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UNIQUE EDUCATIONS Mangaldai & Baihata Chariali Centre

SOLUTIONS OF AHSEC MATHEMATICS PAPER 2015

TOTAL MARKS : 100 TIME : 3 HOURS

1. Answer the following questions: 1 x 10 = 10

a. If A={0,1,3}, what is the number of relations on A?

Solution: Number of relation on A is 23𝑥3 = 29 = 512.

b. Find the principal value of 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧𝟑𝝅

𝟐).

Solution: Principal value of sin−1 sin3𝜋

5 is

2𝜋

5.

c. If [𝟓 𝟔 𝟕] A = [𝟏𝟑 𝟐𝟑], what is the order of the matrix A?

Solution: Order of A is 3 x 2.

d. If A is a non-singular matrix such than 𝑨𝟐 + 𝑨 − 𝟏 = 𝟎, what is 𝑨−𝟏 ?

Solution: A2+A-I=0,

OR A2+A = I,

OR A-1

AA+ A-1

A= A-1

I, (Multiplying both sides by A-1

)

OR A-1

=A+I

e. What is the co-factor of 7 in the determinant 𝟒 𝟓 𝟔𝟓 𝟔 𝟕𝟏𝟑 𝟏𝟓 𝟏𝟕

?

Solution: Co-factor of 7 in is −1 2+3 4 5

13 15 = - (60 - 65) = 5

f. Is the derivative of an even function even?

Solution: No, derivative of an even function is an odd function.

g. Is the function 𝒇 𝒙 = 𝒙𝟐, 𝒙 ∈ 𝑹 increasing?

Solution: No, because 𝑓 ′ 𝑥 =2x < 0 when x<0 i.e f(x) is not increasing in all x∈ R.

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h. What are the direction cosines of the vector𝒂 = 𝒊 + 𝟐𝒋 + 𝟑𝒌 ?

Solution: Direction ratios of 𝑎 are 1, 2, 3. So direction cosines of 𝑎 are 1

12+22+32,

2

12+22+32,

3

12+22+32 i.e

1

14,

2

14,

3

14.

i. If the distance of a plane from the origin be ‘d’ and the direction cosines of the normal to

the plane through origin be (l ,m, n), what are the co-ordinates of the foot of the normal?

Solution: Co-ordinates of the foot of the normal are ld, md, nd.

j. What are the equations of the planes parallel to xz-plane and at a distance ‘a’ from it?

Solution: Equation of the planes parallel to xz plane and at a distance ‘a’ from it is y = a.

2. A function f:R→ 𝑹 is defined by 𝒇 𝒙 = 𝟐𝒙𝟐. Is the function f one- one, and onto? Justify your

answer. 2+2=4

Solution: f: R→R & f(x) =2x2

Now f(-1) = f(1) = 2

But -1≠1, So f is not one one

Again – 4 ∈ R

But there is no x ∈ R such that 2x2

= - 4

So f is not onto.

OR

Let L be the set of all lines in the 𝒙𝒚 −plane and R be the relation in L defined by

𝑹 = 𝒍𝒊, 𝒍𝒋 𝒍𝒊 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒍𝒋, ∀𝒊, 𝒋}. Show that R is an equivalence relation. Find the set of all lines

related to the line = 𝟕𝒙 + 𝟓 . 3+1=4

Solution: Given R= {(li,lj): li parallel to lj, ∀ i, j }

R is reflexive as any line l1 is parallel to itself i.e. (l1 , l1) ∈ R

Now (l1 , l2) (l1 , l2)

l1 is parallel to l2

l2 is parallel to l1

(l2, l1) ∈ R

So R is symmetric

Again,

Let (l1 , l2)& (l2 , l3) ∈ R

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l1 is parallel to l2 & l2 is parallel to l3

l1 is parallel to l3

(l1, l3) ∈ R

So R is transitive

Hence R is an equivalence relation.

