WBJEE 2017 Maths Paper with Solutions

WBJEE 2017 Maths question paper along with its solutions are provided on this page. We have prepared step by step solutions for each question that will significantly help students to get a better grip of the concepts as well as learn the correct answer to the problems. These WBJEE question paper and solutions will further enable students to practice productively, learn about the paper pattern, understand the difficulty level of the questions and ultimately score higher ranks in upcoming WBJEE exams. Students can download the question paper and solutions in a PDF format instantly at the bottom.

WBJEE 2017 - Maths

Question 1. Transforming to parallel axes through a point (p, q), the equation 2x2 + 3xy + 4y2 + x + 18y + 25 = 0 becomes 2x2 + 3xy + 4y2 = 1. Then

1. a. p = –2, q = 3
2. b. p = 2, q = – 3
3. c. p = 3, q = – 4
4. d. p = – 4, q = 3

Solution:

Given 2x2 + 3xy + 4y2 + x + 18y + 25 = 0 …(i)

After transforming eq.(i) through (p, q) it becomes

2x2 + 3xy + 4y2 = 1 …(ii)

Then, the point (p, q) satisfies (i) and (ii)

Differentiate (i) with respect to x and y respectively, we get

4x + 3y + 1 = 0 …(iii)

3x + 8y + 18 = 0 …(iv)

Because (p, q) satisfies (i) and (ii)

Therefore, Point (p, q) satisfies (iii) and (iv)

So, we get

4p + 3q + 1 = 0 …(v)

3p + 8q + 18 = 0 …(vi)

Solving eq.(v) and (vi) simultaneously we get

p = 2, q = -3

Question 2. Let A(2, –3) and B (–2, 1) be two angular points of ΔABC. If the centroid of the triangle moves on the line 2x + 3y = 1, then the locus of the angular point C is given by

1. a. 2x + 3y = 9
2. b. 2x – 3y = 9
3. c. 3x + 2y = 5
4. d. 3x – 2y = 3

Solution:

Given: A(2, -3), B(-2, 1) of ΔABC

Centroid moves on 2x + 3y = 1

Let C(α, β) be the third point on ΔABC

Let G ->Centroid of ΔABC.

Then, G = [(2-2+α)/3, (-3+1+β)/3] = [α/3, (β-2)/3] …(i)

Since, G moves on the line 2x + 3y = 1

Points on this line are (x, (1-2x)/3)

This implies, G = (t, (1-2t)/3)

From eq.(i) α/3 = t and (β-2)/3 = (1-2t)/3

α= 3t and β = 3 – 2t

Therefore, C = (3t, 3 – 2t)

Locus of point C is given by,

=>(3 – 2t)2 + (3 - 2t + 3)3 = (3t + 2)2 + (3 – 2t – 1)2

=>9t2 + 4 – 12t + 36 + 4t2 – 24t = 9t2 + 4 + 12t + 4 + 4t2 – 8t

=>40t = 32

=> t = 32/40 = 4/5

C = (12/5, 7/5) so the equation of Locus of C is given by 2x + 3y = d

Substituting C in the given equation 2x + 3y = d we get d = 9.

Thus, the Locus of C is 2x + 3y = 9

Question 3. The point P (3, 6) is first reflected on the line y = x and then the image point Q is again reflected on the line y = -x to get the image point Q’. Then the circumcentre of the ΔPQO' is

1. a. (6, 3)
2. b. (6, – 3)
3. c. (3, –6)
4. d. (0, 0)

Solution:

Given P(3, 6) and y = x, Q -> y = –x to get Q’ Circumcenter of ΔPQQ’ = ?

When the point P(3, 6) is reflected on the line y = x the x and y coordination interchange, we get point Q(6, 3). Again reflecting Q’ on the line y = –x we get the image point Q’(–3, –6).

To find out the circumcenter, we need to solve any two bisector equations and then point out the intersection point.

So, midpoint of PQ’ = ((3-3)/2, (6-6)/2) = (0, 0)

Slope of PQ’ is m1 = (-6-6)/(-3-3) = 2

Slope of bisector is the negative reciprocal of the given slope so, slope (m1) of bisector of PQ’ is –1/2

Equation of PQ’ with slope (m1) – 1/2 and point (0, 0) is given by

y - y1 = m(x - x1)

y – 0 = (-1/2)(x – 0)

y = (-1/2)x

2x + y = 0 …(1)

Now, mid-point of PQ = [(3+6)/2, (6+3)/2] = (9/2, 9/2)

Slope of PQ is m2 = (3-6)/(6-3) = -1

Slope of bisector of PQ is 1.

Equation of PQ with slope (m2) = 1 and point (9/2, 9/2) is given by

y - 9/2 = 1(x - 9/2)

y = x

x – y = 0 .....2)

Solving equation (1) & (2), we get

x = 0 & y = 0

Therefore, the circumcenter of triangle PQQ’ is (0, 0).

Question 4. Let d1 and d2 be the lengths of the perpendiculars drawn from any point of the line 7x - 9y + 10 = 0 upon the lines 3x + 4y = 5 and 12x + 5y = 7 respectively. Then

1. a. d1> d2
2. b. d1 = d2
3. c. d1< d2
4. d. d1> 2d2

Solution:

Given: Perpendiculars drawn from any point of

7x – 9y + 10 = 0 …(i)

3x + 4y = 5 …(ii)

12x + 5y = 7 …(iii)

Any point on the line (i) will be of the form (x , (7x+10)/9)

d1 and d2 are the lengths at any point of eq.(i) upon eq.(ii) and eq.(iii) distance from (a, (7a+10)/9) to eq.(ii) is

Question 5. The common chord of the circles x2 + y2 – 4x – 4y = 0 and 2x2 + 2y2 = 32 subtends at the origin an angle equal to

1. a. π/3
2. b. π/4
3. c. π/6
4. d. π/2

Solution:

Given: Circles are

x2 + y2 – 4x – 4y = 0 …(a)

2x2 + 2y2 = 32 …(b)

To find the common chord we need to find the intersection points of the circles.

Divide eq.(b) by 2 and subtract from eq.(a), we get

-4x - 4y = -16

i.e.x + y = 4 …(c)

Substitute x = y - 4 in (b), we get

(y - 4)2 + y2 = 16

y2- 8y + 16 + y2 = 16

y(y - 4) = 0

y = 0, 4

At y = 0, x = 4 and at y = 4, x = 0

So, the points of intersection of these two circles is (4, 0) and (0, 4) thus, the chord drawn from (0, 4) to (4, 0).

