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HSEB Model Question - I (2068)

Mathematics Grade: XII Full Marks: 100 Time: 3 hrs Pass Marks: 35

Candidates are required to give their answer in their own words as far as practicable. The figures in the margin indicate full marks. Attempt ALL questions of group A and group B or C.

Group A

1. a) It is required to seat 5 boys and 4 girls in a row so that the girls occupy the even places. How many such arrangements are possible? [2]

Solution

Suppose the seats in a row are numbered from 1 to 9. Then the 4 girls can occupy 2nd, 4th, 6th and 8th seats. Hence they can occupy their seats in P (4, 4) ways.

The 5 boys can occupy the remaining 5 seats in P (5, 5) ways. Therefore, by the multiplication rule of counting, the girls and boys can be arranged in

P (4, 4)� P (5, 5) = 4!

(4 ñ 4)! � 5!

(5 ñ 5)! = 4� 3� 2� 1

0! � 5� 4� 3� 2� 1

0!

= 2880 ways.

b) Prove that: 1

1.3 + 1

2.5 + 1

3.7 + 1

4.9 + . . . = 2 (1 ñ ln 2) . [2]

Solution

11.3 +

12.5 +

13.7 +

14.9 + . . . = 2 ⎝

⎛⎠⎞1

2.3 + 1

4.5 + 1

6.7 + 1

8.9 + . . .

2 HSEB Model Question I 2068

= 2 ⎣⎡

⎦⎤

⎝⎛

⎠⎞1

2 ñ 13 + ⎝

⎛⎠⎞1

4 ñ 15 + ⎝

⎛⎠⎞1

6 ñ 17 + ⎝

⎛⎠⎞1

8 ñ 19 + . . .

= 2 ⎣⎡

⎦⎤1 ñ ⎝

⎛⎠⎞1

2 + 13 ñ

14 +

15 ñ

16 +

17 ñ

18 +

19 + . . .

= 2 (1 ñ ln 2)

c) Let a * b = 3a + 2b for a, b ∈ . Verify that * is a commutative binary operation on . [2]

Solution

Let a, b ∈ . Then a ∗ b = 3a + 2b is also an integer. i.e. 3a + 2b ∈ . This implies that for each ordered pair (a, b) ∈ � , the number a∗b = 3a + 2b ∈ uniquely. Hence ∗ is binary operation.

Take a = 1, b = 5 in . Then

a ∗ b = 1 ∗ 5 = 3⋅1 + 2⋅5 = 13

But b ∗ a = 5 ∗ 1 = 3⋅5 + 2⋅1 = 17

∴ a ∗ b ≠ b ∗ a

This shows that ∗ is not commutative.

2. a) Find the equation of a hyperbola in standard position such that the length of transverse axis is 6 and it passes through (4, 2). [2]

Solution

Let the equation of hyperbola be

x2

a2 ñ y2

b2 = 1 . . . (i)

Given that the length of the transverse axis is

2a = 6

⇒ a = 3.

if (i) passes through the point (4, 2), then

(4)2

(3)2 ñ (2)2

b2 = 1

⇒ 169 ñ

4b2 = 1

⇒ 4b2 =

169 ñ1 =

79

∴ b2 = 367 .

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Putting the value of a and b in (i), we get

x2

9 ñ y2

36/7 = 1

⇒ x2

9 ñ 7y2

36 = 1

⇒ 4x2 ñ 7y2

36 = 1

∴ 4x2 ñ 7y2 = 36,

which is the required equation.

b) Find the locus of points which are equidistant from the points (1, 2, 3) and (3, 2, ñ1). [2]

Solution

Let A(1, 2, 3) and B(3, 2,ñ1) be given points and P(x, y, z) be the point on the locus. If AP = BP, then

AP2 = BP2

⇒ (x ñ 1)2 + (y ñ 2)2 + (z ñ 3)2 = (x ñ 3)2 + (y ñ 2)2 + (z + 1)2

⇒ x2 ñ 2x + 1 + z2 ñ 6z + 9 = x2 ñ 6x + 9 + z2 + 2z + 1

⇒ 4x ñ 8z = 0

Therefore, the required equation of the locus is x ñ2z = 0.

c) Find the cosines of the angle between the vectors:

a→

= (1, ñ 2, ñ 2), b→

= (2, 1, ñ 2). [2]

Solution

We have

a→ = (1, ñ2, ñ2), b→ (2, 1, ñ2).

Then

a→ . b→ = 2 ñ 2 + 4 = 4

a = | a→| = 1 + 4 + 4 = 3

b = | b→| = 4 + 1 + 4 = 3.

Let θ be the angle between the vectors. Then

cos θ = a→ . b→

ab = 4

3 � 3 = 49 .

∴ cos θ = 49 .


3. a) Find the derivative of (ln x)sinh x. [2]

Solution

Let y = (ln x)sinh x. Then

ln y = sinh x ln (ln x) Differentiating both sides w.r.t. x, we get

ddx ln y =

ddx sinh x ln (ln x)

⇒ 1y

dydx = sinh x

ddx ln (ln x) + ln (ln x)

ddx sinh x

⇒ dydx = y ⎣

⎡⎦⎤sinh x �

1ln x �

1x + ln (ln x) cosh x

∴ dydx = (ln x)sinh x ⎣

⎡⎦⎤sinh x

x ln x + cosh x ln (ln x) .

b) Find the integral ⌡⌠ dx

1 + 2 sin x . [2]

Solution

We have

⌡⌠ dx

1 + 2 sinx .

Put y = tan x/2. Then

dy = 12 sec2 x/2 dx ⇒ dx =

2dy1 + y2 .

Also

sin x = 2 tan y/2

1 + tan2 y/2 =

2y1 + y2 .

Now,

⌡⌠ dx

1 + 2 sinx = 2

⌡⎮⌠

dy1 + y2

1 + 2 ⎝⎛

⎠⎞2y

1 + y2

dx = 2 ⌡⌠ dy

1 + y2 + 4y

= 2 ⌡⌠ dy

(y + 2)2 ñ 3 = 2 � 1

2 3 ln

⎪⎪⎪

⎪⎪⎪y + 2 ñ 3

y + 2 + 3 + C

= 13

ln ⎪⎪⎪

⎪⎪⎪tan x/2 + 2 ñ 3

tan x/2 + 2 + 3 + C.

