which

1

1

1

1

1

1+ − −x x x , , are in A.P. is given by

(a) (–∞, ∞) (b) (1, ∞)(c) [0, ∞) – {1} (d) [0, ∞)

20. Te ratio o nth terms o two A.P.’s is

(14n – 6) : (8n + 23). Ten the ratio o their sum o first

m terms is

(a)4 4

7 24

m

m

++

(b)7 1

4 24

m

m

−+

(c)7 1

4 24

m

m

−− (d)

7 1

4 27

m

m

++

21. I A = −

1 1

1 1, then A16 =

(a)0 256

256 0

(b)

256 0

0 256

(c)−

−

16 0

0 16 (d)

0 16

16 0

22. angents are drawn to the circle x 2

+ y 2

= 12 at thepoint where it is met by the circle x 2 + y 2 – 5x + 3 y – 2 = 0.

Te point o intersection o these tangents is

(a)18

56,

(b) 618

5, −

(c) (4, –5) (d) (5, –3)

23. Let a, b and c be positive real numbers. Ten the

ollowing system o equations in x, y and z

x

a

y

b

z

c

x

a

y

b

z

c

x

a

y

b

z

c

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

21 1 1+ − = − + = − + + =, ,

has(a) no solution (b) unique solution(c) infinitely many solutions(d) finitely many solutions

24. I the line lx + my + n = 0 cuts the ellipsex

a

y

b

2

2

2

21+ =

in points whose eccentric angles differ byp2

, then

a l b m

n

2 2 2 2

2

+=

(a) 1 (b) 2 (c) 4 (d) 3/2

25. I F x n x x x n

n n( ) ( ),/ /( )= − >→∞

+lim 2 1 1 1 0 then

xF x dx ( ) ∫ is equal to

(a)x

x c2

2+ +ln (b) − + + + +

x x x

x c

2 2

4 2ln

(c)x

x x c3

3+ +ln (d)

x x

x c

2 2

2 4. ln − +

26. Te number o zero’s at the end o 60! is(a) 14 (b) 15 (c) 16 (d) 10



MATHEMATICS TODAY

| APRIL ‘15 33

27. Te number ormed by last two digits o the

number (17)256 is(a) 81 (b) 82 (c) 91 (d) 93

28. An ellipse o eccentricity2 2

3 is inscribed in a

circle and a point with in the circle is chosen at random.

Te probability that this point lies outside the ellipse, is(a) 1/9 (b) 4/9 (c) 1/3 (d) 2/3

29. In the binomial distribution 3

5

2

5

5

+

i variance

is V and standard deviation S, then(a) S = V 2 (b) 4(S – V 2) = V 2 (c) 5(V 2 – S2) = S2 (d) none o these

30. I C r stands or nC r , then the sum o the series

(where n is an even positive integer)

22 2 2 3 1 10

212

22 2

n n

nC C C n C n

n

− + + + − +! !

![ ..... ( ) ( ) ]

is equal to(a) 0 (b) (–1)n/2(n + 1)(c) (–1)n(n + 2) (d) None o these

31. Te coefficient o x 3l + 2, m the expansion o

(a + x )l (b + x )l + 1 (c + x )l + 2 (l is a positive integer) is(a) l (a + b + c) (b) l (a + b + c) + b + 2c(c) l (a + b + c) + a + 2b + 3c (d) None o above

32. I n is a natural number , then nn – nC 1(n – 1)n +nC 2(n – 2)n – nC 3(n – 3)n + ....... must be equal to(a) n! (b) (n!)n (c) nn! (d) 0

33. Tree normals are drawn to the curve y 2 = x rom

a point (c, 0). Out o three, one is always on x -axis. I

two other normals are perpendicular to each other,

then the value o c is(a) 3/4 (b) 1/2 (c) 3/2 (d) 2

34. I area o triangle ormed by tangents rom the

point (x 1

, y 1

) to the parabola y 2 = 4ax and their chord

o contact is

(a)( ) / y ax

a

12

13 2

2

4

2

− (b)

( ) / y ax

a

12

13 3

2

4−

(c) ( ) / y ax

a12

13 24

2

− (d) none o these

35. A normal to the parabola y 2 = 4ax with slope ‘m’

touches the rectangular hyperbola x 2 – y 2 = a2 i,(a) m6 + 4m4 – 3m2 + 1 = 0(b) m6 – 4m4 + 3m2 – 1 = 0

(c) m6

+ 4m4

+ 3m2

+ 1 = 0(d) m6 – 4m4 – 3m2 + 1 = 0

36. Te number o ways o selecting 10 balls out o an

unlimited number o white, red, blue and green balls is(a) 270 (b) 84 (c) 286 (d) 86

37. A five digit number divisible by 3 is to be ormed

using the numerals 0, 1, 2, 3, 4 and 5 without repetition.

