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 Multiple Choice QuestionsMultiple Choice Questions

1.

Let A (2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line

  • 2x + 3y = 9

  • 2x – 3y = 7

  • 3x + 2y = 5

  • 3x – 2y = 3


A.

2x + 3y = 9

If C be (h, k) then centroid is (h/3, (k – 2)/3) it lies on 2x + 3y = 1.
⇒ locus is 2x + 3y = 9.

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2.

The curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -

  • {–2, 2}

  • {1}

  • {-2}

  • {2}


C.

{-2}

for point of intersection at exactly one point
λx + 3 = (λ + 1)x2 + 2
(λ + 1)x2 – λx – 1 = 0
Δ = 0
λ2 + 4(λ + 1) = 0
λ2 + 4λ + 4 = 0
(λ + 2)2 = 0 λ = – 2

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3.

The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that

  • it passes through the origin

  • it makes angle π/2 + θ with the x-axis

  • it passes through (aπ/2 ,-a)

  • it is at a constant distance from the origin


D.

it is at a constant distance from the origin

Clearly dy/dx = an θ
⇒ slope of normal = - cot θ
Equation of normal at ‘θ’ is
y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

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4.

If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is

  • 12

  • 8

  • 4

  • 6


C.

4

The tangent of slope m must be of the form y = m(x + 2) + a/m

so comma space 2 straight m space plus 2 over straight m space equals space straight c
rightwards double arrow space straight c space equals 2 open parentheses straight m plus 1 over straight m close parentheses greater or equal than space 2 space straight x space 2 space
therefore straight c subscript min space equals space 4

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5.

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

  • 3/2

  • 5/2

  • 7/2

  • 9/2


C.

7/2

The distance between 4x + 2y + 4z - 16 = 0 and 4x + 2y + 4z + 5 = 0 is

open vertical bar fraction numerator 5 plus 16 over denominator square root of 16 plus 4 plus 16 end root end fraction close vertical bar space equals space open vertical bar fraction numerator 21 over denominator square root of 36 end fraction close vertical bar space equals space 21 over 6 space equals space 7 over 2

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6.

If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/ 2 , y = 1 + t, z = 2 – t with parameters s and t respectively, are co-planar then λ equals

  • –2

  • –1

  • -1/2

  • 0


A.

–2

Given fraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y plus 3 over denominator negative straight lambda end fraction space equals space fraction numerator straight z minus 1 over denominator straight lambda end fraction space equals space straight s space space and space fraction numerator straight x over denominator 1 divided by 2 end fraction space equals fraction numerator straight y minus 1 over denominator 1 end fraction space equals fraction numerator straight z minus 2 over denominator negative 1 end fraction space equals space straight t  are coplanar then plan passing through these lines has normal perpendicular to these lines
⇒ a - bλ + cλ = 0 and a/2 +b -c =0 (where a, b, c are direction ratios of the normal to the plan) On solving, we get λ = -2.

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7.

A line makes the same angle θ, with each of the x and z-axis. If the angle β, which it makes with y-axis, is such that sin2β = 3sin2θ , then cos2θ equals 

  • 2/3

  • 1/5

  • 3/5

  • 2/3


C.

3/5

A line makes angle θ with x-axis and z-axis and β with y-axis.
∴ l = cosθ, m = cosβ,n = cosθ
We know that, l2+ m2+ n2= 1

cos2θ + cos2β +cos2θ =1
2cos2θ = 1- cos2β
2cos2θ = sin2β
But sin2β = 3 sin2θ
therefore from equation (i) and (ii)
3sin2θ = 2cos2θ
3(1-cos2θ) = 2cos2θ
3-3cos2θ = 2cos2θ
3 = 5cos2θ

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8.

The intersection of the spheres x2 +y2 +z2 + 7x -2y-z =13 and x2 +y2 +z2 -3x +3y +4z = 8 is the same as the intersection of one of the sphere and the plane

  • x-y-z =1

  • x-2y-z =1

  • x-y-2z=1

  • 2x-y-z =1


D.

2x-y-z =1

Required plane is S1 – S2 = 0 where
S1 = x2 + y2 + z2 + 7x – 2y – z – 13 = 0 
and S2 = x2 + y2 + z2 – 3x + 3y + 4z – 8 = 0
⇒ 2x – y – z = 1.

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9.

The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:

  • x – 2y + 11 = 0

  • x + 2y + 11 = 0

  • x + 2y – 11 = 0

  • x – 2y – 11 = 0


B.

x + 2y + 11 = 0

Equation of the required plane is
(x + y + z – 6) + λ(2x + 3y + z + 5) = 0
i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0
This plane is perpendicular to xy plane whose
equation is z = 0
i.e. 0 . x + 0 . y + z = 0
∴ By condition of perpendicularity
0 .(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0
i.e. λ = –1
∴ Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) =0
or x + 2y + 11 = 0

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10.

A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by

  • (3a, 3a, 3a), (a, a, a)

  • (3a, 2a, 3a), (a, a, a)

  • (3a, 2a, 3a), (a, a, 2a)

  • (2a, 3a, 3a), (2a, a, a)


B.

(3a, 2a, 3a), (a, a, a)

Any point on the line straight x over 1 space equals fraction numerator straight y plus straight a over denominator 1 end fraction space equals straight z over 1 space equals space straight t subscript 1 space left parenthesis say right parenthesis space is space left parenthesis straight t subscript 1 comma space straight t subscript 1 minus straight a comma space straight t subscript 1 right parenthesis and any point on the line fraction numerator straight x plus straight a over denominator 2 end fraction space equals space straight y over 1 space equals straight z over 1 space equals space straight t subscript 2 space left parenthesis say right parenthesis space is space left parenthesis 2 straight t subscript 2 minus straight a comma space straight t subscript 2 comma space straight t subscript 2 right parenthesis

Now direction cosine of the lines intersecting the above lines is proportional to (2t2 – a – t1, t2 – t1 + a, t2 – t1).
Hence 2t2 – a – t1 = 2k , t2 – t1 + a = k and t2 – t1 = 2k

On solving these, we get t1 = 3a , t2 = a. Hence points are (3a, 2a, 3a) and (a, a, a).

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