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UPSC NDA Math Model Paper 2

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Question : 4 of 120

Marks: +1, -0

The slope of the function

f(x)=\begin{cases} x^{2} \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}

?

1
0
-1
None

Solution:

Here,

f(x)=\begin{cases} x^{2} \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}

Slope

m=\frac{dy}{dx}=\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}

\lim\limits_{x\to 0}\frac{x^{2}\sin\left(\frac{1}{x}\right)-0}{x}

\lim\limits_{x\to 0}x \sin\left(\frac{1}{x}\right)

As we know,

-1 \leq \sin\left(\frac{1}{x}\right) \leq 1

Multiplying by x, we get

\Rightarrow -x \leq x \sin\left(\frac{1}{x}\right) \leq x

\Rightarrow \lim\limits_{x\to 0}(-x) \leq \lim\limits_{x\to 0}x \sin\left(\frac{1}{x}\right) \leq \lim\limits_{x\to 0}x

\Rightarrow 0 \leq \lim\limits_{x\to 0}x^{2}\sin\left(\frac{1}{x}\right) \leq 0

\therefore \lim\limits_{x\to 0}x^{2}\sin\left(\frac{1}{x}\right)=0

Hence, option (2) is correct.

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