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1

### WB JEE 2009

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.

2

### WB JEE 2009

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

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

3

### WB JEE 2009

The equation of the tangent to the conic $${x^2} - {y^2} - 8x + 2y + 11 = 0$$ at (2, 1) is

A
x + 2 = 0
B
2x + 1 = 0
C
x + y + 1 = 0
D
x $$-$$ 2 = 0

## Explanation

Equation of conic section $${x^2} - {y^2} - 8x + 2y + 11 = 0$$

Differentiating w.r.t. x, we get

$$2x - 2yy' - 8 + 2y' = 0 \Rightarrow y' = {{4 - x} \over {1 - y}}$$

Slope of tangent at (2, 1) is $${(y')_{(2,1)}} = {{4 - 2} \over {1 - 1}} = {2 \over 0}$$

$$\therefore$$ Equation of tangent at (2, 1) is

$$y - 1 = {2 \over 0}(x - 2) \Rightarrow x - 2 = 0$$

4

### WB JEE 2008

A particle is projected vertically upwards and is at a height h after t1 seconds and again after t2 seconds then

A
$$h = g{t_1}{t_2}$$
B
$$h = {1 \over 2}g{t_1}{t_2}$$
C
$$h = {2 \over g}{t_1}{t_2}$$
D
$$h = \sqrt {g{t_1}{t_2}}$$

## Explanation

Let the initial velocity is u

$$\therefore$$ $$h = ut - {1 \over 2}g{t^2}$$

$$\Rightarrow g{t^2} - 2ut + 2h = 0$$

This is a quadratic in t,

let it has two roots t1 and t2

$$\therefore$$ $${t_1}{t_2} = {{2h} \over g}$$ ( Product of roots )

$$\Rightarrow h = {1 \over 2}g{t_1}{t_2}$$.

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