Thus, the angle subtended at the origin by this chord is a right angle.

Hence, the chord subtends an angle of π/2 at the origin.

Question 6. The locus of the mid-points of the chords of the circle x2 + y2 + 2x – 2y – 2 = 0 which make an angle of 90 degrees at the centre is

1. a. x2 + y2 – 2x – 2y = 0
2. b. x2 + y2 – 2x + 2y = 0
3. c. x2 + y2 + 2x – 2y = 0
4. d. x2 + y2 + 2x – 2y – 1=0

Solution:

Given Circles is x2 + y2 + 2x - 2y - 2 = 0

Which can be written as (x + 1)2 + (y - 1)2 = 4, so C(-1, 1) and radius is 2.

Let O be the centre and AB be the chord, angle = 90°

Draw OA and OB

& ∠AOB = 90°

Now OA and OB are radius of the circle.

OA = OB

& ∠OAB = & ∠OBA [Angle opposite to equal sides are equal]

But ∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB = 180 – 90

∠OAB = 45° = ∠OBA

OP perpendicular to C so P(h, k) is the mid-point of chord AB.

Join OP which makes an angle of 90° with the chord AB.

Now, triangle APO is a right angled triangle such that ∠ APO = 90°

This implies, sin 45o = OP/OA = OP/2 => OP = √2

Question 7. Let P be the foot of the perpendicular from focus S of hyperbola x2/a2 - y2/b2 = 1 on the line bx – ay = 0 and let C be the centre of the hyperbola. Then the area of the rectangle whose sides are equal to that of SP and CP is

1. a. 2ab
2. b. ab
3. c. (a2 + b2)/2
4. d. a/b

Solution:

Given hyperbola,x2/a2 - y2/b2 = 1, Center (C) = (0, 0)

Let the coordinates of the focus s be (s, 0) and assume that a > b

Therefore, s = √(a2+b2)

Also, perpendicular from focus to asymptote [bx – ay = 0]

y = (b/a)x slope (-a/b) we have equation of normal

As y = (-a/b)x +K

by + ax = bk =>a√(a2+b2) (passing through the focus)

CP = a

Hence, area of rectangle with sides CP and SP is ab.

Question 8. B is an extremity of the minor axis of an ellipse whose foci are S and S’. If angle SBS’ is a right angle, then the eccentricity of the ellipse is

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

Solution:

Given: ∠SBS' is a right angle

By Pythagoras theorem

⇒SS`2 = SB2 + S’B2

⇒(ae + ae)2 = (ae2) + b2 + (ae)2 + b2

⇒(2ae)2 = 2(ae)2 + 2b2

⇒= 4(ae)2 = 2(ae)2 + 2b2

⇒ae = b

⇒e = b/a

But e2 = 1 – b2/a2

⇒e2 =1 – e2

⇒2e2 = 1

⇒e = 1/(√2)

Question 9. The axis of the parabola x2 + 2xy + y2 – 5x + 5y – 5 = 0 is

1. a. x + y = 0
2. b. x + y – 1 = 0
3. c. x – y + 1 = 0
4. d. x - y = 1/√(2)

Solution:

Given parabola x2 + 2xy + y2 – 5x + 5y – 5 = 0

(x + y)2 = 5x – 5y + 5

(x + y)2 = 5 (x – y + 1)

For a parabola (Ax + Cy)2 + Dx + Ey + F = 0

The axis is Ax + Cy + (AD+CE)/(2(A2 + C2)) = 0

So, axis is (x + y) + [(-5+5)/2(1+1)] = 0

Therefore, axis is x + y = 0

Question 10. The line segment joining the foci of the hyperbola x2 – y2 + 1 = 0 is one of the diameters of a circle. The equation of the circle is

1. a. x2 + y2 = 4
2. b. x2 + y2 =√(2)
3. c. x2 + y2 = 2
4. d. x2 + y2 = 2√(2)

Solution:

Given hyperbola x2- y2+ 1 = 0

(x – 0)2 + (y - 0)2 = 0

Foci of the given hyperbola = (0, $\pm \sqrt{2}$)

C (0, 0) diameter of the circle is the distance b + w foci

Diameter (D) = $\sqrt{(0+0)^2 + (\sqrt {2} +\sqrt {2})^2} = 2\sqrt {2}$)

Centre of the circle will be same as that of hyperbola centre (c) = (0, 0) and Radius (R) = √2

Equation of the circle is x2 + y2 = 2.

Question 11. The equation of the plane through (1, 2, -3) and (2, -2, 1) and parallel to X-axis is

1. a. y – z + 1 = 0
2. b. y – z – 1 = 0
3. c. y + z – 1 = 0
4. d. y + z + 1 = 0

Solution:

Given equation plane (1, 2, –3) and (2, –2, 1) points

Question 12. Three lines are drawn from the origin O with direction cosines proportional to (1, –1, 1), (2 –3, 0) and (1, 0, 3). The three lines are

1. a. not coplanar
2. b. coplanar
3. c. perpendicular to each other
4. d. coincident

Solution:

Given: The lines are drawn from origin 0 with dcs proportional to (1, –1, 1) (2, –3, 0) and (1, 0, 3)

Consider $\begin{vmatrix} 1 &-1 &1 \\ 2& -3 &0 \\ 1& 0 &3 \end{vmatrix}$

=>1(–9 –0) + 1 (6 – 0) + 1 (0 + 3)

=>– 9 + 9 = 0

Therefore, Δ = 0

Hence, the lines are co-planar.

Question 13. Consider the non-constant differentiable function f of one variable which obeys the relation f(x)/f(y) = f(x-y). If f’(0) = p and f’(5)=q, then f’(-5) is

1. a. p2/9
2. b. q/9
3. c. p/q
4. d. q

Solution:

Question 14. If f(x) = log5log3x, then f’(e) is equal to

1. a. e loge5
2. b. b. e loge3
3. c. 1/(eloge5)
4. d. 1/(eloge3)

Solution:

Question 15. Let F(x) = ex, G(x) = e-x and H(x) = G(F(x)), where x is a real variable. Then dH/dx at x = 0 is

1. a. 1
2. b. -1
3. c. -1/e
4. d. -e

Solution:

f(x) = ex

G(x) = e-x

G(f(x) = e-(f(x))

H(x) = e-ex

$\frac{dH}{dx}(x)=-e^xe^{-e^x}$

Question 16. If f’’(0) = k, k ≠ 0, then the value of

1. a. K
2. b. 2k
3. c. 3k
4. d. 4k

Solution:

Question 17.

