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1

WB JEE 2008

MCQ (Single Correct Answer)

The function $$f(x) = {e^{ax}} + {e^{ - ax}},a > 0$$ is monotonically increasing for

A
$$-$$1 < x < 1
B
x < $$-$$1
C
x > $$-$$1
D
x > 0

Explanation

$$f(x) = {e^{ax}} + {e^{ - ax}}$$

$$f'(x) = a{e^{ax}} - a{e^{ - ax}} = a({e^{ax}} - {e^{ - ax}})$$

If f(x) is monotonically increasing

then $$f'(x) > 0$$

$$ \Rightarrow a({e^{ax}} - {e^{ - ax}}) > 0 \Rightarrow ({e^{ax}} - {e^{ - ax}}) > 0$$

$$ \Rightarrow {e^{ax}} > {e^{ - ax}} \Rightarrow {e^{2ax}} > 1 = {e^0}$$

$$ \Rightarrow 2ax > 0 \Rightarrow x > 0$$ ($$because$$ $$a > 0$$)

$$\therefore$$ function f(x) is increasing when x > 0.

2

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.

3

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.

4

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.

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