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1

WB JEE 2009

MCQ (Single Correct Answer)

If the rate of increase of the radius of a circle is 5 cm/sec., then the rate of increase of its area, when the radius is 20 cm, will be

A
10$$\pi$$
B
20$$\pi$$
C
200$$\pi$$
D
400$$\pi$$

Explanation

Let radius r $$\therefore$$ $${{dr} \over {dt}} = 5$$ cm/sec (given)

Area of circle $$(A) = \pi {r^2}$$

$$\therefore$$ $${{dA} \over {dt}} = {{d(\pi {r^2})} \over {dt}} = \pi (2r){{dr} \over {dt}}$$

when r = 20 then $${{dA} \over {dt}} = \pi \,.\,2\,.\,20\,.\,5 = 200\pi $$.

2

WB JEE 2009

MCQ (Single Correct Answer)

The distance covered by a particle in t seconds is given by x = 3 + 8t $$-$$ 4t2. After 1 second its velocity will be

A
0 unit/second
B
3 units/second
C
4 units/second
D
7 units/second

Explanation

x = 3 + 8t $$-$$ 4t2

$${{dx} \over {dt}} = 8 - 8t$$

$$\therefore$$ Velocity at $$t = 1 = {\left( {{{dx} \over {dt}}} \right)_{t = 1}} = 8 - 8\,.\,1 = 0$$

3

WB JEE 2009

MCQ (Single Correct Answer)

The Rolle's theorem is applicable in the interval $$-$$1 $$\le$$ x $$\le$$ 1 for the function

A
f(x) = x
B
f(x) = x2
C
f(x) = 2x3 + 3
D
f(x) = |x|

Explanation

(a) f(x) = x

$$f'(x) = {{df(x)} \over {dx}} = 1$$ which is greater than zero

$$\therefore$$ f (x) is strictly increasing in [$$-$$1, 1].

So Rolle's theorem is not applicable.

(b) $$\because$$ f($$-$$1) = f(1) = 1

Also f(x) = x2 is continuous in [$$-$$1, 1] and differentiable in ($$-$$1, 1)

$$\therefore$$ Rolle's theorem is applicable.

(c) f(x) = 2x3 + 3 $$\Rightarrow$$ f'(x) = 6x2 > 0

$$\therefore$$ f(x) is strictly increasing in [$$-$$1, 1].

So, Rolle's theorem is not applicable.

(d) f(x) = |x| = x, x $$\ge$$ 0 and $$-$$x, x < 0

f(1) = f($$-$$1) = 1, also f(x) is continuous but f(x) is not differentiable at x = 0 $$\in$$ ($$-$$1, 1). So all conditions of Rolle's theorem is not satisfied.

4

WB JEE 2009

MCQ (Single Correct Answer)

A particle is moving in a straight line. At time t, the distance between the particle from its starting point is given by x = t $$-$$ 6t2 + t3. Its acceleration will be zero at

A
t = 1 unit time
B
t = 2 unit time
C
t = 3 unit time
D
t = 4 unit time

Explanation

$$x = t - 6{t^2} + {t^3}$$

velocity $$ = {{dx} \over {dt}} = 1 - 12t + 3{t^2}$$

acceleration $$ = {d \over {dt}}\left( {{{dx} \over {dt}}} \right) = - 12 + 6t$$

If acceleration is zero, then

$$ - 12 + 6t = 0 \Rightarrow t = 2$$.

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