Also,

The set of lines all lines related to y = 7x+5 is the set of all lines parallel to it i.e. y = 7x+k; k∈

3. If 𝐭𝐚𝐧−𝟏 𝟏+𝒙𝟐− 𝟏−𝒙𝟐

𝟏+𝒙𝟐+ 𝟏−𝒙𝟐 = 𝜶, prove that 𝒙𝟐 = 𝐬𝐢𝐧 𝟐𝜶. 4

Solution: Given tan−1 1+𝑥2− 1−𝑥2

1+𝑥2+ 1−𝑥2= 𝛼

Or tan 𝛼 = 1+𝑥2− 1−𝑥2

1+𝑥2+ 1−𝑥2 =

sin 𝛼

cos 𝛼

Or sin 𝛼+cos 𝛼

sin 𝛼 −cos 𝛼=

2 1+𝑥2

−2 1−𝑥2

Or sin 𝛼+cos 𝛼

sin 𝛼 −cos 𝛼

2 =

1+𝑥2

1−𝑥2

Or 1+sin 2𝛼

1−sin 2𝛼 =

1+𝑥2

1−𝑥2

Or 1+sin 2𝛼+1−sin 2𝛼

1+sin 2𝛼−1+sin 2𝛼 =

1+𝑥2+1−𝑥2

1+𝑥2−1+𝑥2

Or 𝑥2 = sin 2𝛼

4. If 𝑨 = 𝟎 −𝟏 𝟐𝟐 −𝟐 𝟎

𝒂𝒏𝒅 𝑩 = 𝟎 𝟏𝟏 𝟎𝟏 𝟏

, find a matrix C such that CAB=I=ABC, where I is the 2x2

matrix. 4

Solution: As A is of 2x3 and B is of 3x2 so C must be of 2x2

Let C= 𝑎 𝑏𝑐 𝑑

Now CAB= 0 + 2𝑏 −𝑎 − 2𝑏 2𝑎 + 00 + 2𝑑 −𝑐 − 2𝑑 2𝑐

0 11 01 1

= 0 − 𝑎 − 2𝑏 + 2𝑎 2𝑏 + 0 + 2𝑎0 − 𝑐 − 2𝑑 + 2𝑐 2𝑑 + 0 + 2𝑐

= 𝑎 − 2𝑏 2𝑎 + 2𝑏𝑐 − 2𝑑 2𝑐 + 2𝑑

Also ABC= 1 2

−2 2

𝑎 𝑏𝑐 𝑑

= 𝑎 + 2𝑐 𝑏 + 2𝑑

−2𝑎 + 2𝑐 −2𝑏 + 2𝑑

As CAB=ABC= I = 1 00 1

So 𝑎 − 2𝑏 = 𝑎 + 2𝑐 = 1, 2𝑎 + 2𝑏 = 𝑏 + 2𝑑 = 0, 𝑐 − 2𝑑 = −2𝑎 + 2𝑐 = 0,

2𝑐 + 2𝑑 = −2𝑏 + 2𝑑 = 1

Solving the equations a = 1, b = -1, c = 2 , d = -1

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Therefore C = 1 −12 −1

OR

Using elementary row operation, find the inverse of the matrix 𝟐 𝟎 −𝟏𝟓 𝟏 𝟎𝟎 𝟏 𝟑

. 4

Solution:

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5. If 𝒚 + 𝐬𝐢𝐧𝒚 = 𝐜𝐨𝐬 𝒙 then find the values of y for which 𝒅𝒚

𝒅𝒙 is valid. 4

Solution: y + siny = cosx

Or 𝑑𝑦

𝑑𝑥 =

− sin 𝑥

1+cos 𝑦

For 𝑑𝑦

𝑑𝑥 to be valid 1 + cos 𝑦 ≠ 0 i.e. cos 𝑦 ≠ −1

We know cos 𝑦 = −1 = cos 𝜋

Or y = (2n+1) 𝜋, 𝑛 ∈ 𝑍

So if y ∈ R − (2n+1) 𝜋, 𝑛 ∈ 𝑍 𝑑𝑦

𝑑𝑥 is valid

6. If a function is differentiable at a point, prove that it is continuous at that point. 4

Solution: Since any function say f is differentiable at a point c, we have

lim𝑥→𝑐

𝑓 𝑥 − 𝑓 𝑐

𝑥 − 𝑐= 𝑓 ′ 𝑐

But for x≠c, we have

𝑓 𝑥 − 𝑓 𝑐 =𝑓 𝑥 −𝑓 𝑐

𝑥−𝑐. 𝑥 − 𝑐

So lim𝑥→𝑐

[𝑓 𝑥 − 𝑓 𝑐 ] = lim𝑥→𝑐

𝑓 𝑥 −𝑓 𝑐

𝑥−𝑐. 𝑥 − 𝑐

Or lim𝑥→𝑐

𝑓 𝑥 − lim𝑥→𝑐

𝑓 𝑐 = lim𝑥→𝑐

𝑓 𝑥 −𝑓 𝑐

𝑥−𝑐 . lim

𝑥→𝑐(𝑥 − 𝑐)