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c) Find the integral ⌡⌠ dx

(x + 7) 2 ñ x . [2]

Solution

Put 2 ñ x = y2. Then

dx = ñ 2y dy, x + 7 = 2 ñ y2 + 7 = 9 ñ y2 Now,

⌡⌠ dx

(x + 7) 2 ñ x = ñ ⌡

⌠ 2y dy(9 ñ y2). y = ñ2⌡

⌠ dy9 ñ y2 = 2⌡

⌠ dy y2 ñ 9

= 2 1

2 � 3 ln⎪⎪

⎪⎪y ñ 3

y + 3 + C = 13 ln

⎪⎪⎪

⎪⎪⎪2 ñ x ñ 3

2 ñ x + 3 + C.

4. a) Solve the differential equation: dydx = ex + y + 3x2 ey. [2]

Solution

We have

dydx = ex + y + 3x2 ey

⇒ dydx = ey(ex + 3x2).

⇒ eñy dy = (ex + 3x2) dx

On integration, we get

ñ ey = (ex + x3) + C

∴ ex + ey + x3 + C = 0.

b) From the following data, calculate the expected value of Y when X = 25,

X Y Average Standard deviation

5.6 3.2

12.5 2.4

and correlation coefficient r = 0.95. [2]

Solution

We have

óX = 5.6, óY = 12.5,

σX = 3.2, σY = 2.4, r = 0.95.


Then the regression coefficient of Y on X is

b = r σY

σX = 0.95 �

2.43.2 . = 0.7125.

Regression equation of Y on X is

Y ñ óY = b (X ñ óX )

⇒ Y ñ 12.5 = 0.7125 (X ñ 5.6).

∴ Y = 0.7125X + 8.51.

Putting X = 25, we get

Y = 0.7125 � 25 + 8.51.

∴ Y = 26.3225.

c) The average percentage of failures in a certain examination is 40. What is the probability that out of 5 candidates, at least 3 will be passed in the examination. [2]

Solution

We have

p = P(failure) = 40% = 0.4, q = P(not failure) = 1 ñ 0.4 = 0.6.

Also, n = 5.

Therefore, the probability that at least 3 will be passed in the examination, i.e. at most 2 will be failed is given by

P(x ≤ 0) = P(x = 0) + P(x = 1) + P(x = 2)

= 5C0 q5 + 5C1 p q4 + 5C2 p2 q3

= (0.6)5 + 5 � 0.4 (0.6)4 + 10 (0.4)2(0.6)3

= 0.07776 + 0.2592 + 0.3456

= 0.683.

5. a) Show that the number of combinations of n different objects taken r at a time is given by

C(n, r) = n!

(n ñ r)! r! .

Also, prove that C(n, n ñ r) = C(n, r). [4]

OR

State the multiplication principle of counting. Prove that the number of circular permutations of n different objects taken all at a time is (n ñ 1)! [4]

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Solution

First part

Let x be the required number of combinations. Each of these combinations has r different objects. So, these r objects can be arranged among themselves in r! ways with regard to order. Thus, each combination will produce r! permutations. As a result, x combinations will produce x � r! permutations. But these are all possible permutations of n objects taken r at a time. Hence

x � r! = P(n, r)

∴ x = P(n, r)

r! = n!

(n ñ r)! r! .

Thus, the number of combinations of n objects taken r at a time is given by

C(n, r) = n!

(n ñ r)! r! .

Second part

We have

C(n, n – r) = n!

(n ñ (n ñ r))! (n ñ r)!

= n!

r! (n ñ r)! = C(n, r).

OR Part

Multiplication Principle of counting: If one task can be accomplished in m ways and, following this task, a second task can be accomplished in n ways, then the first task followed by the second task can be accomplished in mn ways.

In circular arrangement the position of each object is determined relative to other objects in the arrangement. Hence the first object can be arranged in only one way, because for the first object there will not be the question: relative to which the first object is to be kept fixed. After arranging the first object, the remaining n ñ 1 objects are arranged relative to the first and so, they can be arranged in (n ñ 1)! ways. Then by the multiplication principle of counting the first object followed by the remaining objects can be arranged in

1 � (n ñ 1)! = (n ñ 1)!

ways. Therefore, the number of circular permutations of n objects taken all at a time is (n ñ 1)!.


b) What is a group? If a binary operation * is defined on a set

S = {a, b, c} by the following Caleyís table

* a b c a a b c b b c a c c a b

show that (S, *) is a group. [4] OR

Let a, b, c and x be elements of a group G. Solve for x if axb = c and x2 b = x añ1 c. [4]

Solution

Group: A group is a set G together with a binary operation * on G such that each of the following axioms is satisfied: G1. Closure: a, b ∈ G ⇒ a * b ∈ G.

G2. Associative: ∀ a, b, c ∈ G a * (b * c) = (a * b) * c

G3 Identity element: ∃ e ∈ G: ∀ a ∈ G a * e= e * a = a.

G4 Inverse element: ∀ a ∈ G ∃ a -1 ∈ G: a * a -1 = a -1 * a = e.

Second Part

(S, *) is a group. In fact, the above table shows that

i. S is closed under the binary operation. ii. We have

(a * a) * x = x = a * (a * x), for any x ∈ S

(a * b) * b = c = a * (b * b),

(a * c) * c = b = a * (c * c)

(a * b) * c = a = a * (b * c)

(b * b) * c = b = b * (b * c)

and so forth. Hence the binary operation is associative.

iii. We also have a * a = a, a * b = b, a * c = c

Hence a is an identity element for the operation.

iv. Moreover, a * a = a ⇒ añ1 = a,

c * b = b * c = a ⇒ (bñ1 = c) ∧ (cñ1 = b)

This shows that each element in S has an inverse. Therefore, (S, *) is a group.

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OR Part

Let a, b, c, x ∈ G. Then

x2b = xañ1c ⇒ xñ1 (x⋅x⋅b) = xñ1 (xañ1c) ⇒ (xñ1x) (xb) = (xñ1x) (añ1c) [by associativity] ⇒ e(xb) = e(añ1c) [by existance of inverse] ⇒ xb = añ1c [by existence of identity] ⇒ (xb)bñ1 = (añ1c)bñ1 ⇒ x(bbñ1) = añ1cbñ1 [by assiciativity] ⇒ xe = añ1cbñ1 [by existence of inverse] ⇒ x = añ1cbñ1 ∴ x = añ1cbñ1 The above value of x satisfies another equation. In fact, axb = a (añ1cbñ1)b = (aañ1) c (bñ1b) [by associativity] = e c e [by existence of inverse] = c = c [by existence of identity]

6. a) Find the integral ⌡⌠ x

x3 + 1 dx . [4]

Solution

We have

⌡⌠ x

x3 + 1 dx = ⌡⌠ x

(x +1) (x2 ñ x + 1) dx .