Te total number o ways in which this can be done is

(a) 216 (b) 240 (c) 600 (d) 3125

38. Te number o ways in which 7 persons can be

seated at a round table i two particular persons are not

to sit together is(a) 120 (b) 480(c) 600 (d) 720

39. A is a set containing ‘n’ elements. A subset P o A

is chosen. Te set A is reconstructed by replacing the

elements o P . A subset Q o A is again chosen. Te

number o ways o choosing P and Q, so that P ∩ Q

contain exactly two elements is(a) 9 nC 2 (b) 3n – nC 2 (c) 2 nC n (d) nC 2 × 3n – 2

40. Te number o +ve integer satisying the inequalityn+1C n–2 – n+1C n–1 ≤ 100 is(a) nine (b) eight(c) five (d) None o these

41. Te solution o the differential equation

dy

dx

y y dx = +

∫ 0

1

, given that y = 1, when x = 0 is

(a) 2 1

3

e e

e

x − +−

(b)2 1

3

e e

e

x + ++

(c)2 1

3

e e

e

x + +−

(d)e e

e

x − +−2 1

3

42. Te polynomial unction f (x ) o degree 6, which

satisfies lim

/

x

x f x

x e

→+

( )

=0 3

121 and has local maximum

at x = 1 and local minimum at x = 0 and x = 2 is

(a) f x x x x ( ) = − +2 122

34 5 6

(b) f x x x x ( ) = − +212

524 5 6

(c) f x x x x ( ) = − +212

5

2

34 5 6

(d) f x x x x ( ) = + −4 5 6125 23



MATHEMATICS TODAY

| APRIL ‘1534

43. I f x x xy x y f y dy ( ) ( ) ( ) ,= + + ∫ 2 2

0

1

then f (x ) is

(a)180

119

80

1192x x + (b)

180

119

80

1192x x −

(c) − +180119

80119

2x x (d) − + 180119

80119

2x x

44. A periodic unction with period ‘1’ is integrable

over any finite interval. Also or two real numbers a, b

and or two unequal non-zero positive integers m & n

f x dx f x dx

b

b m

a

a n

( ) ( )=++

∫ ∫ , then the value o f x dx

m

n

( ) ∫ is

(a) 0 (b) 1/2 (c) 2 (d) mn

45. I y arc x

x

dy dx

=

cos cos

cos,3

3then is

(a) 3

4 2cos cosx x + (b)

2

4 2cos cosx x +

(c)6

4 2cos cosx x + (d) 6

4 2cos cosx x −

SOLUTIONS

1. (b) : 9000 = 23 × 53 × 32 and 10800 = 24 × 33 × 52.

A common divisor is o the orm 2 3 51 2 3a a a , where0 ≤ a1 ≤ 3, 0 ≤ a2 ≤ 2, 0 ≤ a3 ≤ 2. Hence, number ocommon divisors is (3 + 1)(2 + 1)(2 + 1) = 36.

2. (c) : reating 2 adjacent rooms as a single unit,number o ways o assigning n rooms to n students is2(n – 1)! and total number o ways o assigning roomsis n!. Hence n! – 2(n – 1)! = 3600 (given)i.e., (n – 2)(n – 1)! = 5(6!) ⇒ n = 7

3. (d) : Tere are totally (3n + 3) points and i noneo them are collinear, we can orm (3n+3)C 3 triangles.

But by joining any 3 on AB (or AC or BC ), we do notget any triangle, hence given(3n+3)C 3 – (nC 3 + (n+1)C 3 + (n+2)C 3) = 205 ⇒ n = 3

4. (c) : angent is perpendicular to normal at any

point and slope o a line = −intercept on -axis

intercept on -axis

y

x

5. (b) : y mx m= + −4 32 is a tangent to 1st hyperbolaand it becomes a common tangent i m = ± 1. Hencecommon tangents are y = ±x + 1. Te points o contact

are (–4, –3) and (3, 4), hence length o common tangentis 7 2

6. (c) : [tan ]; tan

;− =

≤ <≥

1 0 1

1x

x

x

0

tan1

[cot ]; cot

;− =

≤ ≤

1 1 1

0x

x

x

0

>cot1

1 1 2 1 1 2 1 2

0

1

1

2dx dx + = − + = + ∫ ∫

cot

tan

tan cot ( cot )

