IIT-JEE 2012 · Paper 1 · Mathematics Q42

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MathematicsSingle correct+3 / -1Question 42 of 60

Let $f(x)=\begin{cases}x^2\left|\cos\dfrac\pi x\right|,&x\ne0\\0,&x=0\end{cases}$ , $x\in\mathbb R$ , then $f$ is

differentiable both at

x=0

and at

x=2

differentiable at

x=0

but not differentiable at

x=2

not differentiable at

x=0

but differentiable at

x=2

differentiable neither at

x=0

nor at

x=2

Answer and worked solution

Official answer: B

At $x=0$ .
$\frac{f(h)-f(0)}h=h\left|\cos\frac\pi h\right|\to0,$
since $|\cos|\le1$ . Differentiable, with $f'(0)=0$ .

At $x=2$ . $\cos\frac\pi x$ vanishes at $x=2$ and changes sign there (its derivative $\frac\pi{x^2}\sin\frac\pi x=\frac\pi4\ne0$ ). So near 2,
$f(x)=x^2\left|\cos\frac\pi x\right|$
has one-sided derivatives $\pm4\cdot\frac\pi4=\pm\pi$ . Not differentiable.

(B)

Topics:Differentiability

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IIT-JEE 2012 · Paper 1, Mathematics question 42