= f′(c).0 = 0

Or lim𝑥→𝑐

𝑓 𝑥 = 𝑓(𝑐)

Hence f is continuous at x = c

OR

Using Rolle’s theorem, find at what points on the curve 𝒚 = 𝐜𝐨𝐬𝒙 − 𝟏 in 𝟎, 𝟐𝝅 the tangent is

parallel to x-axis. 4

Solution: Given curve is y = f(x) = cosx−1

Here a = 0 and b = 2π

As f(0)=0=f(2 π

Now f′ 𝑥 = −𝑠𝑖𝑛𝑥

According to Rolle’s theorem if the tangent is parallel to x axis at any point x= c in 0,2π

Then f’ c =0

Or –sinc=0

Or sinc = 0

Or c = 0, π, 2 π

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7. Evaluate any one of the integrals : 4

i. 𝒙−𝒙𝟑

𝟏𝟑

𝒙𝟒

𝟏𝟏

𝟑 𝒅𝒙

Solution: 𝑥−𝑥3

13

𝑥4

11

3 𝑑𝑥 =

1

𝑥3 1

𝑥2 − 1

1

3𝑑𝑥

11

3

Let 1

𝑥2 − 1 = 𝑡

−21

𝑥3 𝑑𝑥 = 𝑑𝑡

Or 1

𝑥3𝑑𝑥=

𝑑𝑡

−2

When x=1

3, t=8, When x=1, t=0

Hence the integral becomes

𝑡 13

𝑑𝑡

−2

0

8

=1

−2

𝑡43

8

0

43

= −3

8𝑥 0 − 16

= 6

ii. 𝒙𝟐 − 𝒂𝟐𝒅𝒙

Solution: Taking const function 1 as the second function and integrating by parts, we

have

I = x 𝑥2 − 𝑎2 − 1

2

2𝑥

𝑥2−𝑎2𝑥dx

= x 𝑥2 − 𝑎2 − 𝑥2

𝑥2−𝑎2dx

= x 𝑥2 − 𝑎2 − 𝑥2−𝑎2+𝑎2

𝑥2−𝑎2dx

= x 𝑥2 − 𝑎2 − 𝑥2 − 𝑎2dx−𝑎2 dx

𝑥2−𝑎2

= x 𝑥2 − 𝑎2 − 𝐼 − 𝑎2 dx

𝑥2−𝑎2

Or 2I = x 𝑥2 − 𝑎2−𝑎2 dx

𝑥2−𝑎2

Or I = 𝑥

2 𝑥2 − 𝑎2−

𝑎2

2𝑙𝑜𝑔 𝑥 + 𝑥2 − 𝑎2 + C

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8. Prove that 𝒇 𝒙 𝒂

−𝒂𝒅𝒙 = 𝟎, when f is an odd function. Hence evaluate 𝐥𝐨𝐠

𝟐−𝒙

𝟐+𝒙

𝟏

−𝟏𝒅𝒙 4

Solution: 𝑓 𝑥 𝑑𝑥 =𝑎

−𝑎 𝑓 𝑥 𝑑𝑥 + 𝑓 𝑥 𝑑𝑥

𝑎

0

0

−𝑎

Let t =−x in the first integral on the R.H.S.

So dt =−dx & when x =−a, t = a and when x = 0, t = 0

Therefore 𝑓 𝑥 𝑑𝑥 = −𝑎

−𝑎 𝑓 −𝑡 𝑑𝑡 + 𝑓 𝑥 𝑑𝑥

𝑎

0=

0

𝑎 𝑓 −𝑥 𝑑𝑥 + 𝑓 𝑥 𝑑𝑥

𝑎

0

0

𝑎…………..(1)

Now as f(x) is an odd function f(-x) = −f(x)

Hence (1) becomes 𝑓 𝑥 𝑑𝑥 = −𝑎

−𝑎 𝑓 𝑥 𝑑𝑥

𝑎

0+ 𝑓 𝑥 𝑑𝑥

𝑎

0= 0

2nd

Part:

Here f(x) = log2−𝑥

2+𝑥

f(-x) = log2+𝑥

2−𝑥 = log

2−𝑥

2+𝑥

−1

= −log2−𝑥

2+𝑥 = − f(x)

So f(x) is an odd function.