Let x

(x + 1) (x2 ñx + 1) = A

x + 1 + Bx + C

x2 ñ x + 1 .

⇒ x = A (x2 ñ x + 1) + (Bx + C) (x + 1)

If x = ñ1, then A = ñ 13 .

If x = 0, then C = 13 .

If x = 1, B = 13 . Thus,

⌡⌠ x

x3 + 1 dx = ñ 13 ⌡⌠ 1

x + 1 dx + 13 ⌡⌠ x + 1

x2 ñ x + 1 dx


= ñ 13 ln |x + 1| +

13 ⌡⌠ x +1

x2 ñ x + 1 dx

= ñ 13 ln |x + 1| +

13 ⎣⎢⎡

⎦⎥⎤1

2 ⌡⌠(2x ñ 1) dx

x2 ñ x + 1 dx + 32 ⌡⌠ 1

x2 ñ x + 1 dx

= ñ 13 ln |x + 1| +

16 ln |x2 ñ x + 1| +

12

⌡⎮⎮⌠

1

(x ñ 1/2)2 + 1 ñ 14

dx

= ñ 13 ln |x + 1| +

16 ln |x2 ñ x + 1| +

12

⌡⎮⎮⌠ 1

⎝⎛

⎠⎞x ñ

12

2

+ ⎝⎜⎛

⎠⎟⎞3

2

2 dx

= ñ 13 ln |x + 1| +

16 ln |x2 ñ x + 1| +

12 .

13

2

tanñ1

⎝⎜⎛

⎠⎟⎞xñ

12

32

+ C

= ñ 13 ln |x + 1| +

16 ln |x2 ñ x + 1| +

13

tanñ1 ⎝⎜⎛

⎠⎟⎞2x ñ 1

3 + C.

b) What is a linear differential equation? Solve:

(x2 + 1) dydx + 2xy = 3x2. [4]

Solution Linear differential Equation: A differential equation of the form

dydx + P(x) y = Q(x)

is called a linear differential equation. Second Part

We have

(x2 + 1) dydx + 2xy = 3x2

⇒ dydx +

2xx2 + 1 y =

3x2

x2 + 1 . . . (i)

I.F. = e ∫Pdx = e ∫ 2x/(x2+ 1) dx = e ln (x2 + 1) = x2 + 1. Multiplying (i) by I.F., we get

(x2 + 1) dydx + 2xy = 3x2

⇒ ddx {(x2 + 1) y} = 3x2

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On integration, we get

(x2 + 1) y = 3 ∫ x2 dx

⇒ (x2 + 1) y = x3 + C

∴ y (x2 + 1) = x3 + C.

7 a) State the first mean value theorem of differential calculus and interpret it geometrically. Using it to f (x) = sin x on [0, x], prove that sin x ≤ x for x ≥ 0. [4]

Solution

First mean value theorem

Let a function f be continuous on [a, b] and differentiable on (a, b). Then

∃ c ∈ (a, b) : f (b) ñ f (a) = f ' (c) (b ñ a).

Geometrical interpretation:

If f satisfies the conditions of the theorem, then the geometrical interpretation of the theorem is that there is a point c ∈ (a, b) such that the tangent at c is parallel to the chord joining (a, f(a)) and (b, f(b)).

Last part

First assume that x > 0. The first mean value theorem applied to f (x) = sin x on the interval [0, x], implies the existence of c ∈ (0, x) such that

f (x) ñ f (0)

x ñ 0 = f ' (c) = cos c.

Then

sin x ñ 0

x ñ 0 = cos c

⇔ sin x

x = cos c < 1

⇔ sin x

x < 1

⇔ sin x < x

Therefore, sin x ≤ x for all x ≥ 0.

O a c b X

Y

f(b)

f(a)


b) An urn contains four white, eight black, six red and two green marbles. If three balls are drawn at random, find the probability of getting (i) all white marbles (ii) 2 red and 1 green marbles. [4]

Solution

The urn contains 4 + 8 + 6 + 2 = 20 marbles in all. Since three marbles are drawn at random, the sample space S for the experiment consists of all possible combinations of 3 marbles. We can select 3 out of 20 marbles in C(20, 3) ways. Thus,

n(S) = C(20, 3) = 20!

(20 ñ 3)! 3! = 1140.

i. Let W be the event of drawing all white marbles. We can select 3 out of 4 white marbles in C(4, 3) = 4 ways. Thus,

n(W) = 4.

Therefore, the probability of drawing two red and two black is given by

P(W) = n(W)n(S) =

41140 =

1285 .

ii. Let A be the event of drawing 2 red and 1 green marbles. We can select 2 out of 6 red marbles in C(6, 2) ways and 1 out of 2 green marbles in 2 ways. By the multiplication rule of counting, we can draw 2 red and 1 green marbles in C(6, 2)� 2 ways. Thus,

n(A) = C(6, 2)� 2 = 6!

(6 ñ 2)! 2! � 2 = 30.

Therefore, the probability of drawing 2 red and 1 green marbles is given by

P(A) = n(A)n(S) =

301140 =

138 .

8 a) What is a conic section? Find the equation of the tangent to the parabola y2 = 8x which is parallel to the straight line 2x ñ 3y + 7 = 0. Also find its point of contact. [4]

Solution

Conic Section: A conic section is defined as the locus of a point which moves in the plane so that the ratio of its distance from a fixed point to its distance from a fixed straight line remains the same throughout the motion.

Second Part

We have

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y2 = 8x.

Comparing it with y2 = 4ax, we get

a = 2

Then a tangent to the parabola is given by the equation

y = mx + am . . . (i)

The slope of the given line 2x ñ 3y + 7 = 0 is 23 . Since the tangent is

parallel to the given line, the slope of tangent is

m = 23 .

Putting m = 23 in (i), we get

y = 23 x +2 �

32

⇒ 3y = 2x + 9

∴ 2x ñ 3y + 9 = 0

which is the required equation of the tangent.

The point of contact = ⎝⎛

⎠⎞a

m2 , 2am = ⎝

⎛⎠⎞2 � 9

4 , 2 � 2 � 3

2 = (9/2, 6).

b) Define linearly independent vectors. Show that the following vectors are linearly dependent.

2 i→

+ j→

ñk→

, 3 i→

ñ 2 j→

+ k→

, i→

+ 4 j→

ñ 3 k→

. [4]

OR

Prove that if θ is the angle between the vectors a→

and b→

, then

a→

. b→

= a b cos θ. [4]

Solution

Linearly Independent Vectors:

Vectors r1→, r2

→, . . . , rn→are said to be linearly independent if

x1 r1→ + x2 r2

→ + . . . + xn rn→ = 0 ⇒ x1 = x2 = . . . = xn = 0.

