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Mathematics
If f(x) = begincases (x log cos x/ log(1+x2)) , x ≠ 0 0 , x = 0 endcases then f(x) is
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Q. If $f(x) = \begin{cases} \frac{x \log \cos x}{\log(1+x^2)} & , x \ne 0 \\ 0 & , x = 0 \end{cases}$ then $f(x)$ is
BITSAT
BITSAT 2016
A
continuous as well as differentiable at x = 0
67%
B
continuous but not differentiable at x = 0
33%
C
differentiable but not continuous at x = 0
0%
D
neither continuous nor differentiable at x = 0
0%
Solution:
$Lf'\left(0\right) =\displaystyle\lim_{h\to0} \frac{f\left(0-h\right)-f\left(0\right)}{-h} =\displaystyle\lim_{h\to0} \frac{-h \log\cos h}{-h \log\left(1+h^{2}\right)} $
$ =\displaystyle\lim_{h\to0} \frac{\log\cos h}{\log\left(1+h^{2}\right)} \left(\frac{0}{0} \text{form}\right) $
$ =\displaystyle\lim_{h\to0} \frac{- \tan h}{2h/\left(1+h^{2}\right)}= - 1/2$
$ Rf'\left(0\right) =\displaystyle\lim_{h\to0} \frac{f\left(0+h\right)-f\left(0\right)}{h} =\displaystyle\lim_{h\to0} \frac{ h \log\cos h}{\log\left(1+h^{2}\right)} $
$ =\displaystyle\lim_{h\to0} \frac{\log\cos h}{\log\left(1+h^{2}\right)} \left(\frac{0}{0} \text{form}\right) $
$ =\displaystyle\lim_{h\to0} \frac{-\tan h}{2h/\left(1+h^{2}\right)}= \frac{-1}{2} $
Since $Lf'(0) = Rf '(0)$,
therefore $f(x)$ is differentiable at $x = 0$
Since differentiability
$\Rightarrow $ continuity, therefore $f(x)$ is continuous at $x = 0$.