1. a. m2
2. b. 2
3. c. -1
4. d. -m2

Solution:

Again, differentiate equation (2) and applying the quotient rule, along with the chain rule we get.

Therefore, (1 – x2)y'' = m2y + xy'

(1 – x2) y'' – xy' = m2y

k = m2

Question 18. The chord of the curve y = x2 + 2ax + b, joining the points where x = α and x = β, is parallel to the tangent to the curve at abscissa x =

1. a. (a+b)/2
2. b. (2a+b)/3
3. c. (2α +β)/3
4. d. (α +β)/2

Solution:

Given curve is y = x2 + 2ax + b, differentiate above equation with respect to x we get.

dy/dx = 2x + 2a ………….(1)

Equation (1) of tangent to the curve

But tangent to the curve is parallel to the chord of the curve which joints the points x = α and x = β

Tangent to this curve = (α + β) + 2a

2x + 2a = (α + β) + 2a

x = (α+β)/2

Question 19. Let f(x) = x13 + x11 + x9 + x7 + x5 + x3 + x + 19. Then f(x) = 0 has

1. a. 13 real roots
2. b. only one positive and only two negative real roots
3. c. not more than one real root
4. d. has two positive and one negative real root.

Solution:

Given f(x) = x13 + x11 + x9 + x7 + x5 + x3 + x + 19 = 0

=> f'(x) = 13x12 + 11x10 + 9x8 + 7x6 + 5x4 + 3x2 + 1

Since f'(x) consists of all even powers of x therefore

f'(x) ≥ 0

That f(x) increasing function. Hence, it intersect the x-axis at only one point. Therefore, f(x) has atmost one real root.

Question 20. If

Then Lagrange’s mean value theorem is applicable to f(x) in closed interval [0, x]

1. a. for all p, q
2. b. only when p > q
3. c. only when p < q
4. d. for no value of p, q

Solution:

Given Lagrange’s mean theorem apply.

f, it is continuous at (a, b) and differential with in the closed interval [a, b]

i.e. limx->a f(x) = f(a)

Here, f(a) = f(0) = 0

f(x) is differentiable at (0, x)

Consider limx->a f(x)

Therefore, limx->a f(x) = 0 only when p > q

Hence, Lagrange’s mean value theorem is applicable to f(x) in [0, x] only when p > q.

Question 21. $\lim_{x\rightarrow \infty} (sin\ x)^{2tanx}$

1. a. is 2
2. b. is 1
3. c. is 0
4. d. does not exist

Solution:

Given y = (sinx)2tanx

log y = 2 tanx.log (sinx)

=>limx->0tanx

tan x = 0

log y = 0

y = e0 = 1

So, limx->0 (sinx)2tanx =1

Question 22. ∫cos(log x) dx = F(x) + c, where c is an arbitrary constant. Here F(x) =

1. a. x[cos(log x) + sin(log x)]
2. b. x[cos(log x) - sin(log x)]
3. c. (x/2)[cos(log x) + sin(log x)]
4. d. (x/2)[cos(log x) - sin(log x)]

Solution:

Given ∫ cos(log x) dx

Let logx = t ⇒ x = et

(1/x) dx = dt ⇒ dx = xdt

dx = et dt

I ⇒ ∫ cos(log x) dx ⇒ ∫ et cos t dt

a = 1, b = 1

I = et/(12+12) * (sin t + cos t) + c

I ⇒ (x/2)[sin(log x)+cos(log x)] + c

Question 23.

Solution:

Dividing by x2 numerator or denominator

Question 24. $I = \int_{10}^{19}\frac{sinx}{1+x^8}dx$, then

1. a. |I| < 10-9
2. b. |I| < 10-7
3. c. |I| < 10-5
4. d. |I| > 10-7

Solution:

Question 25. Let $I_1 = \int_{0}^{n}[x]dx$ and $I_2 = \int_{0}^{n}(x)dx$, where [x] and (x) are integral and fractional parts of x and n ∈ N - {1}. Then I1/I2 is equal to

1. a. 1/(n-1)
2. b. 1/n
3. c. n
4. d. n - 1

Solution:

$I_1 = \int_{0}^{n}[x]dx= \int_{0}^{1}0dx + \int_{1}^{2}1dx +....+ \int_{n-1}^{n}(n-1))dx$

⇒ 0 + 1 (2 – 1) + 2 [3– 2] + 3[4–3]+ ..... +(n–1)[n–(n–1)]

⇒ 1 + 2 + 3 + 4 + .... + (n – 1)

I1 = (n-1)n/2 = [(n(n-1)]/2

$I_2 = \int_{0}^{n}(x)dx\Rightarrow\int_{0}^{n}(x-[x])dx$

Question 26. The value of

1. a. nπ/4
2. b. π/4
3. c. π/4n
4. d. π/2n

Solution:

Question 27. The value of the integral

1. a. is less than 1
2. b. is greater than 1
3. c. is less than or equal to 1
4. d. lies in the closed interval [1, e]

Solution:

a = 0, b = 1

f(x) = ex^2

m = f(0) = e0 =1

m f(1) = e1 = e

Now,

Question 28: $\int_{0}^{100}e^{x - [x]}dx=$

Solution:

$\int_{0}^{100}e^{x - [x]}dx \Rightarrow \int_{0}^{100}e^{[x]}dx$

Here x = [x] + {x} and x - [x] = {x}

0 ≤ {x} < 1 period = 1

$\int_{0}^{mT}f(x) dx = m \int_{0}^{T}f(x) dx$

Here T is time period

$\int_{0}^{100}e^x dx \Rightarrow 100[e^x]_0^1$

100(e1 - e0) = 100(e - 1)

Question 29: Solution of (x+y)2dy/dx = a2 (a’ being a constant) is

1. a. (x+y)/a = tan[(y-c)/a], c is an arbitrary constant
2. b. xy = a tan cx, c is an arbitrary constant
3. c. x/a = tan(y/c), c is an arbitrary constant
4. d. xy = tan(x+c), c is an arbitrary constant

Solution:

(x+y)2dy/dx = a2

Let x + y = t, 1 + dy/dx = dt/dx and dy/dx = dt/dx - 1

t2(dt/dx – 1) = a2

t2 dt/dx = a2 + t2

Question 30: The integrating factor of the first order differential equation x2(x2 – 1)dy/dx + x( x2+ 1)y = x2- 1 is