Second Part

Let x, y, z be scalars such that


x (2 i→ + j→ ñ k→) + y (3 i→ ñ 2 j→ + k→) + z ( i→ + 4 j→ ñ 3 k→) = 0

⇒ (2x + 3y + z) i→ + (x ñ 2y + 4z) j→ + (ñx + y ñ 3z) k→ = 0

Since i→, j→, k→ are linearly independent, so that

2x + 3y + z = 0

x ñ 2y + 4z = 0

ñx + y ñ 3z = 0

⇒ AX = 0

where A =

⎣⎢⎡

⎦⎥⎤2 3 1

1 ñ2 4

ñ1 1 ñ3

, X = ⎝⎜⎛

⎠⎟⎞x

yz

Now, |A| =

⎪⎪⎪

⎪⎪⎪2 3 1

1 ñ2 4

ñ1 1 ñ3

= 2 (6 ñ 4) ñ 3 (ñ3 + 4) + (1 ñ 2) = 0

Since X ≠ 0, so the system has non-zero solution i.e. scalars x, y, z not all zero. Hence the given vectors are linearly dependent.

OR Part

Let OA→= a

→, OB→= b

→and ∠AOB = θ, as shown in the figure. Then

|OA→| = a, |OB

→| = b→,

|AB→| = |OB

→ ñ OA→| = | b

→ñ a→| = | a

→ñ b→|.

Applying the law of cosines to ∆AOB in the figure, we obtain

| a→ñ b

→|2 = a2 + b2 ñ 2ab cosθ [∴ AB2 = OA2 + OB2 ñ 2(OA) (OB) cos θ)

Observe that the law of cosines still applies in the limiting case when

θ = 0 or π or a→ = 0 or b

→ = 0.

On the other hand, using the properties of the dot product we obtain

| a→ñ b

→|2 = ( a→ñ b

→) ( a→ñ b

→)

= a→. a

→ ñ a→. b

→ ñ b→. a

→ + b→. b

→

⇒ a→2 ñ 2ab cosθ + b

→2 = a→2 ñ 2 a

→. b→ + b

→2.

Z

B

θ

O X

A

θ

b

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Thus, ñ2 a→. b

→ = ñ2ab cos θ

⇒ a→. b

→ = ab cos θ.

9. For any positive integer n, prove that

(a + x)n = C(n, 0)an + C(n, 1)anñ1 x + C(n, 2)anñ2 x2 + . . . + C(n, n) xn.

Find the term containing x2, if any, in the expansion of ⎝⎛

⎠⎞2x

3 ñ 32x

6

. [6]

Solution

First part

Let n be a positive integer. We observe that

(a + x)1 = a + x = C(1, 0)a1 + C(1, 1)x1.

This shows that the theorem holds for n = 1. Assume that the theorem is true for n = k. Then

(a + x)k = C(k, 0)ak + C(k, 1)akñ1 x + . . . + C(k, k)xk.

We wish to show that the theorem is true for n = k + 1 as well. Now,

(a + x)k + 1 = (a + x) (a + x)k

= (a + x) (C(k, 0)ak + C(k, 1)akñ1 x + . . . + C(k, k)xk)

= C(k, 0)ak+1 + [C(k, 1) + C(k, 0)]akx + [(C(k, 2) + C(k, 1)]akñ1 x2

+ . . . + [C(k, k) + (C(k, k ñ 1)] axk + C(k, k) xk+1.

= C(k + 1, 0)ak+1 + C(k + 1, 1)ak x + C(k + 1, 2)akñ1 x2 + . . .

+ C(k + 1,k) axk + C(k + 1, k + 1)xk+1

using that C(k + 1, 0) = 1 = C(k, 0), C(k + 1, k + 1) = 1 = C(k, k), C(k, r) + C(k, r ñ 1) = C(k + 1, r).

Thus, the theorem is true for n = k + 1, whenever it is true for n = k. But the theorem is true for n = 1. So, it must be true for n = 2 = 1 + 1 and then for n = 3 = 2 + 1, and so on. Therefore the theorem is true for any positive integer n.

Second part

The general term is

tr+1 = C (6, r) ⎝⎛

⎠⎞2x

3

6 ñ r

⎝⎛

⎠⎞ñ3

2x

r

= (ñ1)r C (6, r) 26 ñ r

36 ñ r 3r

2r x6 ñ r ñ r = (ñ1)r C (6, r) 26 ñ 2r

36 ñ 2r x6 ñ 2r


For the term containing x2, 6 ñ 2r = 2 ⇒ 2r = 4 ⇒ r = 2.

Therefore the required term is given by

t3 = t2 + 1 = (ñ1)2 C (6, 2) 26 ñ 4

36 ñ 4 = 15 � 4

9 = 203 .

10. Find the direction cosines of two lines which are connected by the relations 2l + 2m ñ n = 0, mn + nl + lm = 0. [6]

OR

Prove that a plane through three points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by

131313

121212

111

zzyyxx

zzyyxx

zzyyxx

−−−−−−−−−

= 0.

Also, find the angle between planes 2xñy+z = 6 and x+y+2z = 3. [6]

Solution

We have

2l + 2m ñ n = 0, i.e., n = 2l + 2m . . . (i)

mn + nl + lm = 0 . . . (ii)

Eliminating n from (i) & (ii), we get

m(2l + 2m) + (2l + 2m)l + lm = 0

⇒ 2lm + 2m2 + 2l2 + 2lm + lm = 0

⇒ 2l2 + 5lm + 2m2 = 0

⇒ (l + 2m) (2l + m) = 0

∴ l

m = ñ 21 or

ñ12

If l

m = ñ 21 , then

lñ2 =

m1 . . . (iii)

From (i), we get,

2 lm + 2 =

nm

i.e. nm = 2.

ñ 21 + 2 =

ñ 21

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∴ n

ñ 2 = m1 . . . (iv)

From (iii) and (iv), we get

l

ñ2 = m1 =

nñ2

i.e. l2 =

mñ 1 =

n2 =

122 + (ñ1)2 + 22

= 13

Hence, direction cosines of one line are,

∴ l = 23 , m =

ñ13 , n =

23

Again if, lm =

ñ12 , we get the direction cosines of the second line are,

l = ñ13 , m =

23 , n =

23 .