7. (b) : 4 6 9 4 9 4e e A e e Bd

dx e ex x x x x x − = − + −− − −( ) ( )

8. (b) : f x x x b a

f x

f x ( ) sin cos

( )

( )

( ) =

1

2 2 2−

′

⇒ 2 22 22

b x x a x x f x

f x sin cos sin cos

( )

( ( ))− =

′

Integrating both sides w.r.t ‘x ’, we get

− − = −

b x a x f x

2 2 2 2 1cos sin

( )

⇒ f x a x b x

( )sin cos

=+

12 2 2 2

9. (a) : (sin ) sinx x − = ⇒ − = ±1 2 1 22

⇒ sin x = −1 2 (since sinx is not greater than 1)

2 solutions in [0, 2p] and two more in [2p, 4p]

10. (c) : Let the circle be x y cx c y c2 2 2 2 0+ − − + =

Since it passes through (a, b), we have

c c a b a b− + + + =2 02 2( )

⇒ c c a b c c a b1 2 1 22 22+ = + = +( );

Te circles are orthogonal

\ 4 1 2 1 2c c c c= + , ⇒ a2 – 4ab + b2 = 0

11. (c) : ∠ = ° ∠ = °+ = °C ASB C 90 902

135,

12. (c) :

1

11

11

3

2

5 2x

x

dx x

t

−

− = ∫ , Let

⇒ − = −−

=

−

+ =−

+− −

∫ 1

2

1

2 4

1

8 1

18 1

54

2

4

8

2 4t dt

t

x

C x

x

C

( )

13. (c) : f (x ) = (x + 1), x ∈ [0, 100]



MATHEMATICS TODAY

| APRIL ‘15 35

14. (d) : D.E. ormed afer eliminating a is

y x

dy

dx

dy

dx = −

−

log

1

Hence degree is not defined.

15. (b) : d dx

y dy dx

x =

⇒ = + \ = + +

\ = + + \ =

y dy

dx

x P

y x Px q

y x

C x C

2 23

23

1 2

2 2

1

6

33l

16. (a) : Let x x b t + + =22

⇒ x b x b

t

2 22

+ − =

⇒ 2 22 22 2

x b t b

t x t

b

t + = + = −; .

\ − + +( )( )2 2 2 2a x x x b

= − +

= − +

= + − − ≤ +

2 22

2 2

2 2 2 2 2

a t b

t t at t b

a b t a a b( ) ( )

17. (c) : I zero is included, it will be at z, then no.o three digit numbers = 9C 2. I zero is excluded,

then

x y z C

x z y C

x y z C

, , !all di ⇒ ×

= < ⇒

< = ⇒

93

92

92

2

\ otal number o ways = 276

18. (d): S kr

r

= − − = −

−

− = −

=

∑ ( . ) .( . )

.

.1 5 1 1 51 5 1

1 5 1

2 5 29

20

29 30

19. (c) :1

1

1

1

2

1+ +

− =

−x x x is true or all x and it

is defined or x ≥ 0, x ≠ 1

20. (d) :T

T

a n d

a n d

a n d

a n d

nn

n′ =

+ −+ −

= + −

+ − =

−1 1

2 2

1 1

2 2

1

1

2 2 2

2 2 2

14 6

8

( )

( )

( )

( ) nn+23

Put 2n – 2 = m –1 in both sides,

⇒2 1

2 1

7 1

4 271 12 2

a m d

a m d

m

m

+ −+ − =

++

( )

( )

21. (b) : A A A2 4 80 2

2 0

4 0

0 4

16 0

0 16=

−

=

−−

=

, , ;

A16 256 0

0 256=

22. (b) : Common chord is 5x – 3 y – 10 = 0. Te chordo contact o required point w.r.t circle x 2 + y 2 = 12 isalso the common chord o given two circles. Chord ocontact o P (x 1, y 1) is xx 1 + yy 1 – 12 = 0

⇒x y 1 1

5 3

12

10=

− =

−−

23. (d) : Letx

a X

y

bY

z

cZ

2

2

2

2

2

2= = =, , ⇒ X + Y – Z = 1,

X – Y + Z = 1, – X + Y + Z = 1. On solving X = Y = Z = 1⇒ x = ± a, y = ± b, z = ± c ⇒ 8 solutions