1. a. ex
2. b. x – 1/x
3. c. x + 1/x
4. d. 1/x2

Solution:

x2(x2 - 1)dy/dx + x( x2+ 1) = x2- 1

dy/dx + [(x2+1)/(x(x2-1)] y = 1/x2

dy/dx + py = θ

A = 2/(-1)(1) = -2

B = 2/2 = 1

C = 2/(-1)(-2) = 1

Question 31: In a G.P. series consisting of positive terms, each term is equal to the sum of next two terms. Then the common ratio of this G.P. series is

1. a. √5
2. b. (√5-1)/2
3. c. √5/2
4. d. (√5+1)/2

Solution:

Each term is sum of next two terms

⇒ tn = tn + 1 + tn + 2

⇒ arn–1 = arn + arn + 1

⇒ 1 = r + r2

⇒ r2 + r–1 = 0

⇒ r = -1±√(1-4(-1)/2(1)

⇒ r = (-1±√5)/2

r = (√5-1)/2 or r = (-√5-1)/2

But common ratio can be negative

r = (√5-1)/2

Question 32: If (log5x)(logx3x)(log3xy) = logxx3, then y equals

1. a. 125
2. b. 25
3. c. 5/3
4. d. 243

Solution:

(log5x)(logx3x)(log3xy) = logxx3

⇒ (log x/log 5)(log 3x/log x)(log y/log 3) = (log x3/log x)

⇒ (log x. log 3x. log y)/(log 5 log x. log 3x) = 3log x/log x

⇒ log y/log 5 = 3

log y = 3 log 5

y = 53 = 125

Question 33: The expression (1+i)n/(1-i)n-2 equals

1. a. -in+1
2. b. -in+1
3. c. -2in+1
4. d. 1

Solution:

(1+i)n/(1-i)n-2 ⇒ (1+i)n/(1-i)n/(1-i)2

⇒ [cos (nπ/2)+i sin (nπ/2)](-2i)

Question 34: Let z = x + iy, where x and y are real. The points (x, y) in the X-Y plane for which (z+i)/(z-i) is purely imaginary lie on

1. a. a straight line
2. b. an ellipse
3. c. a hyperbola
4. d. a circle

Solution:

Let z = x + iy

x2 + y2 –1 = 0

x2 + y2 = 1 is circle.

Question 35: If p, q are odd integers, then the roots of the equation 2px2+(2p+q)x+q = 0 are

1. a. rational
2. b. irrational
3. c. non-real
4. d. equal

Solution:

Given equation 2px2 +(2p + q)x + q = 0 …..(1)

Now D = (2p + q)2 – 8pq

⇒ 4p2 + 4pq + q2– 8pq

⇒ 4p2–4pq + q2

⇒ (2p – q)2 always a perfect square

So roots are given by

x = [-(2p-q)+ √(2p+q)2-8pq]/4p

x = [-2p-q ±(2p-q)]/4p

x = [-2p-q +(2p-q)]/4p = -q/2p (rational p, q are odd integers)

Or x = [-2p-q -(2p-q)]/4p = -1 which is also rational.

Question 36: Out of 7 consonants and 4 vowels, words are formed each having 3 consonants and 2 vowels. The number of such words that can be formed is

1. a. 210
2. b. 25200
3. c. 2520
4. d. 302400

Solution:

No of ways of selecting = 7C3 × 4C2 =210

No. of group, each heavy 3 consonants and 2 vowels = 210

Each group contains 5 letters.

No of ways of arranging 5 letters = 5! = 210

Required no. of ways = 210×120 = 25200

Question 37: The number of all numbers having 5 digits, with distinct digits is

1. a. 99999
2. b. 9×9P4
3. c. 10P5
4. d. 10P4

Solution:

5 digits 9 9 8 7 6

Total no. (1 to 9)

Formula = npr = n!/(n-r)! (since 0 can’t be used at first place)

9×(9×8×7×6×5!)/5! = 9×9!/5!

Apply formula 9×9!/(9-4)! = 9×9P4

Question 38: The greatest integer which divides (p + 1)(p + 2)(p + 3)......(p + q) for all p ∈N and fixed q ∈N is

1. a. p!
2. b. q!
3. c. p
4. d. q

Solution:

(p + 1), (p + 2), (p + 3)…..(p + q) are q consecutive numbers.

(p + 1) (p + 2) (p + 3) …. (p + q) is a product of ‘q’ consecutive natural numbers

The product will be divisible by all its factors and since it is a product of consecutive natural numbers, so it will always be divisible by q!

Also q! is the greatest integer amongst all their divisors.

Hence, the greatest integer which divides the product is q!

Question 39: Let ((1 + x) + x2)9 = a0 + a1x + a2x2 +..... + a18x18. Then

1. a. a0 + a2 +..... + a18 = a1 + a3 + ...... + a17
2. b. a0 + a2 +..... + a18 is even
3. c. a0 + a2 +..... + a18 is divisible by 9
4. d. a0 + a2 +..... + a18 is divisible by 3 but not by 9

Solution:

Given, ((1+ x)+x2)9 = a0 + a1x + a2x2 + ……….+ a18 x18…….(1)

Put x = 1 in eq. (1), we get.

⇒ (1+1+1)9 = a0 + a1 + a2 + ……..+ a18

⇒ (3)9 = a0 + a1 + a2 + ……. + a18…….(2)

Put x = –1 in eq. (1) we get.

⇒ (1)9 = a0 –a1 + a2 – a3 + …….a17 + a18 …….(3)

Adding eq. (2) and (3), we get.

⇒ 39 + 1 = a0 + a1 + …. + a18 + a0 – a1 + a2 – ……. – a17 + a18

⇒2a0 + 2a2 + 2a4 + …… + 2a18

⇒ 2(a0 + a2 + a4 + …….+ a18) = 39 +1

⇒ a0 + a2 + …….a18 = (39+1)/2 ⇒ even

Hence, a0 + a2 + a4 + ……….. + a18 is even

Question 40: The linear system of equations

has

1. a. only ‘zero solution’
2. b. only a finite number of non-zero solutions
3. c. no non-zero solution
4. d. infinitely many non-zero solutions

Solution:

Given 8x – 3y – 5 z = 0

5x – 8y + 3z = 0

3x + 5y – 8z =0

Use Cramer’s rule

x= ∆x/ ∆ , y = ∆y/ ∆ , z = ∆z/ ∆

∆x = ∆y = ∆z

∆ = $\begin{vmatrix} 8 & -3 &-5 \\ 5 & -8 & 3\\ 3 & 5 & -8 \end{vmatrix}$

∆ = (512 – 125 – 27) – (120 + 120 + 120)

∆ = 360 – 360 = 0

So ∆x/ ∆ = 0/0 , ∆y/ ∆ = 0/0, ∆z/ ∆ = 0/0

So infinitely many non-zero solutions.