OR Part

First part

Let ( x1, y1, z1), ( x2, y2, z2) and ( x3, y3, z3) be any non collinear points. A plane passing through point ( x1, y1, z1 ) is given by the equation

a ( x ñ x1 ) + b ( y ñ y1 ) + c ( z ñ z1 ) = 0 Ö (i)

If points ( x2, y2, z2) and ( x3, y3, z3) lie on it, then

a ( x2 ñ x1 ) + b ( y2 ñ y1 ) + c ( z2 ñ z1 ) = 0 Ö (ii)

a ( x3 ñ x1 ) + b ( y3 ñ y1 ) + c ( z3 ñ z1 ) = 0 Ö (iii)

Solving (ii) and (iii) for a, b and c, from (i) we get

(x ñ x1)1313

1212

zzyy

zzyy

−−−−

+ (y ñ y1 )1313

1212

zzxx

zzxx

−−−−

+ ( z ñ z1 )

1313

1212

yyxx

yyxx

−−−−

= 0.

∴

131313

121212

111

zzyyxx

zzyyxx

zzyyxx

−−−−−−−−−

= 0,

which represents the equation through three points ( x1, y1, z1), ( x2, y2, z2) and ( x3, y3, z3).

Second part

Given two equation of planes are

2x ñ y + z = 6 . . . (1)


x + y + 2z = 3 . . . (2)

Here, a1 = 2, b1 = ñ1, c1 = 1 and a2 = 1, b2 = 1, c2 = 2.

Let θ be the angle between the planes, then

cos θ = a1a2 + b1b2 + c1c2

a12 + b12 + c12 a22 + b22 + c22

= 2 � 1 + (ñ1) � 1 + 1 � 2

22 + (ñ1)2 + 12 12 + 12 + 22 =

12 = cos

π3

∴ θ = π3 .

11. Lives of two models of refrigerators turned in for new models in a recent survey are

No. of refrigerators No. of years

Model A Model B 0ñ2 2ñ4 4ñ6 6ñ8 8ñ10

10ñ12

5 16 13 7 5 4

2 7 12 19 9 1

What is the average life of each model of these refrigerators? Which model has more uniformity? [6]

Solution

Mid value Model A Model B No. of year (X) f fx fx2 f fx fx2

0-2 2-4 4-6 6-8

8-10 10-12

1 3 5 7 9 11

5 16 13 7 5 4

5 48 65 49 45 44

5 144 325 343 405 484

2 7

12 19 9 1

2 21 60

133 81 11

2 63

300 931 729 121

Σf = 50 Σfx = 256 Σfx2 = 1706 Σf = 50 Σfx = 308 Σfx2= 2146

For Model A:

óX (A) =

ΣfxN =

25650 = 5.12

σ (A) = Σfx2

N ñ ⎝⎛

⎠⎞Σfx

N

2

= 170650 ñ ⎝

⎛⎠⎞256

50

2

= 2.81

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For Model B

óX (B) =

ΣfxN =

30850 = 6.16

σ (B) = Σfx2

N ñ ⎝⎛

⎠⎞Σfx

N

2

= 214650 ñ ⎝

⎛⎠⎞308

50

2

= 2.23

Average life of mode (A) = 5.12 year

Average life of model (B) = 6.16 year

Therefore, Model B is the average life.

C.V. of A = σ (A)X(A) � 100% =

2.815.12 � 100% = 54.88%

C.V. of B = σ (B)X(B) � 100% =

2.336.16 � 100% = 36.20%

Since C.V. of A > C.V. of B, So model B has more uniformity.

Group B 12. a) Three forces P, Q and R acting on a particle are in equilibrium, the

angle between the P and Q is 60� and that between Q and R is 150� . Find the ratios of the forces. [2]

Solution

We have the angle between P and Q is 60� and that between Q and R is 150� .

∴ Angle between R and P = 360� ñ 60� ñ 150� = 150� .

Then

P

sin 150� = Q

sin 150� = R

sin 60�

⇒ P12

= Q12

= R3

2

∴ P : Q : R = 1 : 1 : 3 .

b) A uniform beam, 4 m long, is supported in a horizontal position by two props which are 3 m apart, so that the beam projects one meter beyond one of the props. Show that the force on one of the props is double of that on the other. [2]

Solution

150°

Q

P

R

60°

150°


Let O be the centre of a uniform beam AB = 4 m supported by two props at A and C. Then

AO = 4/2 m = 2 m, CO = BO ñ BC = 4/2 m ñ 1 m = 1 m.

If w is the weight of the beam, and P and Q are the forces on the props at A and C respectively, by the theorem on parallel forces, we have

P

CO = Q

AO = w

AC .

⇒ P � AO = Q � CO

⇒ 2P = Q.

Therefore, the force on the prop at C is double of that on the other.

c) A pump having a power of 392 W pumps water at the rate of 100 litres per minute. Find the height to which the water is raised. (g = 9.8 m/s2, 1 litre of water = 1 kg). [2]

Solution

We have

Mass (m) = 100 Time (t) = 60 sec,

Power (P) = 392 W, g = 9.8 m/s2.

We know that

P = mgh

t

⇒ h = Ptmg =

392 � 60100 � 9.8 m = 24 m.

∴ Required height = 24 m.

13. a) A body of weight w is suspended by strings of length 3 m and 4 m attached to two points in the same horizontal line whose distance apart is 5 m. Find the tensions along the strings. [4]

Solution

Let A and B be two fixed points in the horizontal line of length AB = 5 m and a weight of w kg be suspended at C with strings CA and CB of lengths CA = 4 m, CB = 3 m. Produce the line of action of the weight which meets AB at D. We have

52 = 42 + 32

i.e. AB2 = AC2 + BC2.

∴ ∠ACB = 90� .

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Let ∠CAD = θ. Then

cos θ = ACAB =

45 , sin θ =

BCAB =

35 .

Let T, and T2 be the tensions along CA and CB. Now three forces T1, T2 and w kg acting at C in equilibrium. Then by Lami's theorem,

T1

sin (180 ñ θ) = T2

sin (90 + θ) = w

sin 90 ,

i.e. T1

sin θ = T2

cos θ = w.

