Let `f(x)""=""x|x|""and""g(x)""=""sinx in 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`

(1) Statement1 is true, Statement2 is true, Statement2 is a correct explanation for statement1
(2) Statement1 is true, Statement2 is true; Statement2 is not a correct explanation for statement1.
(3) Statement1 is true, statement2 is false.
(4) Statement1 is false, Statement2 is true

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

We have `f(x)=x|x| and g(x)=sin x`
`therefore gof(x)=sin (x|x|)={{:(,-sin x^(2),x lt 0),(,sin x^(2),x ge 0):}`
`therefore ("LHD of gof at x=0")=underset(x to 0^(-))lim (gof (x)-gof(0))/(x-0)`
`Rightarrow ("LHD of gof at x=0")=underset(x to 0^(-))lim (-sin x^(2))/(x)=underset(x to 0)lim x ((sinx^(2))/(x^(2)))=0` and `Rightarrow ("RHD of gof at x=0")=underset(x to 0^(+))lim (gof(x)-gof(0))/(x-0)`
`Rightarrow ("RHD of gof at x=0")=underset(x to 0^(+))lim (sin x^(2))/(x)=-underset(x to 0)lim x((sinx^(2))/(x^(2)))=0`
So, gof is differentiable at x=0 such that `(gof)'(x)={{:(,-2x cos x^(2),xlt 0),(,2x cos x^(2),x ge 0):}`
Clearly, `underset(x to 0^(-))lim (gof)'(x)=underset(x to 0^(+))lim (gof)'(x)=(gof)'(0)`
So, (gof)' is continuous at x=0.
Hence, statement-1 is true.
`"Let "phi(x)=(gof)'(x)="Then"`
`("LHD of "phi(x)" at x"=0)=underset(x to 0^(-))lim (phi(x)-phi(0))/(x-0)=underset(x to 0^(-))lim (-2x cos x^(2))/(x)=2` `and ("RHD of "phi(x)"at x=0")=underset(x to 0^(+))lim (phi(x)-phi(0))/(x-0)=underset(x to 0^(+))lim (2x cos x^(2))/(x)=2`
`therefore ` (gof)' (x) is not differentiable at x=0. ie. gof is not twice differentiable at x=0.
So, statement-2 is not true.