Question 41: Let P be the set of all non-singular matrices of order 3 over R and Q be the set of all orthogonal matrices of order 3 over R .Then,

1. a. P is a proper subset of Q
2. b. Q is a proper subset of P
3. c. Neither P is a proper subset of Q nor Q is a proper subset of P
4. d. P⋂Q the void set

Solution:

P = {P1, P2, P3 ……Pn} (non –singular matrix)

Q = {Q1, Q2, Q3 ……Qn} (orthogonal matrix)

Q, QT = I

Det (Q) = ±1

P contains matrix whose value is non zero

Q is proper sub-sets of P.

Question 42: Let A = $\begin{pmatrix} x+2 & 3x\\ 3 & x+2 \end{pmatrix}$ , B = $\begin{pmatrix} x &0 \\ 5 & x+2 \end{pmatrix}$ . Then all solutions of the equation det (AB) = 0 is

1. a. 1, –1, 0, 2
2. b. 1, 4, 0, –2
3. c. 1, –1, 4, 3
4. d. –1, 4, 0, 3

Solution:

Given A = $\begin{pmatrix} x+2 & 3x\\ 3 & x+2 \end{pmatrix}$ , B = $\begin{pmatrix} x &0 \\ 5 & x+2 \end{pmatrix}$

det |AB| = $\begin{vmatrix} x^{2}+17x &3x^{2}+6x \\ 8x+10& (x+2)^{2} \end{vmatrix}$

⇒ x(x + 2)(x-4)(x–1) = 0

⇒ x = 0, -2, 1, 4

Question 43: The value of det A, where A = $\begin{pmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1& -\cos \theta & 1 \end{pmatrix}$ lies

1. a. in the closed interval [1, 2]
2. b. in the closed interval [0, 1]
3. c. in the open interval (0, 1)
4. d. in the open interval (1, 2)

Solution:

Given A = $\begin{pmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1& -\cos \theta & 1 \end{pmatrix}$

R1 -> R1+R3

det A ⇒ $\begin{vmatrix} 0 & 0 &1 \\ -\cos \theta & 1 &\cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$

Question 44: Let f : R -> R be such that f is injective and f(x) f(y) = f(x + y) for all x,y belongs to R . If f(x), f(y), f(z) are in G.P, then x, y, z, are in

1. a. A.P always
2. b. G.P always
3. c. A.P depending on the value of x, y, z
4. d. G.P depending on the value of x, y, z

Solution:

Given: f : R-> R injective

f(x)f(y) = f(x + y)

Let f(x) = ax then f(x) f(y) = ax ay = ax+y

Also f(x) = f(y)

⇒ ax = ay

⇒ ax–y = a0

⇒ x – y = 0

⇒ x = y

Hence an is injective

Now ax, ay, az are in G.P.

⇒ ay = ax . r where r is common ratio

Also az = ayr

⇒ ay/az = ax/ay

⇒a2y= ax+z

y = (x+z)/2

Hence x, y, z are in A.P.

Question 45: On the set R of real numbers we define xPy if and only if xy ≥ 0. Then the relation P is

1. a. reflexive but not symmetric
2. b. symmetric but not reflexive
3. c. transitive but not reflexive
4. d. reflexive and symmetric but not transitive

Solution:

Let p be the relation on the set real number R such that xpy if an only if xy ≥ 0

(a) we know that, for any real number x, x2 ≥ 0

⇒xx ≥ 0 ⇒ xpx

So, p is reflexive

(b) let (x, y) ∈ p i.e. xpy

⇒xy ≥ 0 yx ≥ 0 ⇒ ypx

So p is symmetric

(c) Let xpy and ypz

⇒ xy ≥ 0 and yz ≥0

But from this, we can’t conclude x Z ≥ 0

For example

(–1, 0), (0, 2) satisfies the relation xy ≥ 0 but (–1, 2)

Doesn’t satisfy relation xy≥ 0

Thus p is not transitive

Hence p is reflexive symmetric but not transitive.

Question 46: On R, the relation ρ be defined by ‘x ρy holds if and only if x–y is zero or irrational’. Then

1. a. ρ is reflexive and transitive but not symmetric.
2. b. ρ is reflexive and symmetric but not symmetric.
3. c. ρ is symmetric and transitive but not reflexive.
4. d. ρ is an equivalence relation.

Solution:

xRy ⇒ x – y is zero or irrational.

xRx ⇒ 0 So reflexive

if xRy ⇒ x – y is zero or irrational.

⇒ y – x is zero or irrational

So yR x symmetric xRy

⇒ x – y is 0 or irrational yRz

⇒ y – z is 0 or irrational then (x – y)+(y–z) = (x–2) may be rational.

So it is not transitive.

Question 47: Mean of n observations x1, x2, ......., xn is $\bar{x}$ . If an observation xq is replaced by xq then the new mean is

1. a. $\bar{x}-x_{q}+x_{q}^{'}$
2. b. $\frac{(n-1)\bar{x}+x_{q}^{'}}{n}$
3. c. $\frac{(n-1)\bar{x}-x_{q}^{'}}{n}$
4. d. $\frac{n\bar{x}-x_{q}+x_{q}{'}}{n}$

Solution:

Given mean of x1, x2, ......., xn is $\bar{x}$.

Question 48: The probability that a non-leap year selected at random will have 53 Sundays is

1. a. 0
2. b. 1/7
3. c. 2/7
4. d. 3/7

Solution:

We know non-leap year has 365 days.