24. (b) : Let the point o intersection o the line andthe ellipse be (a cosq, bsinq) and

a bcos , sin .p

q p

q2 2

+

+

Since they lie on the

given line lx + my + n = 0,la cosq + mb sinq +n = 0 ⇒ la cosq +mb sinq = n and–la sinq + mb cosq + n = 0 ⇒ la sinq – mb cosq = n.Squaring and adding, we get

a l b m na l b m

n

2 2 2 2 22 2 2 2

22 2+ = ⇒

+=

25. (d) : F x n x x n

n n n( )

/ ( )= −

→∞

+( ) +lim 2 1 1

1

11

=−

+ ×

+ =

→∞

+ +( )

limn

n n nx x

n n

n n

n

x

1 1 1 1

2

1

1

1

1

/( ) /

( )

( )log

Hence,

xF x dx x x dx x

x x c( ) log log= = − + ∫ ∫ 2

2

2

1

4

26. (a) : Number o zero’s at the end o60! = exponent o 10 in 60!

= min {E2(60!), E5(60!)} = E5(60!) = 14

27. (a) : (17)256 = (289)128 = (300 – 11)128

= 128C0(–11)128 + 100m, or some integer m

= 11128 + 100m = (10 + 1)128 + 100m

= 128C 01128 + 128C 110 + 100m1 + 100m or some

integer m1

= 1 + 1280 + 100k, (m + m1 = k)

= 1281 + 100kHence the required number is 81.



MATHEMATICS TODAY

| APRIL ‘1536

28. (d) : Let the radius o the circle be a, then the majoraxis o the inscribed ellipse is o length 2a.

Te required probability = − −p p

p

a a e

a

2 2 2

2

1

= − −1 1 2e = − − =1 1

8

9

2

3(area o ellipse = = ⋅ −p pab a a e1 2 ,

‘e’ being eccentricity)

29. (c) : V npq= = =53

5

2

5

6

5. .

S =6

5 and 5 5

36

25

6

52 2( )V S− = −

= −

5

36

25

30

25 = =

6

5

2S

30. (d) : Let us test the choice or n = 2For n = 2, the series

= − ⋅ ⋅ = −2 1 1

240

22

( !)( !)

![ ]2 2

12 2 2C 2 C +3 C

Now observe the table

Choice Value at n = 2

(a) 0

(b) –3

(c) 4Since none o the choices become – 4, the correctchoice must be (d).

31. (b) : Given expression = (x + a) (x + a) . . . l times(x + b) (x + b) ………( l + 1) times × (x + c) (x + c). . . ( l + 2) times

= + + + +x al b l c l x l l 3 3

1+3 + 2+ + 2[ ( ) ( )] ...

⇒ co-efficient o x 3l + 2 is (a + b + c)l + b + 2c Choice (b) is correct.

32. (a) : nn

–n

C 1(n – 1)n

+n

C 2(n – 2)n

+ ....Number o ways o districting n objects in n districtcells such that no cell remains empty = n!

33. (a) : Normal at (at 2, 2at ) is y + tx = 2at + at 3 a =

1

4

I this passes through (c, 0)

We have, ct = 2at + at 3 =t t

2+

4

3

⇒ t = 0 or t 2 = 4c – 2I t = 0, the point at which the normal is drawn is

(0, 0). I t ≠ 0, then the two values o t represents slopeo normals through (c, 0)

I these normals are perpendicular, then

( )( ) ( )( )− − = − ⇒ = − ⇒ − − − = −t t t t c c1 2 1 21 1 4 2 4 2 1

⇒ c =3

4

34. (c) : Let A(x 1, y 1) be any point outside the parabola

and B(a, b), C (a′, b′) be the points o contact o tangentsrom point A. Eq. o chord BC , yy 1 = 2a(x + x 1)Lengths o ^ AL rom A to BC

ALa x x y y

y a

y ax

y a=

−=

−2 +

+ 4

4

+ 4

1 1

2

12

12 2

( )1 1

12

1

Area o =1

2D ABC AL BC ×

We get,41

21( ) / y ax

a

− 3 2

2

35. (c) : Equation o the normal to the parabola y 2 = 4ax with slope m is y = mx – 2am – am3. It touchesthe rectangular hyperbola x 2 – y 2 = a2 i(–2am – am3)2 = a2(m2 – 1)

⇒ a2m2(2 + m2)2 = a2(m2 – 1)

⇒ m2(m4 + 4m2 + 4) = m2 – 1

⇒ m6 + 4m4 + 3m2 + 1 = 0

36. (c) : Let x 1, x 2, x 3 and x 4, be the no. o white, red,blue and green balls that are selected. Ten x 1 + x 2 +x 3 + x 4 = 10. Te required no. o ways

= coefficient o y 10 in (1 + y + y 2 + y 3 + ….)4= coefficient o y 10 in (1 – y )–4

= coefficient o y 10 in (1 + 4C 1 y + 5C 2 y 2 + 6C 3 y 3 + ...+ 13C 10 y 10....)

