Solve this

Question:

If $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 10 x}{x^{2}} & , \quad x<0 \\ a & , \quad x=0 \\ \frac{\sqrt{x}}{\sqrt{625+\sqrt{x}-25}}, & x>0\end{array}\right.$

then the value of a so that f (x) may be continuous at x = 0, is
(a) 25
(b) 50
(c) −25
(d) none of these

Solution:

(b) 50

If $f(x)$ is continuous at $x=0$, then

$\lim _{\mathrm{x} \rightarrow 0^{-}} f(x)=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} f(-h)=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} \frac{(1-\cos (-10 h))}{(-h)^{2}}=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} \frac{(1-\cos (10 h))}{h^{2}}=f(0)$

$\Rightarrow \lim _{h \rightarrow 0} \frac{\left(2 \sin ^{2}(5 h)\right)}{h^{2}}=a$

$\Rightarrow \lim _{h \rightarrow 0} \frac{2 \times 25\left(\sin ^{2}(5 h)\right)}{25 h^{2}}=a$

$\Rightarrow 50 \lim _{h \rightarrow 0} \frac{\left(\sin ^{2}(5 h)\right)}{(5 h)^{2}}=a$

$\Rightarrow 50 \lim _{h \rightarrow 0}\left(\frac{\sin (5 h)}{5 h}\right)^{2}=a$

 

$\Rightarrow a=50$

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