365 days, number of weeks = 52 and 1 day remaining. In 52 weeks, there will be 52 Sundays, 1 day remaining can be either Sunday, Mon, Tue, Wed, Thus, Fri and Saturday. out of these 7 total outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 Sundays = 1/7

Question49: The equation sin x (sin x + cos x) = k has real solutions, where k is a real number. Then

1. a. 0 ≤k ≤(1+ √2)/2
2. b. 2-√3 ≤k≤2+√3
3. c. 0≤k≤2-√3
4. d. (1-√2)/2≤k≤(1+ √2)/2

Solution:

Given: sin × (sin x + cos x) = k

sin2 x + sin x cos x = k

(1-cos2x)/2 +(sin 2x)/2 = k

1 – cos2 x + sin 2x = 2k

sin 2x – 2 cos2 x = 2k –1

-√2≤2k-1≤ √2 (since -√2 ≤sin2x-cos2x≤√2)

1-√2≤2k≤1+√2

(1-√2)/2 ≤k≤(1+√2)/2

Question 50: The possible values of x, which satisfy the trigonometric equation tan-1(x-1)/(x-2) + tan-1(x+1)/(x+2) = π/4 are

1. a. ±1/√2
2. b. ±√2
3. c. ±1/2
4. d. ±2

Solution:

Given tan-1(x-1)/(x-2) + tan-1(x+1)/(x+2) = π/4

Question 51: On set A = {1, 2, 3}, relations R and S are given by

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

S = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)} Then

1. a. R ⋃ S is an equivalence relation
2. b. R ⋃ S is reflexive and transitive but not symmetric
3. c. R ⋃ S is reflexive and symmetric but not transitive
4. d. R ⋃ S is symmetric and transitive but not reflexive

Solution:

Given A = {1, 2, 3}

R = {(1,1), (2,2), (3,3), (1,2), (2,1)}

S = {(1,1), (2,2), (3,3), (1,3), (3,1)}

(i) Check reflexive

Then (a,a) ∈ R for every a ∈ {1, 2, 3}

Since R⋃S = {(1,1), (2,2), (3,3), (1,2), (2,1), (1, 3), (3,1)}

R is reflexive.

(ii) Check symmetric

If (a,b) ∈ R and S, then (b, a) ∈ R and S

Here (1,2) ∈ R, but (2, 1) ∈ R

So (1, 3) ∈ R, but (3, 1) ∈ S

So R and S symmetric

(iii) Check transitive

IF (a,b) ∈ R and S & (b, c) ∈ R & S, then (a, c) ∈ R & S

Here R & S not transitive

R⋃ S is reflexive and symmetric but not transitive.

Question 52: If one of the diameters of the curve x2 + y2 – 4x – 6y + 9 = 0 is a chord of a circle with centre (1, 1), the radius of this circle is

1. a. 3
2. b. 2
3. c. √2
4. d. 1

Solution:

Given equation of circle is

x2 + y2 – 4x – 6y + 9 = 0

{ x2 + y2 + 2gx + 2fy + c = 0}

So, g = –2, f = –3, c = 9

Centre (–g, –f) ⇒ (2, 3)

r = √(-32+(-2)2-9)

r = √(9+2-9)

r = √4 = 2

Chord (0’, 0) = √(1-2)2+(1-3)2

= √5

We apply Pythagoras theorem

In ΔOO’A ⇒ O’O2 + O’A2 = OA2 = R2

R = √(√5)2+(2)2 = √(5+4)

R = 3

Question 53: Let A (–1, 0) and B (2, 0) be two points. A point M moves in the plane in such a way that ∠MBA = 2∠MAB. Then the point M moves along

1. a. a straight line
2. b. a parabola
3. c. an ellipse
4. d. a hyperbola

Solution:

Given A(–1, 0), B (2, 0)

We know that ∠ MBA = 2∠MAB

Also tan θ = | (m2-m1)/(1+m1m2) |

Where m1 and m2 are the slopes of two lines and θ is the angle between two lines.

Question 54: If f(x) = $\int_{-1}^{x}\left | t \right |dt$ , then for any x ≥0, f(x) is equal to

1. a. ½ (1-x2)
2. b. 1-x2
3. c. ½ (1+x2)
4. d. 1+x2

Solution:

Given f(x) = $\int_{-1}^{x}\left | t \right |dt$ , x ≥0

Apply formula $\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$

$f(x)=\int_{-1}^{0}-t\: dt+\int_{0}^{x}t\: dt$

$f(x)=\left [ \frac{-t^{2}}{2} \right ]_{-1}^{0}+\left [ \frac{t^{2}}{2} \right ]_{0}^{x}$

f(x) ⇒ 0-(-1/2)+(x2/2)-0

⇒ (1/2)+(x2/2)

= (1/2)(1+x2)

Question 55: Let for all x > 0, f(x) = $\lim_{x\to \infty }n(x^{\frac{1}{n}}-1)$ , then

1. a. f(xy)=f(x)-f(y)
2. b. f(xy)=f(x)+f(y)
3. c. f(xy)=x f(y) + yf(x)
4. d. f(xy) = x f(x) + y f(y)

Solution:

f(x) = $\lim_{x\to \infty }n(x^{\frac{1}{n}}-1)$

let n = 1/t

n-> ∞, t->0

$\lim_{t\to 0 }\frac{1}{t}(x^{t}-1)$

$\lim_{t\to 0 }\frac{(x^{t}-1)}{t}$

= (x0-1)/0

= 0/0

L hospital

$\lim_{t\to 0 }\frac{(x^{t}ln \: x-0)}{1}$ = x0ln x

= ln x

f(x) = ln x

⇒f(xy) = ln (xy) = ln x+ln y

⇒f(xy) = f(x)+f(y)

Question 56: Let I = $\int_{0}^{100\pi }\sqrt{(1-\cos 2x)}dx$, then

1. a. I =0
2. b. I = 200√2
3. c. I = π√2
4. d. I =100

Solution:

Given I = $\int_{0}^{100\pi }\sqrt{(1-\cos 2x)}dx$

cos 2x = 1 – 2sin2x

1–cos2x = 2sin2x

Question 57: The area of the figure bounded by the parabolas x = -2y2 and x = 1- 3y2 is

1. a.4/3 square units
2. b. 2/3 square units
3. c. 3/7 square units
4. d. 6/7 square units

Solution:

Question 58: Tangents are drawn to the ellipse (x2/9)+(y2/5)= 1 at the ends of both latus rectum. The area of the quadrilateral so formed is

1. a. 27 sq. units
2. b. 13/2 sq. units
3. c. 15/4 sq. units
4. d. 45 sq. units

Solution:

Now e2–1 –5/9 = 4/9

So e = 2/3

Equation of tangent at [2, (5/3)] is given as

(2x/9)+(y/3) = 1

Let F and f’ be foci

area of ΔCPQ = (1/2)×(9/2) ×3 = 27/4

area of quadrilateral PQRS = 4 ×27/4 = 27 sq. units.