∴ T1 = 45 w, T2 =

35 w.

b) A body of mass 49 kg is falling freely under gravity at the rate of 20 m/s. What is the uniform force that will stop it (i) in 2 sec (ii) in 50 cm? (g = 9.8 m/s2.) [4]

OR

A bullet fired into a target losses half its velocity after penetrating 3 cm. How much further will it penetrate? [4]

Solution

We have

Mass of body (m) = 49 kg, Initial velocity (u) = 20 m/s

Time taken (t) = 2 sec, Final velocity (v) = 0 m/s

i. If t = 2 sec, then using v = u ñ at, we get

a = u ñ v

t = 202 = 10 m/s2

⇒ Retardation (a) = 10 m/s2

Therefore, the resistance force is

F = mg + ma

= 49 � 9.8 + 49 � 10 = 480.2 + 490 = 970.2 N

= 970.29.8 kg-wt = 99 kg-wt.

ii. If s = 50 cm, then using v2 = u2 ñ 2as , we get

a = u2 ñ v2

2s = 202 ñ 02 � 0.5 = 400 m/s2

Therefore, the resistance force is

F = ma + mg


= 49 � 400 + 49 � 9.8

= 19600 + 480.2

= 20080.2N

= 20080.2

9.8 = 2049 kg-wt.

OR Part

Let u be the velocity of the bullet just before the penetration and a be its acceleration. Then Initial velocity (u) = u cm/s

Final velocity (v) = u2 cm/s

Distance (s) = 3 cm Using v2 = u2 + 2as, we get

⇒ u2

4 = u2 + 2� a� 3

⇒ u2 = ñ 8a. . . . (i)

Second Part

We have

Initial velocity (óu ) = u2 cm/s

Final velocity (óv ) = 0 cm/s

Acceleration (óa ) = a cm/s2

Distance (ós ) = ?

Using óv 2 = óu 2 + 2óa ós , we get

02 = ⎝⎛⎠⎞u

2

2

+ 2 � a � ós

⇒ ñ u2

4 = 2aós [using (i)]

⇒ 2a = 2aós

⇒ ós = 1.

∴ ós = 1 cm.

14. The resultant of two like parallel forces P and Q acting on a rigid body is a force of magnitude P + Q in the same direction as P and Q are. If A and B are any points on the lines of action of P and Q respectively,

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prove that the resultant divides line segment AB internally in the inverse ratio of the forces. [6]

OR

Define the moment of a force. Forces 1, 2, 4, 5 kg-wts act along the sides of a square taken in order. Prove that their resultant is parallel to a diagonal and find where it cuts the side along which the first force acts. [6]

Solution

M N A B

E G F H

P Q

C

O

P Q +

S →

→ →

S →

S →

S →

→ →

Let A and B be any points on the lines of action of two like parallel forces P and Q respectively. We may assume that the forces act at A and B by the principle of transmissibility. At A and B we may suppose two additional forces, each equal to S, to act in opposite directions along line segment AB, as in the figure. By the principle of transmissibility, the resultant of the additional forces will be zero and so they will not affect the equilibrium of the body. Complete parallelograms AEGM and BFHN with AE = P, AM = S, BF = Q, BN = S. Clearly, AG represents the resultant of P and S and BH that of Q and S by the parallelogram of forces. Produce the diagonals AG and BH to meet at O. Resolving each of the resultants into their original components, acting at O, we obtain two forces, each equal to S, in the opposite directions, and two forces P and Q acting along OC that is parallel to AE and BF. The forces in opposite directions, each of which is equal to S, vanish. Therefore, there remains a single force P + Q acting along OC. Now, since triangles AOC and GAE are similar by construction, we must have

OCAC =

AEGE =

PS

⇒ AC � P = OC � S.


Likewise, by the similarity of triangles BOC and HBF, we must have

OCBC =

BFHF =

QS ⇒ BC � Q = OC � S

∴ AC � P = BC � Q i.e. AC : BC = Q : P. Thus, C divides AB internally in the inverse ratio of forces P and Q.

OR Part

Moment of a force: The moment of a force about a point is defined as the product of the force and the force arm about the point.

Second Part

Let forces 1, 2, 4, 5 kg-wt. act along the sides AB, BC, CD and DA respectively. Since forces 1 kg-wt. and 4 kg-wt. are horizontal parallel forces,

RX = 1 kg-wt. ñ 4 kg-wt. = ñ 3 kg-wt.

Ry = 2 kg-wt. ñ 5 kg-wt. = ñ 3 kg-wt.

Therefore, the resultant is given by

R = 9 + 9 = 3 2 kg.wt.

If the resultant makes an angle of θ with AB, then

tan θ = Ry

Rx =

ñ3ñ3 = 1 = tan (180 + 45).

Since RX = ñ 3 kg-wt. < 0 and RY = ñ 3 kg-wt., θ = 225� .

Hence the resultant is parallel to the diagonal CA.

Suppose resultant cuts AB produced at E and AE = x. Then moment of the resultant about E is zero. Hence

2 � (AB + AE) + 4 � AD ñ 5 � EA = 0

2 (x + a) + 4a ñ 5x = 0

where ëaí is the length of a side of the square ABCD. Then

ñ 3x + 6a = 0

x = 2a

Therefore, point E divides the side AB in the ratio

ABBE =

xx + a =

2a2a + a =

23 i.e. 2:3.

15. A man travels from A to B in 45 minutes. At C, somewhere between A and B, it attains its maximum velocity of 45 m per hr. If he travels with uniform acceleration from A to C and uniform retardation from C to B,

A B

C D 4

2

1

5 R

E

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find the distance between A and B, it being supposed that the man starts from rest at A and comes to rest at B. [6]

Solution

Let a, s and t denote the acceleration, the distance and time taken from A to C respectively. And let a', s' and t' be the retardation, the distance and the time taken from C to B respectively. Then

t + t' = 45 min = 45 � 60 sec. = 2700 sec.

For travelling from A to C, we have

Initial velocity (u) = 0,

Final velocity (v) = 45 m/hr = 45

3600 m/s = 180 m/s.

Using v = u + at, we get

180 = at.

Using s = ut + 12 at2, we get

s = 12 �

180 t

⇒ s = 1

160 t. . . . (i)

Now, for travelling from C to B, we have

u = 180 m/s v = 0 m/s

Using v = u ñ a' t, we get

0 = 180 ñ a' t'

⇒ a' t' = 180

Using s = ut ñ 12 at2, we get

s' = 180 t' ñ

12 a' t'2

⇒ s' = 180 t' ñ

12 �

180 t'.

C BA 0 m/s 45 0 m/s


⇒ s' = 1

160 t' . . . (ii)

From (i) and (ii), we obtain

s + s' = 1

160 ( t + t') = 1

160 � 2700 = 135

8 =16 78 .

Therefore, the total distance from A to B is

s + s' = 16 78 m.

