Mathematics Test 158

Result

Mathematics Test 158

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Weekly Quiz Competition

Question 1
4 / -1

The area (in sq. units) of the region {(x, y) : y² ≥ 2x and x² + y² ≤ 4x, x ≥ 0, y ≥ 0} is :-

A

B

C

D

Solution
Question 2
4 / -1

A line y = mx + 1 intersects the circle (x – 3)² + (y + 2)² = 25 at the points P and Q. If the midpoint of the line segment PQ has x-coordinate -3/5, then which one of the following options is correct?

A
– 3 < m < – 1

B
6 < m < 8

C
4 < m < 6

D
2 < m < 4

Solution
Question 3
4 / -1

The axis of parabola is along the line y = x and the distance of vertex from origin is √2 and that of origin from its focus is 2√2. If vertex and focus both lie in the 1^st quadrant, then the equation of the parabola is :

A
(x + y)² = (x – y – 2)

B
(x – y)² = (x + y – 2)

C
(x – y)² = 4(x + y – 2)

D
(x – y)² = 8(x + y – 2)

Solution
Question 4
4 / -1

The common tangents to the circle x² + y² = 2 and the parabola y² = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is :

A
3

B
6

C
9

D
15
Question 5
4 / -1

If chords of the hyperbola x² – y² = a² touch the parabola y² = 4ax then the locus of the middle points of these chords is the curve :

A
y²(x + a) = x³

B
y²(x – a) = x³

C
y²(x + 2a) = 3x³

D
y²(x – 2a) = 2x³

Solution
Question 6
4 / -1

L₁ and L₂ are two lines whose vector equations are

where λ and μ are scalars. If 'α' is the acute angle between L₁ and L₂, which is independent of 'θ', then α =

A
π/6

B
π/3

C
π/4

D
5π/12

Solution
Question 7
4 / -1

Then the range of a, so that f(x) has maximum at x = –2, is :-

A
|a| ≥ 1

B
|a| < 1

C
a > 1

D
a < 1

Solution

f(x) will have maxima at x = –2 only if a² + 1 ≥ 2 or or |a| ≥ 1.
Question 8
4 / -1

A
Periodic with period π/2

B
Periodic with period π

C
Periodic with period 2π

D
Aperiodic
Question 9
4 / -1

exists and is equal to 0, then

A
a = –3 and b = 9/2

B
a = 3 and b = 9/2

C
a = –3 and b = –9/2

D
a = 3 and b = –9/2

Solution
Question 10
4 / -1

Let g : R → R be a differentiable function satisfying g(x) = g(y) g(x – y) x, y ∈ R and g'(0) = a and g'(3) = b then g'(–3) is :-

A
a²/b

B
a/b

C
b/a

D
None of these

Solution

Differentiating partially w.r.t. x keeping y as a constant g'(x) = g(y) [g'(x – y)]

Put y = x

g'(x) = g(x) · g'(0) = a · g(x)

⇒ g(x) = ae^x (∵ g(0) = a)

Now g' (x) = ae^x, g'(3) = ae³ = b

and g'(–3) = ae^–3 = a²/b