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1

### WB JEE 2008

MCQ (Single Correct Answer)

The equation ex + x $$-$$ 1 = 0 has, apart from x = 0

A
One other real root
B
Two real roots
C
No other real root
D
Infinite number of real roots

## Explanation ex + x $$-$$ 1 = 0

$$\Rightarrow$$ ex = 1 $$-$$ x

Let y = ex, y = 1 $$-$$ x

graph of y = ex and y = 1 $$-$$ x intersect only at point A(0, 1), so they have no other real root except x = 0.

2

### WB JEE 2008

MCQ (Single Correct Answer)

Select the correct statement from (a), (b), (c), (d). The function $$f(x) = x{e^{1 - x}}$$

A
strictly increases in the interval (1/2, 2)
B
increases in the interval (0, $$\infty$$)
C
decreases in the interval (0, 2)
D
strictly decreases in the interval (1, $$\infty$$)

## Explanation

$$f(x) = x{e^{1 - x}}$$

$$f'(x) = x{d \over {dx}}{e^{1 - x}} + {e^{(1 - x)}}{d \over {dx}}x$$

$$= - x{e^{1 - x}} + {e^{1 - x}} = (1 - x){e^{1 - x}}$$

If $$f'(x) < 0$$

$$\Rightarrow (1 - x){e^{1 - x}} < 0$$ ($$\because$$ $${e^{1 - x}} > 0$$)

$$\Rightarrow 1 - x < 0 \Rightarrow x > 1$$

$$\Rightarrow x \in (1,\infty )$$ so function is strictly decreasing.

3

### WB JEE 2008

MCQ (Single Correct Answer)

The area included between the parabolas y2 = 4x and x2 = 4y is

A
$${8 \over 3}$$ sq. units
B
8 sq. units
C
$${16 \over 3}$$ sq. units
D
12 sq. units

## Explanation Equation of curves are

y2 = 4x ..... (i)

and x2 = 4y .... (ii)

From (i) and (ii)

$${\left( {{{{x^2}} \over 4}} \right)^2} = 4x \Rightarrow {x^4} - 64x = 0$$

$$\Rightarrow x({x^3} - 64) = 0 \Rightarrow x = 0$$ or $$x = 4$$

$$\therefore$$ Required area $$= \int\limits_0^4 {\left( {\sqrt {4x} - {{{x^2}} \over 4}} \right)dx = \left[ {2.{{2{x^{3/2}}} \over 3} - {{{x^3}} \over {12}}} \right]_0^4}$$

$$= {{32} \over 3} - {{64} \over {12}} = {{32} \over 3} - {{16} \over 3} = {{16} \over 3}$$ sq. unit.

4

### WB JEE 2008

MCQ (Single Correct Answer)

The line which is parallel to x-axis and crosses the curve y = $$\sqrt x$$ at an angle 45$$^\circ$$ is

A
y = $${1 \over 4}$$
B
y = $${1 \over 2}$$
C
y = 1
D
y = 4

## Explanation Since equation of any line parallel to y-axis is y = k where k is any real constant. We draw a tangent which makes an angle 45$$^\circ$$ with the line parallel to x-axis.

Given equation of curve is

$$y = \sqrt x$$ ..... (1)

Taking derivative

$${{dy} \over {dx}} = {1 \over {2\sqrt x }}$$

Since $${{dy} \over {dx}} = \tan 45^\circ = 1$$

$$\therefore$$ $${1 \over {2\sqrt x }} = 1$$ or $$x = {1 \over 4}$$

Substituting $$x = {1 \over 4}$$ in (1)

$$y = \sqrt {{1 \over 4}} = {1 \over 2}$$

$$\therefore$$ Point $$P\left( {{1 \over 4},{1 \over 2}} \right)$$, it also passes through the line parallel to x-axis.

$$\therefore$$ $$y = {1 \over 2}$$ is required equation of line.

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