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Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability1 Mark
MCQ
The set points where the function $f(x)$ given by $\text{f(x)=}|\text{x}-3|\cos\text{x}$ is diffrentiable, is:
  • ✓
    $R$
  • B
    $R - \{3\}$
  • C
    $(0,\infty)$
  • D
    None of these.
✓

Answer

Correct option: A.
$R$
$(\text{LHL}$ at $x = 3) =\lim\limits_{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$(\text{LHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{3-\text{h}-3}$
$(\text{LHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{-\text{h}}$
$(\text{LHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{|3-\text{h}-3|\cos(3-\text{h})-\text{f}(3)}{-\text{h}}$
$(\text{LHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3-\text{h})-0}{-\text{h}}=-\cos3$
$(\text{RHL}$ at $x = 3) =\lim\limits_{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$
$(\text{RHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{3+\text{h}-3}$
$(\text{RHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{\text{h}}$
$(\text{RHL}$ at $x = 3) =\lim\limits_{\text{h}\rightarrow0}\frac{|3+\text{h}-3|\cos(3+\text{h})-\text{f}(3)}{\text{h}}$
$(\text{RHL}$ 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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