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Let $$f:R \to R$$ be a continuous function defined by $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$$Statement - 1 : $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$. Statement - 2 : $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$ A Statement - 1 is true, Statement -2 is true; Statement - 2 is not a correct explanation for Statement - 1. B Statement - 1 is true, Statement - 2 is false. C Statement - 1 is false, Statement - 2 is true. D Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1. ## Explanation $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$ $$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$ $$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$ $${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2$$ maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$ $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$ Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow$$ for some $$c \in R$$ $$f\left( c \right) = {1 \over 3}$$ 2 ### AIEEE 2010 MCQ (Single Correct Answer) The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that is parallel to the $$x$$-axis, is A $$y=1$$ B $$y=2$$ C $$y=3$$ D $$y=0$$ ## Explanation Since tangent is parallel to $$x$$-axis, $$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$ Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$ 3 ### AIEEE 2010 MCQ (Single Correct Answer) Let $$f:R \to R$$ be defined by $$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,if} & {x \le - 1} \cr {2x + 3,\,\,if} & {x > - 1} \cr } } \right.$$$

If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is

A
$$0$$
B
$$- {1 \over 2}$$
C
$$-1$$
D
$$1$$

## Explanation

$$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr {2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr } } \right.$$

This is true where $$k=-1$$
4

### AIEEE 2009

Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $$x=0$$ is the only
real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$
A
$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$
B
$$P(-1)$$ is the minimum but $$P(1)$$ is not the maximum of $$P$$
C
Neither $$P(-1)$$ is the minimum nor $$P(1)$$ is the maximum of $$P$$
D
$$P(-1)$$ is the minimum and $$P(1)$$ is the maximum of $$P$$

## Explanation

We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$

$$\Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$

But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$

$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$

As given that $$P\left( { - 1} \right) < P\left( a \right)$$

$$\Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$

Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$

As $$P'\left( x \right) = 0,$$ there is only one solution $$x = 0,$$

therefore $$4{x^2} + 3ax + 2b = 0$$ should not have any real roots i.e. $$D < 0$$

$$\Rightarrow 9{a^2} - 32b < 0$$

$$\Rightarrow b > {{9{a^2}} \over {32}} > 0$$

Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$

$$\forall x > 0$$

$$\therefore$$ $$P(x)$$ is an increasing function on $$\left( {0,1} \right)$$

$$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$

Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$

$$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$

So we can conclude that

Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$

$$\Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $$P.$$

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