Group C 16. a) Determine graphically the solution set of the following system of

inequalities:

2x + y ≥ 2, 3x + 2y ≤ 4, x ≥ 0, y ≥ 0. [2]

Solution We have

x + 2y ≥ 0, 2x + y ≤ 4

The equation associated with given inequalities are

x + 2y = 0 . . . (i)

2x + y = 4 . . . (ii)

The boundary line (i) passes through

x 0 2 y 0 ñ 1

Taking (1, 0) as a testing point, we get

1 ≥ 0

which is true. So the graph (i) contains the point (1, 0)

The boundary line (ii) passes through

x 0 2 y 4 0

Taking origin as a testing point, we get

0 ≤ 4

which is true. So the graph (ii) contains the origin.

The inequalities x ≥ 0 represents the half plane on the right of the y-axis.

The inequalities y ≥ 0 represents the half plane above the x-axis.

The shaded portion in the figure represents the required solution set.

O X

Y

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b) Write a short note on accuracy of a numerical method. [2]

Solution

If α is the exact solution and xn its estimate or approximation, then

|α � xn|

is called the absolute error.

Every method of numerical computation produces results with some errors. They may be either due to

ñ using an approximate representation of a mathematical object under study (known as truncation errors) or

ñ rounding of numbers (known as round-off errors).

These errors affect the accuracy of the results. The results we obtain must have the degree of accuracy as required. Therefore, choice of a method depends very much on the particular problem.

c) Apply the Simpons's rule to approximate the value of ∫4

1ex ln x dx

with n = 3. [2]

Solution

Given that f (x) = ex ln x, a = 1, b = 3 and n = 4, then h = 4 ñ 1

3 = 1.

Divide the interval [1, 4] into 3 equal subintervals i.e.

[1, 2], [2, 3], [3, 4].

Applying the Simpson's 3/8 rule, we get

⌡⌠1

3

ex ln x dx ≈ 3h8 [f (1) + 3{f (2) + f (3)} + f (4)]

≈ 3 � 1

8 [0 + 3 (5.1217 + 22.0662) + 75.6891]

≈ 58.9698.

17. a) Using the simplex method, maximize p = 6x ñ 9y subject to

2x ñ 3y ≤ 6, x + y ≤ 20, x ≥ 0, y ≥ 0. [4]

Solution

We have to maximize p = 6x ñ 9y subject to 2x ñ 3y ≤ 6, x + y ≤ 20. This is a linear programming problem in standard form. In order to solve it by simplex method, we introduce slack variables s and t by setting


s = 6 ñ (2x ñ 3y)

t = 20 ñ (x + y)

Clearly s, t ≥ 0. Now the given LP problem can be restated in the canonical form as follows:

2x ñ 3y + 1s + 0t + 0p = 6

x + y + 0s + 1t + 0p = 20

ñ 6x ñ 9y + 0s + 0t + 1p = 0

Then initial tableau (first tableau) is

Basic variables x y s t p RHSs t

2 1

-3 1

1 0

0 1

0 0

6 20

p ñ6 9 0 0 1 0

In the last row, there is a negative number ñ6. This indicates that the initial solution given by the initial tableau is not optimal solution. Since ñ6 is the only negative number the s column is the pivot column and s is the entering variable. Dividing the last element in each row by the corresponding element in the pivot column, we obtain

6/2 = 3, 20/1 = 20

Since 3 < 20, then s row is the pivot row and t is the departing variable. We see that 2 lies in the intersection of pivot column and pivot row. Then dividing the pivot row by 2 we get,

Basic variables x y s t p RHSx t

1 1

-3/21

1/20

01

00

3 20

p -6 9 0 0 1 0

Changing R3 into R3 ñ R2 and R4 into R4 + 6R2, we obtain

Basic variables x y s t p RHSx t

10

-3/25/2

1/2 -1/2

01

00

3 17

p 0 0 3 0 1 18

In last row there is no negative numbers, so that the second tableau gives the optimal solution. Hence maximum value p = 18 at (3, 0).

b) Use Bisection method to find solution accurate to within 10 ñ2 for x3 ó 7x2 + 14x ó 6 = 0 on the interval [1, 3.2]. [4]

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OR

Write three methods for measuring error. Approximate 11 by Newton-Raphson's method with accuracy 0.00001. [4]

Solution

For the interval [1, 3.2], we have

f (3.2) = ñ 0.112 < 0, f (1) = 2 > 0.

Clearly,

f (3.2) � f (1) < 0.

Set

[a0, b0] = [1, 3.2], x0 = a0 + b0

2 = 2.1

Then f (x0) = f(2.1) = 1.791.

On computation we obtain the following table of approximations.

n Interval xn f (xn) 0 [1, 3.2] 2.1 1.791 1 [2.1, 3.2] 2.65 0.552125 2 [2.65, 3.2] 2.925 0.0858281 3 [2.925, 3.2] 3.0625 ñ0.0544434 4 [2.925, 3.2] 2.99375 0.00632788

Now,

f (x4) = f (2.99375) = 0.00632788 < 10ñ2

Hence x4 = 2.99375 can be taken as an approximate solution with error less than 10ñ2.

OR Part Measuring Error

If we are trying to find a numerical solution of an equation f (x) = 0, then there are few different ways we can measure the error of our approximation. The most direct way to measure the error would be as:

Error at step n = En = xn � α

where xn is the n-th approximation and α is the true value. However, we usually do not know the value of α, or we wouldnít be trying to find it. This makes it impossible to know the error directly, and so we must be more clever. We know that after n steps the root lies in an


interval of length (b - a)/2n. The midpoint of this last interval, xn,

satisfies

| xn - true root | ≤ (b - a)/ 2

n.

For example, we can select an error tolerance ε > 0 and generate x1, x2, x3, Ö until one of the following conditions is met:

| xn ñ xn-1| < ε Ö (1)

| xn ñ xn-1|

| xn| < ε Ö (2)

|f (xn)| < ε Ö (3)

Without additional knowledge about f or the true root α, Inequality (2) is the best stopping criterion to apply because it comes closest to testing relative error.

To approximate 11

Start with x0 = 3 to get

x1 = 12 ⎝⎛

⎠⎞x0 +

11x0

= 12 ⎝⎛

⎠⎞3 +

113 = 3.333

x2 = 12 ⎝⎛

⎠⎞3.333 +

113.333 = 3.31667

x3 = 12 ⎝⎛

⎠⎞3.31667 +

113.31667 = 3.31662

x4 = 12 ⎝⎛

⎠⎞3.31662 +

113.31662 = 3.31662

Here, error ⎪⎪

⎪⎪x4 ñ x3

x4 < 0.00001.

