The set points where the function f(x) given by f(x)=|x-3| is diffrentiable, is: 1. R 2. R - 3 3. (0, ) 4. None of these.

1. R - 3 Solution: (LHL at x = 3) = _ x 3^ - f(x)-f(3) x-3 (LHL at x = 3) = _ h 0 f(3-h)-f(3) 3-h-3 (LHL at x = 3) = _ h 0 f(3-h)-f(3) -h (LHL at x = 3) = _ h 0 |3-h-3| (3-h)-f(3) -h (LHL at x = 3) = _ h 0 h (3-h)-0 -h =- 3 (RHL at x = 3) = _ x 3^ + f(x)-f(3) x-3 (RHL at x = 3) = _ h 0 f(3+h)-f(3) 3+h-3 (RHL at x = 3) = _ h 0 f(3+h)-f(3) h (RHL at x = 3) = _ h 0 |3+h-3| (3+h)-f(3) h (RHL at x = 3) = _ h 0 h (3+h)-0 h = 3 So, f(x) is not diffrentiable at x = 3. Also,f(x) is diffrentiable at all other points because both modulus and cosine function are differentiable and the product of two differentiable function is differentiable.

The set points where the function f(x) given by f(x)=|x-3| is dif…

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Question

The set points where the function f(x) given by $\text{f(x)=}|\text{x}-3|\cos\text{x}$ is diffrentiable, is:

R
R - {3}
$(0,\infty)$
None of these.

✓

Answer

R - {3}

Solution:
(LHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{3-\text{h}-3}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{-\text{h}}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3-\text{h}-3|\cos(3-\text{h})-\text{f}(3)}{-\text{h}}$
(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3-\text{h})-0}{-\text{h}}=-\cos3$
(RHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{3+\text{h}-3}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{\text{h}}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3+\text{h}-3|\cos(3+\text{h})-\text{f}(3)}{\text{h}}$
(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3+\text{h})-0}{\text{h}}=\cos3$
So, f(x) is not diffrentiable at x = 3.
Also,f(x) is diffrentiable at all other points because both modulus and cosine function are differentiable and the product of two differentiable function is differentiable.

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