Hence log2−𝑥

2+𝑥𝑑𝑥 = 0

1

−1

9. Solve (any one): 4

i. 𝟏 + 𝒙𝟐 𝒅𝒚

𝒅𝒙+ 𝒚 = 𝐭𝐚𝐧−𝟏 𝒙

Solution: (1+𝑥2)𝑑𝑦

𝑑𝑥+ 𝑦 = tan−1 𝑥

Or 𝑑𝑦

𝑑𝑥+ 𝑦.

1

1+𝑥2 =tan −1 𝑥

1+𝑥2 which is of the form 𝑑𝑦

𝑑𝑥+ 𝑃𝑦 = 𝑄

Here P = 1

1+𝑥2, Q = tan −1 𝑥

1+𝑥2

I.F.= 𝑒 1

1+𝑥2𝑑𝑥= 𝑒tan −1 𝑥

So the solution is y.I.F.= (𝑄𝑥𝐼. 𝐹)𝑑𝑥 + 𝐶

Or y. 𝑒tan −1 𝑥 = tan −1 𝑥

1+𝑥2 𝑒tan −1 𝑥𝑑𝑥+ C……………….(1)

Let I = tan −1 𝑥

1+𝑥2 𝑒tan −1 𝑥𝑑𝑥

Substituting tan−1 𝑥 = t so that 1

1+𝑥2𝑑𝑥 = dt

I = 𝑡𝑒𝑡𝑑𝑡 = 𝑡. 𝑒𝑡 − 1. 𝑒𝑡𝑑𝑡 = 𝑡. 𝑒𝑡 − 𝑒𝑡 = 𝑒𝑡(𝑡 − 1)

= 𝑒tan −1 𝑥(tan−1 𝑥 − 1)

(1) => y. 𝑒tan −1 𝑥 = 𝑒tan −1 𝑥 tan−1 𝑥 − 1 + 𝐶

Or y = tan−1 𝑥 − 1 + 𝐶𝑒− tan −1 𝑥

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(ii) x𝒅𝒚

𝒅𝒙 = y – x tan

𝒚

𝒙

Solution: x𝑑𝑦

𝑑𝑥 = y – x tan

𝑦

𝑥

Or 𝑑𝑦

𝑑𝑥=

y – x tan 𝑦

𝑥

𝑥 ……………………….(1)

Let F(x,y) = y – x tan

𝑦

𝑥

𝑥

F(λx, λy) = λy – λx tan

λ𝑦

λ𝑥

λ𝑥 =

y – x tan 𝑦

𝑥

𝑥= 𝜆𝑜 F(x,y)

Therefore the given differential equation is a homogeneous equation,

To solve it, we make it substitution as, y = vx

So 𝑑𝑦

𝑑𝑥 =

𝑑𝑣𝑥

𝑑𝑥= 𝑣 + 𝑥

𝑑𝑣

𝑑𝑥

Eq(1) => 𝑣 + 𝑥𝑑𝑣

𝑑𝑥 =

𝑣𝑥−𝑥𝑡𝑎𝑛 (𝑣)

𝑥= 𝑣 − 𝑡𝑎𝑛𝑣

Or −−𝑑𝑣

𝑡𝑎𝑛𝑣=

𝑑𝑥

𝑥

Or cotv dv= −−𝑑𝑥

𝑥

Integrating both sides we get

log 𝑠𝑖𝑛𝑣 = −logx + logC

Or sin( 𝑦

𝑥 ) =

𝐶

𝑥

Or x sin( 𝑦

𝑥 ) = 𝐶

10. Find the equation of a curve passing through the origin, given that the slope of the tangent to the

curve at any point 𝒙, 𝒚 is equal to the sum of the co-ordinates of the point. 4

Solution :

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11. Using vectors prove that angle in a semicircle ia a right angle. 4

Solution:

Let O(0,0) be the centre of the semicircle of radius a. The co-ordinate of any point P on the

semicircle can be written as (acosq,asinq) where q is the angle formed by OP with positive

direction of X- axis.

