Let f( x ) = x| x | and g( x ) = \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 .

Statement-1 is true, Statement-2 is false

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

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

Limits Continuity and Differentiability PYQ — JEE Main 2009 · Shift 0 · Q37 with Solution

View written solutionFree

Correct answer: B

Form the composite function

Given $f(x)=x|x|,$ $g(x)=\sin x.$ So $(g\circ f)(x)=g(f(x))=\sin(x|x|).$

Let $h(x)=\sin(x|x|).$

We must examine differentiability and second differentiability at $x=0$ .

Write $x|x|$ piecewise

Since

x,& x\ge 0,\\ -x,& x<0, \end{cases}$$ we get $$x|x|=\begin{cases} x^2,& x\ge 0,\\ -x^2,& x<0. \end{cases}$$ Hence $$h(x)=\sin(x|x|)=\begin{cases} \sin(x^2),& x\ge 0,\\ \sin(-x^2)=-\sin(x^2),& x<0. \end{cases}$$ --- 3. **Check differentiability at $x=0$** We use the definition: $$h'(0)=\lim_{x\to 0}\frac{h(x)-h(0)}{x}.$$ Now $$h(0)=\sin 0=0,$$ so $$h'(0)=\lim_{x\to 0}\frac{\sin(x|x|)}{x}.$$ Since $x|x|=O(x^2)$, for small $x$, $$\sin(x|x|)\sim x|x|.$$ Thus $$\frac{\sin(x|x|)}{x}\sim \frac{x|x|}{x}=|x|\to 0.$$ Therefore $$h'(0)=0.$$ So $g\circ f$ is differentiable at $0$. --- 4. **Find derivative for $x\neq 0$ and test continuity at $0$** For $x\ne 0$, by chain rule, $$h'(x)=\cos(x|x|)\cdot \frac{d}{dx}(x|x|).$$ Now $$\frac{d}{dx}(x|x|)=\begin{cases} 2x,& x>0,\\ -2x,& x<0, \end{cases}$$ which is simply $$\frac{d}{dx}(x|x|)=2|x| \quad (x\ne 0).$$ So $$h'(x)=2|x|\cos(x|x|), \quad x\ne 0.$$ Also $h'(0)=0$. Now check continuity of $h'$ at $0$: $$\lim_{x\to 0} h'(x)=\lim_{x\to 0} 2|x|\cos(x|x|)=0=h'(0).$$ Hence $h'$ is continuous at $x=0$. Therefore **Statement-1 is true**. --- 5. **Check twice differentiability at $x=0$** We need to see whether $$h''(0)=\lim_{x\to 0}\frac{h'(x)-h'(0)}{x}$$ exists. Since $h'(0)=0$, $$h''(0)=\lim_{x\to 0}\frac{2|x|\cos(x|x|)}{x}.$$ Now examine one-sided limits: - For $x>0$: $$\frac{2|x|\cos(x|x|)}{x}=2\cos(x^2)\to 2.$$ - For $x<0$: $$\frac{2|x|\cos(x|x|)}{x}=-2\cos(-x^2)=-2\cos(x^2)\to -2.$$ Left-hand and right-hand limits are different, so $h''(0)$ does **not** exist. Therefore **Statement-2 is false**. --- 6. **Conclusion** - Statement-1: **True** - Statement-2: **False** So the correct option is $$\boxed{\text{B}}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** They agree.

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View written solutionFree

Correct answer: B

Form the composite function

Given $f(x)=x|x|,$ $g(x)=\sin x.$ So $(g\circ f)(x)=g(f(x))=\sin(x|x|).$

Let $h(x)=\sin(x|x|).$

We must examine differentiability and second differentiability at $x=0$ .

Write $x|x|$ piecewise

Since

x,& x\ge 0,\\ -x,& x<0, \end{cases}$$ we get $$x|x|=\begin{cases} x^2,& x\ge 0,\\ -x^2,& x<0. \end{cases}$$ Hence $$h(x)=\sin(x|x|)=\begin{cases} \sin(x^2),& x\ge 0,\\ \sin(-x^2)=-\sin(x^2),& x<0. \end{cases}$$ --- 3. **Check differentiability at $x=0$** We use the definition: $$h'(0)=\lim_{x\to 0}\frac{h(x)-h(0)}{x}.$$ Now $$h(0)=\sin 0=0,$$ so $$h'(0)=\lim_{x\to 0}\frac{\sin(x|x|)}{x}.$$ Since $x|x|=O(x^2)$, for small $x$, $$\sin(x|x|)\sim x|x|.$$ Thus $$\frac{\sin(x|x|)}{x}\sim \frac{x|x|}{x}=|x|\to 0.$$ Therefore $$h'(0)=0.$$ So $g\circ f$ is differentiable at $0$. --- 4. **Find derivative for $x\neq 0$ and test continuity at $0$** For $x\ne 0$, by chain rule, $$h'(x)=\cos(x|x|)\cdot \frac{d}{dx}(x|x|).$$ Now $$\frac{d}{dx}(x|x|)=\begin{cases} 2x,& x>0,\\ -2x,& x<0, \end{cases}$$ which is simply $$\frac{d}{dx}(x|x|)=2|x| \quad (x\ne 0).$$ So $$h'(x)=2|x|\cos(x|x|), \quad x\ne 0.$$ Also $h'(0)=0$. Now check continuity of $h'$ at $0$: $$\lim_{x\to 0} h'(x)=\lim_{x\to 0} 2|x|\cos(x|x|)=0=h'(0).$$ Hence $h'$ is continuous at $x=0$. Therefore **Statement-1 is true**. --- 5. **Check twice differentiability at $x=0$** We need to see whether $$h''(0)=\lim_{x\to 0}\frac{h'(x)-h'(0)}{x}$$ exists. Since $h'(0)=0$, $$h''(0)=\lim_{x\to 0}\frac{2|x|\cos(x|x|)}{x}.$$ Now examine one-sided limits: - For $x>0$: $$\frac{2|x|\cos(x|x|)}{x}=2\cos(x^2)\to 2.$$ - For $x<0$: $$\frac{2|x|\cos(x|x|)}{x}=-2\cos(-x^2)=-2\cos(x^2)\to -2.$$ Left-hand and right-hand limits are different, so $h''(0)$ does **not** exist. Therefore **Statement-2 is false**. --- 6. **Conclusion** - Statement-1: **True** - Statement-2: **False** So the correct option is $$\boxed{\text{B}}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** They agree.

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