Question 59: The value of K in order that f(x) = sin x-cos x-Kx+5 decreases for all positive real values of x is given by

1. a. K<1
2. b. K ≥1
3. c. K>√2
4. d. K<√2

Solution:

Given: f(x) = sin x – cos x – kx + 5

f(x) decrease for all x if f’(x) < 0

f’(x) = cos x + sin x – k < 0

So k > cos x + sin x

We know, cos x + sin x = √2((1/√ 2)cos x+(1/ √2) sin x)

⇒ √2 (sin π/4 cos x+cos π/4 sin x)= √2 sin(π/4 +x) and

–1 ≤ sin (π/4 +x) ≤ 1

⇒ -√2≤ √2 sin (π/4 +x) ≤ √2

So- √2≤ cos x + sin x≤ √2

Max (cos x +sin x) = √2

Hence k >√2

Question 60: For any vector x, the value of $(\vec{x}\times \hat{i})^{2}+(\vec{x}\times \hat{j})^{2}+(\vec{x}\times \hat{k})^{2}$ is equal to

1. a. $\left | \vec{x} \right |^{2}$
2. b. $2\left | \vec{x} \right |^{2}$
3. c. $3\left | \vec{x} \right |^{2}$
4. d. $4\left | \vec{x} \right |^{2}$

Solution:

Let x = x1i + x2j + x3k

Then x × i = –x2k + x3j

x × j = x1k–x3i

x × k = –x1j +x2i

So (x × i)2 + (x × j)2 + (x × k)2

⇒ x22+x32+x12+x32+x12+x22

⇒ 2(x12+x22+x32) ⇒ 2x2

⇒ 2|x|2

Question 61: If the sum of two unit vectors is a unit vector, then the magnitude of their difference is

1. a. √2 units
2. b. 2 units
3. c. √3 units
4. d. √5 units

Solution:

Question 62: Let α and β be the roots of x2+x+1= 0. If n be a positive integer, then αnn is

1. a. 2 cos 2nπ/3
2. b. 2 sin 2nπ/3
3. c. 2 cos nπ/3
4. d. 2 sin nπ/3

Solution:

x2+ x+1 = 0 ….(1)

Roots of this equation is

x = -1± √(1-4 ×1 ×1)/2 = (-1±√3i)/2

α and β are the roots of eq.(1)

So α = (-1+ √3i)/2 and β = (-1- √3i)/2

Consider, αn + βn

= ((-1+ √3i)/2)n + ((-1- √3i)/2 )n

= (-1/2)+i√3/2))n+((-1/2)-i√3/2))n

= (e i2π/3)n+(e -i2π/3)n ( since sin (2π/3) = √3/2 and cos (2π/3) = -1/2)

= e i2nπ/3+e -i2nπ/3

= cos (2nπ/3)+i sin (2nπ/3)+cos (2nπ/3)-i sin (2nπ/3)

= 2 cos(2nπ/3)

Hence αnn = 2cos(2nπ/3)

Question 63: For real x, the greatest value of (x2+2x+4)/(2x2+4x+9) is

1. a. 1
2. b. -1
3. c. 1/2
4. d. ¼

Solution:

y = (x2+2x+4)/(2x2+4x+9)

⇒2x2y + 4xy + 9y = x2 + 2x + 4

⇒ x2(2y–1) + x (4y–2) + 9y–4 = 0

Δ≥ 0

⇒ (4y – 2)2 – 4 (2y–1) (9y–4) ≥ 0

⇒ 4(2y–1)2–4(2y–1) (9y–4) ≥0

⇒ 4(2y–1) [2y–1–9y + 4] ≥ 0

⇒ 4(2y–1) (3–7y) ≥ 0

⇒ 4(2y–1) (7y–3) ≥ 0

⇒ 3/7 ≤y≤1/2

y greatest = 1/2

Question 64: Let A = $\begin{pmatrix} 1 & 1 & 1\\ 0& 1& 1\\ 0 & 0 &1 \end{pmatrix}$ . Then for positive integer n, An is

Solution:

Let A = $\begin{pmatrix} 1 & 1 & 1\\ 0& 1& 1\\ 0 & 0 &1 \end{pmatrix}$

An = $\begin{pmatrix} 1 & n & n(\frac{n+1}{2})\\ 0& 1& n\\ 0 & 0 &1 \end{pmatrix}$

This statement is denoted by P(n).

P(n):An = $\begin{pmatrix} 1 & n & n(\frac{n+1}{2})\\ 0& 1& n\\ 0 & 0 &1 \end{pmatrix}$

P(1) = A1 = $\begin{bmatrix} 1 & 1 &1 \\ 0& 1 &1 \\ 0& 0 & 1\end{bmatrix}$

P(1) is true.

P(2) = A2 = $\begin{bmatrix} 1 & 2 &3 \\ 0& 1 &2 \\ 0& 0 & 1\end{bmatrix}$

= $\begin{bmatrix} 1 & 2 &2(\frac{2+1}{2}) \\ 0& 1 &2 \\ 0& 0 & 1\end{bmatrix}$

So P(2) is true.

Let P(m) be true.

Question 65. Let a, b, c be such that b(a+c) ≠ 0 then the value of n is

1. a. any integer
2. b. zero
3. c. any even integer
4. d. any odd integer

Solution:

= (–1)nD

Therefore, R13, R23 and taking transpose

(1+(-1)n)D1 = 0 for any odd integer

Because, D1 ≠ 0 since b(a + c) ≠ 0

CATEGORY - III (Q66 to Q75)

One or more answer(s) is (are) correct. Correct answer(s) will fetch full marks 2. Any combination containing one or more incorrect answer will fetch 0 marks. Also, no answer will fetch 0 marks. If all correct answers are not marked and also no incorrect answer is marked then score = 2 times number of the correct answer(s) marked divided by the actual number of correct answers.