This means the angle between PA and PB is 900.

OR

Using vectors prove that 𝐜𝐨𝐬 𝜶 − 𝜷 = 𝐜𝐨𝐬 𝜶 𝐜𝐨𝐬𝜷 + 𝐬𝐢𝐧𝜶 𝐬𝐢𝐧𝜷. 4

Solution: Let 𝑎 and 𝑏 be two unit vectors making angle α and β with X axis respectively as

shown in figure.

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Resolving 𝑎 =cosα𝑖 + sinα𝑗

𝑏 =cosβ𝑖 + sinβ𝑗

Taking dot product

𝑎 . 𝑏 = 1.1. cos 𝛼 − 𝛽 = cosα𝑖 + sinα𝑗 ). ( cosβ𝑖 + sinβ𝑗 )

Or cos 𝛼 − 𝛽 =cosα cosβ + sinαsinβ

12. Find the vector equation of a plane in normal form. 4

Solution: Let us consider a plane whose perpendicular distance from the origin is d(d≠0).

If 𝑂𝑁 is the normal from the origin to the plane, and 𝑛 is the unit normal vector along 𝑂𝑁 .Then

𝑂𝑁 = 𝑑𝑛 .

Let P be any point on the plane. Therefore, 𝑁𝑃 is perpendicular to 𝑂𝑁 . Hence 𝑁𝑃 . 𝑂𝑁 = 0…….(1)

Let 𝑟 be the position vector of the point P, then 𝑁𝑃 =𝑟 − 𝑑𝑛

(1) => (𝑟 − 𝑑𝑛 ). 𝑑𝑛 = 0

(𝑟 − 𝑑𝑛 ). 𝑛 = 0

𝑟 . 𝑛 = 𝑑 ………………. (2) which is the vector equation of the

plane.

Again if P(x, y, z) is the point on the plane and (l, m, n) be the direction cosines of 𝑛 then

𝑂𝑃 = 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 and 𝑛 = 𝑙𝑖 + m𝑗 + n𝑘

(2) => (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 ).(l𝑖 + m𝑗 + n𝑘 ) = 𝑑

=> lx + my +nz = d which is the cartesian equation of the plane

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OR

Find the equation of a plane passing through a given point and perpendicular to a given vector in

vector form. 4

Solution: Let a plane passes through a given point A with position vector 𝑎 and perpendicular to the

vector 𝑁 .

Let 𝑟 be the position vector any point P(x, y, z) on the plane.

Then the point P lies in the plane if and only if 𝐴𝑃 is perpendicular to 𝑁 .

i.e. 𝐴𝑃 . 𝑁 = 0

But 𝐴𝑃 = 𝑟 − 𝑎

Therefore (𝑟 − 𝑎 ). 𝑁 = 0 is the vector equation of the plane.

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13. Assume that each child born is equally likely to be a boy or a girl. If a family has two children,

what is the conditional probability that both are girls, given that

i. The youngest is a girl,

ii. Atleast one is a girl? 4

Solution:

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OR

The probability of a shooter hitting a target is 𝟑 𝟒 . How many minimum numbers of times must

he/she fire so that the probability of hitting the target at least once is more than 0.99? 4

Solution: Let the shooter fire n times. Obviously, n fires are n Bernoulli trials.

In each trial, p = probability of hitting the target = 3

4 and

q = probability of not hitting the target =.1

4

Then P(X = x) = 𝑛𝐶𝑥 1

4 𝑛−𝑥

3

4

𝑥

.

Now, given that,

P(hitting the target at least once) > 0.99

i.e. P(x ≥ 1) > 0.99

Therefore, 1 – P (x = 0) > 0.99

or 1− 𝑛𝐶0 1

4

𝑛

> 0.99

or 𝑛𝐶0 1

4

𝑛

< 0.01

or 1

4

𝑛

< 0.01

or 4𝑛>1

0.01 = 100 ...…………………….(1)

The minimum value of n to satisfy the inequality (1) is 4.

Thus, the shooter must fire 4 times.

14. If x, y, x are all different and

𝒙 𝒙𝟐 𝟏 + 𝒙𝟑

𝒚 𝒚𝟐 𝟏 + 𝒚𝟑

𝒛 𝒛𝟐 𝟏 + 𝒛𝟑

= 𝟎, 6

Prove that xyz=-1.

