If fx - xk sin1x- x00- x - 0 is differentiable at x-0- then where k is an integer

At x = 0 R.H.D. =lim_h→ 0 f(0+h) f(0)h = lim_h→ 0 h^k sin1h 0h = lim_h→ 0 h^k 1sin1h For this limit to exist, exponent of h must be positive. i.e., k 1 gt; ...

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Byju's Answer

Standard XIII

Mathematics

Differentiability

If fx = xk si...

Question

If $f (x) = ⎧ ⎨ ⎩ \begin{matrix} x^{k} sin (\frac{1}{x}), & x \neq 0 0, & x = 0 \end{matrix}$ is differentiable at $x = 0,$ then (where $k$ is an integer)

k > 0

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k > 1

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k \geq 1

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k \geq 0

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Solution

The correct option is B $k > 1$
At $x = 0$
R.H.D. $= lim h \to 0 \frac{f (0 + h) - f (0)}{h}$
= $lim h \to 0 \frac{h^{k} sin \frac{1}{h} - 0}{h}$
= $lim h \to 0 h^{k - 1} sin \frac{1}{h}$

For this limit to exist, exponent of $h$ must be positive.
i.e., $k - 1 > 0$
$\Rightarrow k > 1$

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If f(x)=⎧⎨⎩xksin(1x),x≠00,x=0 is differentiable at x=0, then (where k is an integer)

The correct option is B k>1At x=0 R.H.D. =limh→0f(0+h)−f(0)h = limh→0hksin1h−0h = limh→0hk−1sin1h For this limit to exist, exponent of h must be positive. i.e., k−1>0 ⇒k>1

If $f (x) = ⎧ ⎨ ⎩ \begin{matrix} x^{k} sin (\frac{1}{x}), & x \neq 0 0, & x = 0 \end{matrix}$ is differentiable at $x = 0,$ then (where $k$ is an integer)

The correct option is B $k > 1$
At $x = 0$
R.H.D. $= lim h \to 0 \frac{f (0 + h) - f (0)}{h}$
= $lim h \to 0 \frac{h^{k} sin \frac{1}{h} - 0}{h}$
= $lim h \to 0 h^{k - 1} sin \frac{1}{h}$

For this limit to exist, exponent of $h$ must be positive.
i.e., $k - 1 > 0$
$\Rightarrow k > 1$