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AIEEE 2009 Solved Paper

Section: Mathematics

Question : 73 of 89

Marks: +1, -0

Let

f(x)=x\lvert x\rvert

g(x)=\sin x

x=0

x=0

Statement- $-1$ is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement-1.
Statement- 1 is true, Statement-2 is false
Statement- 1 is false, Statement- 2 is true
Statement $-1$ is true, Statement- 2 is true Statement- 2 is a correct explanation for Statement-1

Solution:

Given that

f(x)=x\lvert x\rvert

and

g(x)=\sin x

So that go

f(x)=g(f(x))

=g(x\lvert x\rvert)=\sin x\lvert x\rvert

=\begin{cases} \sin(-x^2) & \text{if } x<0 \\ \sin(x^2) & \text{if } x\ge 0 \end{cases}.

=\begin{cases} -\sin x^2 & \text{if } x<0 \\ \sin x^2 & \text{if } x\ge 0 \end{cases}.

\therefore (g\circ f)'(x)=\begin{cases} -2x\cos x^2 & \text{if } x<0 \\ 2x\cos x^2 & \text{if } x\ge 0 \end{cases}.

Here we observe

L(\text{gof})'(0)=0=R(\text{gof})'(0)

\Rightarrow

f

is differentiable at

x=0

and

(g\circ f)'

is continuous at

x=0

Now

(\text{gof})''(x)=\begin{cases} -2\cos x^2+4x^2\sin x^2 & x<0 \\ 2\cos x^2-4x^2\sin x^2 & x\ge 0 \end{cases}.

Here

L(g\circ f)''(0)=-2

and

R(g\circ f)''(0)=2

L(g\circ f)''(0) \neq R(g\circ f)''(0)

\Rightarrow g\circ f(x)

is not twice differentiable at

x=0

\therefore

Statement - 1 is true but statement

-2

is false.

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