Solution:Let ∆ = 𝑥 𝑥2 1 + 𝑥3

𝑦 𝑦2 1 + 𝑦3

𝑧 𝑧2 1 + 𝑧3

= 𝑥 𝑥2 1𝑦 𝑦2 1

𝑧 𝑧2 1

+ 𝑥 𝑥2 𝑥3

𝑦 𝑦2 𝑦3

𝑧 𝑧2 𝑧3

= −1 2 1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

+ xyz 1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

(Using C3↔C2 and then C1↔C2)

16 | P a g e

= 1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

(1 + 𝑥𝑦𝑧)

=(1 + 𝑥𝑦𝑧) 1 𝑥 𝑥2

0 𝑦 − 𝑥 𝑦2 − 𝑥2

0 𝑧 − 𝑥 𝑧2 − 𝑥2

(Using R2 → R2 – R1 and R3 → R3 – R2)

Taking out common factor (y-x) from R2 and (z-x) from R3 , we get,

∆= 1 + 𝑥𝑦𝑧 𝑦 − 𝑥 (𝑧 − 𝑥) 1 𝑥 𝑥2

0 1 𝑥 + 𝑦0 1 𝑧 + 𝑥

= 1 + 𝑥𝑦𝑧 𝑦 − 𝑥 𝑧 − 𝑥 (𝑧 − 𝑦) (Expanding along C1 )

Since ∆=0 and x, y, z are all different, i.e. x – y ≠ 0, y – z ≠ 0, z-x ≠ 0, we get 1 + 𝑥𝑦𝑧=0

OR

If 𝒂 ≠ 𝒑; 𝒃 ≠ 𝒒; 𝒄 ≠ 𝒓 and 𝒑 𝒃 𝒄𝒂 𝒒 𝒄𝒂 𝒃 𝒓

= 𝟎,

Then find the value of 𝒑

𝒑−𝒂+

𝒒

𝒒−𝒃+

𝒓

𝒓−𝒄. 6

Solution: Given 𝑝 𝑏 𝑐𝑎 𝑞 𝑐𝑎 𝑏 𝑟

= 0

Applying R1 → R1 – R3 and R2 → R2 – R3

𝑝 − 𝑎 0 𝑐 − 𝑟

0 𝑞 − 𝑏 𝑐 − 𝑟𝑎 𝑏 𝑟

= 0

Taking (p – a), (q – b), (c – r) common from C1, C2, C3;

(p – a) (q – b) (c – r)

1 0 10 1 1𝑎

𝑝−𝑎

𝑏

𝑞−𝑏

𝑟

𝑐−𝑟

= 0

Expanding the determinant

𝑟

𝑐 − 𝑟−

𝑏

𝑞 − 𝑏+ 0 + 1. 0 −

𝑎

𝑝 − 𝑎 = 0

𝑟

𝑟−𝑐+

𝑏

𝑞−𝑏+

𝑎

𝑝−𝑎= 0

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Adding 2 on both sides 𝑟

𝑟−𝑐+

𝑏

𝑞−𝑏+ 1 +

𝑎

𝑝−𝑎+ 1 = 2

=> 𝑝

𝑝−𝑎+

𝑞

𝑞−𝑏+

𝑟

𝑟−𝑐= 2

15. Find the maximum amd minimum value of the following functions; if exist. 3+3=6

i. 𝒇 𝒙 =𝒙𝟐−𝒙+𝟏

𝒙𝟐+𝒙+𝟏; 𝒙 ∈ 𝑹.

Solution:

By second derivative test, f is the minimum at x = 1 and the minimum value is given by

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ii. 𝒇 𝒙 = 𝐥𝐨𝐠 𝒙, 𝒙 > 0.

Here f′(x)= 1

𝑥

Since logx is defined for a positive number, f′(x)>0 for any x.

Therefore there does not exist any c ∈ R such that f′(c)= 0

Hence f(x) does not have maxima or minima

OR

Find the maximum area of an isosceles triangle inscribed in the ellipse 𝒙𝟐

𝒂𝟐 +𝒚𝟐

𝒃𝟐 = 𝟏 with its vertex at

one end of the major axis. 6

Solution:

The given ellipse is 𝑥2

𝑎2 +𝑦2

𝑏2 = 1

Let the major axis be along the x −axis.

Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0).

Since the ellipse is symmetrical with respect to the x−axis and y −axis, we can assume

the coordinates of A to be (−x1, y1) and the coordinates of B to be (−x1, −y1).

Now, we have . y1=±𝑏

𝑎 𝑎2 − 𝑥1

2

∴Coordinates of A are (−x1, 𝑏

𝑎 𝑎2 − 𝑥1

2 ) and the coordinates of B are (x1,− 𝑏

𝑎 𝑎2 − 𝑥1

2 )

As the point (x1, y1) lies on the ellipse, the area of triangle ABC (A) is given by

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But, x1 cannot be equal to a.

Also, when x1 = 𝑎

2, then

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Thus, the area is the maximum when x1 = 𝑎

2

Maximum area of the triangle is given by,

16. Evaluate :

𝐭𝐚𝐧−𝟏 𝟐𝒙−𝟏

𝟏+𝒙−𝒙𝟐 𝒅𝒙𝟏

𝟎. 6

Solution:

Adding (1) and (2), we obtain

21 | P a g e

OR

Evaluate (𝒙𝟐 + 𝒙)𝒅𝒙𝟑

𝟏 as the limit of a sum.

Solution: Let I = (𝑥2 + 𝑥)𝑑𝑥3

1

= 𝑥2𝑑𝑥 + 𝑥𝑑𝑥3

1

3

1

= I1 + I2 (say)

We know,

For I1 = 𝑥2𝑑𝑥3

1,

a = 1, b = 3 and f(x) = x2

3−1

𝑛=

2

𝑛

I1 = 𝑥2𝑑𝑥3

1= (3 − 1) lim𝑛→∞

1

𝑛 𝑓 1 + 𝑓 1 + 𝑕 + … . . … . +𝑓 1 + (𝑛 − 1)𝑕

= 2 lim𝑛→∞1

𝑛 12 + 1 +

2

𝑛

2+ 1 + 2.

2

𝑛

2+ … … + 1 + (𝑛 − 1)

2

𝑛

2

= 2 lim𝑛→∞1

𝑛 (12 + … . +12) +

2

𝑛

2 12 + 22 + … . + 𝑛 − 1 2 + 2.

2

𝑛 1 + 2 + ⋯ . (𝑛 − 1)

= 2 lim𝑛→∞1

𝑛 𝑛 +

4

𝑛2 𝑛−1 𝑛 (2𝑛−1)

6 +

4

𝑛 𝑛−1 𝑛−2

2

= 2 lim𝑛→∞ 1 +4

6 1 −

1

𝑛 2 −

1

𝑛 + 2 1 −

1

𝑛 1 −

2

𝑛

= 2 [1+8

6 + 2]

= 26

3

For I2 = 𝑥𝑑𝑥3

1,

a = 1, b = 3 and f(x) = x

3−1

𝑛=

2

𝑛

I2 = 𝑥𝑑𝑥3

1= (3 − 1) lim𝑛→∞

1

𝑛 𝑓 1 + 𝑓 1 + 𝑕 + … . . … . +𝑓 1 + (𝑛 − 1)𝑕

= 2 lim𝑛→∞1

𝑛 1 + 1 +

2

𝑛 + 1 + 2.

2

𝑛 + … … + 1 + (𝑛 − 1)

2

𝑛

= 2 lim𝑛→∞

1

𝑛 (1 + ⋯ + 1) +

2

𝑛 1 + 2 + ⋯ . (𝑛 − 1)

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= 2 lim𝑛→∞1

𝑛 𝑛 +

2

𝑛 𝑛−1 𝑛−2

2

= 2 lim𝑛→∞1

𝑛 𝑛 + 1 −

1

𝑛 1 −

2

𝑛

= 2 .2 = 4

Hence I = 26

3 + 4 =

38

3

17. Find the area bounded by 6

𝒚 = 𝒙𝟐 and 𝒚 = 𝒙 .

Solution:

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OR

Find the ratio in which the area bounded by the curves 𝒚𝟐 = 𝟏𝟐𝒙 & 𝒙𝟐 = 𝟏𝟐𝒚 is divided by the

line 𝒙 = 𝟑.

Solution: As shown in figure the line x= 3 cuts the area between 𝑦2 = 12𝑥 & 𝑥2 = 12𝑦 i.e OACBO in

two parts OABO and ACBA.

