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1

### AIEEE 2010

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 $$k=-1$$
2

### 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 $$P.$$
3

### AIEEE 2009

Let $$f\left( x \right) = x\left| x \right|$$ and $$g\left( x \right) = \sin x.$$
Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point.
Statement-2: gof is twice differentiable at $$x=0$$.
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

Given that $$f\left( x \right) = x\left| x \right|\,\,$$ and $$\,\,g\left( x \right) = \sin x$$

So that go

$$f\left( x \right) = g\left( {f\left( x \right)} \right)$$

$$= g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$$

$$= \left\{ {\matrix{ {\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr } } \right.$$

$$= \left\{ {\matrix{ { - \sin \,{x^2},} & {if\,\,\,x < 0} \cr {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr } } \right.$$

$$\therefore$$ $$\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{ { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr } } \right.$$

Here we observe

$$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$$

$$\Rightarrow$$ go $$f$$ is differentiable at $$x=0$$

and $$\left( {go\,f} \right)'$$ is continuous at $$x=0$$

Now $$\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{ { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr } } \right.$$

Here $$L\left( {gof} \right)''\left( 0 \right) = - 2$$ and $$R\left( {go\,f} \right)''\left( 0 \right) = 2$$

As $$L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$$

$$\Rightarrow go\,f\left( x \right)$$ is not twice differentiable at $$x=0.$$

$$\therefore$$ Statement - $$1$$ is true but statement $$-2$$ is false.
4

### AIEEE 2008

Suppose the cubic $${x^3} - px + q$$ has three distinct real roots
where $$p>0$$ and $$q>0$$. Then which one of the following holds?
A
The cubic has minima at $$\sqrt {{p \over 3}}$$ and maxima at $$-\sqrt {{p \over 3}}$$
B
The cubic has minima at $$-\sqrt {{p \over 3}}$$ and maxima at $$\sqrt {{p \over 3}}$$
C
The cubic has minima at both $$\sqrt {{p \over 3}}$$ and $$-\sqrt {{p \over 3}}$$
D
The cubic has maxima at both $$\sqrt {{p \over 3}}$$ and $$-\sqrt {{p \over 3}}$$

## Explanation

Let $$y = {x^3} - px + q$$

$$\Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$

For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$

$$\Rightarrow x = \pm \sqrt {{p \over 3}}$$

$${{{d^2}y} \over {d{x^2}}} = 6x$$

$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$$ and

$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$$

$$\therefore$$ $$y$$ has ninima at $$x = \sqrt {{p \over 3}}$$

and maxima at $$x = - \sqrt {{p \over 3}}$$

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