Hence, 3.31662 is approximate value of 11 as required.

18. Find the approximate solution of the following system of equations by matrix inversion method:

2x ñ y + z = ñ2, x + y ñ 2z = ñ9, z + 2y + z = 9. [6]

Solution

We have

2x ñ y + z = ñ2

x + y ñ 2z = ñ9

z + 2y + z = 9.

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Let A = ⎣⎢⎢⎡

⎦⎥⎥⎤2 ñ1 1

1 1 ñ2

1 2 1 , X =

⎣⎢⎢⎡

⎦⎥⎥⎤x

y

z , B =

⎣⎢⎢⎡

⎦⎥⎥⎤ñ2

ñ9

9

Then, the given system becomes AX = B and therefore,

X= Añ1B . . . (1)

Now,

[A | I] = ⎣⎢⎢⎡

⎦⎥⎥⎤2 ñ1 1

1 1 ñ2

1 2 1 ⎪⎪⎪⎪ 1 0 0

0 1 0

0 0 1

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 ñ0.5 0.5

1 1 ñ2

1 2 1 ⎪⎪⎪⎪ 0.5 0 0

0 1 0

0 0 1 changing R1 to 1/2 R1

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 ñ0.5 0.5

0 1.5 ñ2.5

0 2.5 0.5 ⎪⎪⎪⎪ 0.5 0 0

ñ0.5 1 0

ñ0.5 0 1

changing R2 to R2 ñ R1

changing R3 to R3 ñ R1

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 ñ0.5 0.5

0 1 ñ1.6667

0 2.5 0.5 ⎪⎪⎪⎪ 0.5 0 0

0.333 0.667 0

ñ0.5 0 1 changing R2 to (1/1.5)R2

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 0 ñ0.3335

0 1 ñ1.6667

0 0 4.6675 ⎪⎪⎪⎪ 0.3335 0.3335 0

-0.333 0.667 0

0.3325 ñ1.6675 1

changing R1 to R1 + 0.5R2

changing R3 to R3 ñ 2.5R2

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 0 ñ0.3335

0 1 ñ1.6667

0 0 1 ⎪⎪⎪⎪ 0.3335 0.3335 0

-0.333 0.667 0

0.0712 ñ0.3573 0.2142 changing R3 to

14.6675R3

= ⎣⎢⎢⎡

⎦⎥⎥⎤1 0 0

0 1 0

0 0 1 ⎪⎪⎪⎪ 0.3572 0.2143 0.0714

-0.2193 0.0715 0.357

0.0712 ñ0.3573 0.2142



∴ X = Añ1B = ⎣⎢⎢⎡

⎦⎥⎥⎤0.3572 0.2143 0.0714

-0.2193 0.0715 0.357

0.0712 ñ0.3573 0.2142 ⎣⎢⎢⎡

⎦⎥⎥⎤ñ2

ñ9

9 =

⎣⎢⎢⎡

⎦⎥⎥⎤ñ2.0005

2.9981

5.0011 .

Hence, x = 2.0005, y = 2.9981, z = 5.0011 is the desired solution, which is closer to the exact solution (x, y, z) = (ñ2, 3, 5).

19. Derive the trapezoidal rule. The capacity of a battery is a measure of ∫ i dt , where i is the current. Estimate, using the Trapezium rule, the


capacity of a battery whose current was measured over an eight hour period with the results shown below

Time/hours 0 1 2 3 4 5 6 7 8 Current/Amps 25.2 29.0 31.8 36.5 33.7 31.2 29.6 27.3 28.6

[6]

OR

Compute an approximate value of ∫01 (1 + x2)� 1 dx by using the

composite trapezoid rule with three points. Then compare with the actual value of the integral. Next, determine the error formula and numerically verify an upper bound on it. [6]

Solution

Trapezoidal rule: The trapezoidal rule can also be derived from geometry. Look at Figure. The area under the curve P1(x) is the area of a trapezoid. The integral

∫ab f (x) dx ≈ Area of the trapezoid

= 12 (Sum of lengths of parallel sides)(Perpendicular

distance between parallel sides)

= 12 [f (b) + f (a) (b ñ a)]

= (b ñ a) ⎣⎡

⎦⎤f (a) + f (b)

2

Second Part

We have

Capacity of battery = ⌡⌠0

8

i (t) dt. where a = 0, b = 8, n = 8

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Using the Trapezium 1/3 rule, we get

⌡⌠0

8

i (t) dt ≈ b ñ a3n [f (0) + 4f (1) + 2f (2) + 4 f (3) + 2 f (4)

+ 4 f (5) + 2f (6) + 4 f (7) + f (8)]

≈ 8 ñ 03 � 8 [25.2 + 4� 29 + 2� 31.8 + 4� 36.5 + 2� 33.7

+ 4� 31.2 + 2� 29.6 + 4� 27.3 + 28.6]

≈ 13 � 740 = 246

23

OR Part

We have

⌡⌠0

1

(1 + x2)ñ1 dx, where a = 0 and b = 1.

For three points n = 2 then h = b ñ a

n = 1 ñ 0

2 = 12

Dividing the internal [0, 1] into two equal subintervals

i.e. [0, 1/2], [1/2, 1].

Using the Trapezoidal rule over each subinterval, we get

⌡⌠0

1

(1 + x2)ñ1 dx ≈ 1

2 � 2 [f (0) + 2f (1/2) + f (1)]

≈ 14 ⎣⎡

⎦⎤1 + 2 �

11 + 1/4 +

12 =

3140 ≈ 0.775

For actual value,

⌡⌠0

1

(1 + x2)ñ1 dx = ⌡⌠0

1

1

1 + x2 dx = [tanñ1x]10

= tanñ11 ñ tanñ10 = π4 ñ 0

= 0.785 (Actual value)

The error in trapezoidal rule = 0.785 ñ 0.775 =0.01

Again, Error formula (Er) = ñ (b ñ a)3

12n2 f "(θ)

We have

f (x) = 1

1 + n2 .


Then

f ' (x) = ñ2x

(1 + n2)2

and f " (x) = 2 (3x2 ñ 1)(1 + n2)3 , i.e. f " (θ) =

2(3θ2 ñ 1)(1 + θ2)3

Max |f "(θ)| ≤ 2.

Hence error formula is

Er = ñ(1 ñ 0)12 � 22 f " (θ)

∴ Er = ñ ñ148 f " (θ)

Absolute error = |Er| ≤ 148 max | f " (θ)|

≤ 148 � 2 =

124 = 0.04167

Hence, upper bound ≈ 0.04167.

***

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