Area of OABO = 𝟏𝟐 𝒙𝒅𝒙𝟑

𝟎−

𝟏

𝟏𝟐𝒙𝟐𝒅𝒙

𝟑

𝟎 = 12 -

𝟑

𝟒 =

𝟒𝟓

𝟒

Similiarly Area of ACBA = 𝟏𝟐 𝒙𝒅𝒙𝟏𝟐

𝟑−

𝟏

𝟏𝟐𝒙𝟐𝒅𝒙

𝟏𝟐

𝟑 = 84 -

𝟏𝟖𝟗

𝟒 =

𝟏𝟒𝟕

𝟒

So required ratio = Area of OABO

Area of ACBA =

𝟒𝟓 𝟒

𝟏𝟒𝟕 𝟒 =

𝟏𝟓

𝟒𝟗

18. Show that the lines

𝒓 = 𝒊 + 𝒋 − 𝒌 + 𝝀(𝟑𝒊 − 𝒋 )

and 𝒓 = 𝟒𝒊 − 𝒌 + µ(𝟐𝒊 + 𝟑𝒌 ) are coplanar.

Also, find the equation of the plane containing both these lines. 6

Solution: Here the given lines are 𝑟 = 𝑖 + 𝑗 − 𝑘 + 𝜆 3𝑖 − 𝑗 ……… …… (1)

and 𝑟 = 4𝑖 − 𝑘 + µ 2𝑖 + 3𝑘 … ………… (2)

Comparing with 𝑟 = 𝑎1 + 𝜆𝑏1 and 𝑟 = 𝑎2 + 𝜆𝑏2

𝑎1 = 𝑖 + 𝑗 − 𝑘 , 𝑎2 = 4𝑖 − 𝑘 & 𝑏1 = 3𝑖 − 𝑗 , 𝑏2

= 2𝑖 + 3𝑘

Now 𝑎2 − 𝑎1 = 3𝑖 − 𝑗

𝑏1 𝑥 𝑏2

= −3𝑖 − 9𝑗 + 2𝑘

But (𝑎2 − 𝑎1 ).( 𝑏1 𝑥 𝑏2

)=−9 + 9 = 0 Which shows that the lines are co-planer.

24 | P a g e

2nd part:

Equation of the plane containing both the lines is (𝑟 − 𝑎1 ).( 𝑏1 x 𝑏2

)=0

Putting the values we get 3x + 9y - 2z = 14

19. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine and 3

hours of craftsman’s time in its making, while a cricket bat takes 3 hours of machine time and 1 hour

of craftsman’s time. In a day the factory has the availability of not more than 42 hours of machine

time and 24 hours of craftsman’s time. If the profit on racket and on a bat is Rs.20 and Rs.10

respectively, find the maximum profit of the factory when it works at full capacity. 6

Solution:

Let the number of rackets and number of bats to be made be x and y respectively

The machine time is not available for more than 42 hours

1.5x+3y≤42 ………………… 1

The craftman’s time is not available for more than 24 hours

3x+y≤24 ………………… 2

25 | P a g e

OR

Minimize and maximize 𝒁 = 𝒙 + 𝟐𝒚

Subject to 𝒙 + 𝟐𝒚 ≥ 𝟏𝟎𝟎 ; 𝟐𝒙 − 𝒚 ≤ 𝟎 ; 𝟐𝒙 + 𝒚 ≤ 𝟐𝟎𝟎 ; 𝒙, 𝒚 ≥ 𝟎.

Solution:

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20. Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a

meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug

reduces its chances by 25%. At a time a patient can choose any one of the two options with equal

probabilities. It is given that after going through one of the two options the patient selected at

random suffers a heart attack. Find the probability that the patient followed a course of meditation

and yoga. 6

Solution:

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OR

In a 20- question true-false examination, suppose a student tosses a fair coin to determine his answer

to each question. If the coin falls head, he answers ‘true’. If it falls tail, he answers false. Find the

probability that he answers at least 12 questions as true.

Solution:

Completed by Nayanmani Sarma, Electrical Engineering, Assam Engineering College, Jalukbari,

Guwahati-13

Big Thanks to Dibyajyoti Bhagawati (Instrumentation Engineering) and Nazimuddin Ahmed (Electrical

Engineering), Assam Engineering College, Guwahati-13