Question 66. Let f :R -> R be twice continuously differentiable. Let f(0) = f(1) = f’(0) = 0. Then

1. a. f''(x) ≠ 0 for all x
2. b. f’’(c) = 0 for some c &isin;R
3. c. f’’(x) ≠ 0if x≠ 0
4. d. f’(x) >0 for all x

Solution:

f(x) is continuous and differentiable

f(0) = f(1) = 0 by Rolle’s theorem

f ‘(a) = 0, a belongs to (0, 1)

Given f’(0) = 0 by Rolle’s theorem

f’’(0) = 0 for some, c belongs to (0, a)

Therefore, answer is f’’(c) = 0 for some ∈ R

Question 67. If f(x) = xn, n being a non-negative integer, then the values of n for which f’(α + β) = f’(α) + f’(β) for all α, β> 0 is

1. a. 1
2. b. 2
3. c. 0
4. d. 5

Solution:

We have f(x) = xn

f ‘(x) = nxn–1

f ‘(α + β) = f’ (α) + f ‘(β) for all α, β > 0

n (α+ β)n–1 = nαn–1 + nβn–1 for all α, β > 0

(α + β)n–1 = αn–1 + βn–1 for all α, β > 0

n –1 = 1 Or n = 2

Also, for n = 0, we have

f(x) = 1 for all x

f’(x) = 0 for all x

f ‘ (α + β) = f ‘(α) + f’(β)

Hence, n = 0, 2

Question 68. Let f be a non-constant continuous function for all x≥0 . Let f satisfy the relation f(x) f(a–x) = 1 for some a ∈ R+. Then$\int_{0}^{a}\frac{dx}{1+f(x)}$ is equal to

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

Solution:

Given: f(x) = f (a – x) =1

Question 69. If the line ax + by + c = 0, ab ≠0, is a tangent to the curve xy = 1-2x, then

1. a. a > 0, b < 0
2. b. a > 0, b > 0
3. c. a < 0, b > 0
4. d. a < 0, b < 0

Solution:

Given: ax + by + c = 0

by = –ax – c

y = –(a/b)x – c/b

m = slope = -a/b

Tangent slope (dy/dx)

-a/b = -1/x2

x2 = b/a

x2> 0

b/a > 0 if

(1) b > 0, a > 0 and

(2) b < 0, a < 0

Question 70. Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity u and the second starts from rest with constant acceleration f. Then

1. a. they will be at the greatest distance at the end of time u/2f from the start
2. b. they will be at the greatest distance at the end of time u/f from the start
3. c. their greatest distance is u2/2f
4. d. their greatest distance is u2/f

Solution:

Let S is distance at time t

Particle 1 is x1 = ut

Particle 2 is x2 = (1/2) ft2

Then S = x1 – x2

= ut + (-1/2) ft2…… (1)

For s being maximum, dS/dt=0

This implies, u – ft = 0

So, t = u/f

(b) Option is correct.

Substituting t in Eqn(1) we get

S = u2/f, this is the greatest distance

(c option correct)

Question 71. The complex number z satisfying the equation |z–i| = |z+1| = 1 is

1. a. 0
2. b. 1 + i
3. c. –1 + i
4. d. 1 – i

Solution:

Given: |Z – i| = |Z + 1| = 1

Let Z = x +iy

|x + iy – i| = 1

x2 + (y – 1)2 = 1

x2 + y2 – 2y + 1 = 1

x2 + y2 – 2y + 1 = 1

x2 + y2 – 2y = 0…….(1)

Also |z + 1| = 1

|x + iy + 1| = 1

(x + 1)2 + y2 = 1

x2 + y2 + 2x = 0……(2)

Eq.(1) and (2) we get

–y = x

For y = 1, x = -1

Therefore, z = –1 + i (c option is correct)

Also y = 0 x = 0

So, z = 0 (a option is correct)

Question 72. On R, the set of real numbers, a relation ρ is defined as ‘aρ b’ if and only if 1 + ab > 0’. Then

1. a. ρ is an equivalence relation
2. b. ρ is reflexive and transitive but not symmetric
3. c. ρ is reflexive and symmetric but not transitive
4. d. ρ is only symmetric

Solution:

Reflexive:If(x, x) ∈ R, where x is domain

Relation = {1 + ab > 0 where a, b ∈ R}

For any a ∈ R, a2≥ 0

So, 1 + a2≥ 1

Or, 1 + a2> 0

1 + a .a> 0

This implies, (a, a) ∈ R, its relation is Reflexive

Symmetric:If (y, x) ∈ R where (x, y) ∈ R

Relation (1 + ab > 0 where (a, b) ∈ R)

Let (a, b)∈ R, a, b ∈ R

So, 1 + ab > 0

1 + ba> 0

(b, a) ∈ R is symmetric

Transitive :If (x, y) ∈ R, where (x, y) ∈Rand (y,z) ∈ R

Relation {1 + ab > 0 where (a, b) ∈ R)

Let’s take (–2, 0) and (0, 1) such that 1 + – 2 × 0 > 0

And 1 + 0 × 1 > 0 so, (–2, 0) and (0, 1) belongs to R.

But (–2, 1) doesn’t belongs to R because 1 + – 2 × 1 < 0

The relation is not transitive.

Question 73. If a, b ∈ {1, 2, 3} and the equation ax2 + bx + 1 = 0 has real roots, then

1. a. a > b
2. b. a ≤ b
3. c. number of possible ordered pairs (a, b) is 3
4. d. a < b

Solution:

ax2 + bx + 1 = 0

a = a, b = b, c = 1

D = b2 – 4a

D ≥ 0

b2 – 4a≥ 0

{because, a, b ∈ (1, 2, 3)}

2≥4a

(a, b) = (1, 2) (2, 3) (1, 3) possible order pairs 3

And a < b

Question 74. If the tangent to y2 = 4ax at the point (at2, 2at) where |t|>1 is a normal to x2- y2 = a2 at the point (a secθ , a tanθ), then

1. a. t = – cosec θ
2. b. t = – sec θ
3. c. t = 2 tanθ
4. d. t = 2 cot θ

Solution:

y2 = 4ax

Tangent y = x/t + at and normal xcosθ + ycotθ = 2a

So (at2, 2at) must satisfy equation of normal

t2cosθ + 2 + xcotθ – 2 = 0

So, the roots are t = –cosecθ and t = 2tanθ

Question 75. The focus of the conic x2 – 6x + 4y + 1 = 0 is

1. a. (2, 3)
2. b. (3, 2)
3. c. (3, 1)
4. d. (1, 4)

Solution:

Given conic is x2 –6x + 4y + 1 = 0

x2 – 6x + 9 – 9 + 4y + 1 = 0

x2 – 3x – 3x + 9 – 9 + 4y + 1 = 0

x(x – 3) –3(x–3) + 4y – 8 = 0

(x – 3)2 + 4y – 8 = 0

(x – 3)2 = 4(–1) (y – 2)

Thus, the focus is (3, 1).