= 133

13 12 11

2 3286C =

× ××

=

37. (a) : Te sum o the numerals 0, 1, 2, 3, 4 and 5is 15. We know that a five digit number is divisibleby 3 i and only i the sum o its digits is divisible by3. Tereore, we should not use either 0 or 3 whileorming the five digit number. I we do not use 0, then

the remaining digits can be arranged in 5P 5 = 5! = 120ways. I we do not use 3, then the remaining digitscan be arranged in 5P 5 – 4P 4 = 5! – 4! = 120 – 24 = 96ways to obtain a five digit number. Tus, the total no.o such 5 digit numbers is 120 + 96 = 216.

38. (b) : 6! – 2! 5! = 480

39. (d) : A = {a1, a2, …., an}

(i) ai ∈ P , ai ∈ Q

(ii) ai ∈ P , ai ∉ Q

(iii) ai

∉ P , ai

∈ Q

(iv) ai ∉ P , ai ∉ Q



MATHEMATICS TODAY

| APRIL ‘15 37

P ∩ Q contains exactly two elements, taking 2 elementsin (i) and (n – 2) elements in (ii) or (iii) or (iv)No. o ways = nC 2 × 3n–2

40. (b) : n+1C n–2 – n+1C n–1 ≤ 100

⇒ − ≤ ⇒ + − −+

≤+ +n n

C C n n n n n1

31

2 100 1

6

1 1 1

2

100( ) ( ) ( )

⇒ + − − ≤ ⇒ + − ≤1

61 1 3 100 1 4 600n n n n n n( ){ } ( )( )

It is true or n = 2, 3, 4, 5, 6, 7, 8, 9

41. (a) : Let y f x f x

f x = ⇒

′′

′ =( )

( )

( )1

⇒ = + f x Ce Dx ( ) ( )

At x = 0, y f x C e

De

= = ⇒ =−

= −−

( ) ,12

31

2

3

42. (c) : f x ax bx cx ( ) = + +4 5 6

a f x x bx cx = ′ = + +2 8 5 63 2, ( ) ( )

b c= − =12 5 2 3/ , /

43. (a) : A y f y dy B y f y dy = = ∫ ∫ 2

0

1

0

1

( ) , ( )

f x x Ax Bx ( ) = + + 2

⇒ = = A B

61

119

80

119,

44. (a) : f (1 + x ) = f (x )

\ = =

∫ ∫ ∫

+

f x dx f x dx n f x dx n

a

a n

( ) ( ) ( )

0

1

0

similarly only m f x dx ( )

0

1

∫

⇒ ( ) ( )n m f x dx − = ∫ 0

1

0 ⇒ = ∫ f x dx ( )

0

1

0

f x dx f m x dx f x dx

m

n n m n m

( ) ( ) ( ) ∫ ∫ ∫ = + =− −

0 0

= − = ∫ ( ) ( )n m f x dx

0

10

45. (c) : coscos

cos

23

3 y

x

x =

⇒dy

dx

x

y =

3 2sec

cos

MTGo f f e r s “ C l a s s r o o mS t u d y M a t e r i a l ” f o r

JEE (Ma in & Advanced) , A IPMT andFOUNDATION MATERIAL for Class 8, 9, 10, 11& 12 with YOUR BRAND NAME & COVERDESIGN.

This study material will save you lots of moneyspent on teachers, typing, proof-reading andprinting. Also, you will save enormous time.Normally, a good study material takes 2 years todevelop. But you can have the material printedwith your logo delivered at your doorstep.

Profit from associating with MTG Brand – themost popular name in educational publishingfor JEE (Main & Advanced)/AIPMT/PMT ....

Order sample chapters on Phone/Fax/e-mail.

Phone : 0124-4951200

09312680856

e-mail : [email protected] | www.mtg.in

ATTENTION COACHING INSTITUTES:

a great offer from MTG

C O N T E N T

PA P E R

P R I N T I N G

�

�

�

� E X C E L L E N

T

Q UA L I T Y

�

CLASSROOM

STUDY MATERIAL

Your logohere

jee main practice

Documents