Q74 / 160 · AP EAMCET 2010 · Mathematics

If $f: R \rightarrow R$ defined by $f(x)=\left\{\begin{array}{cc}\frac{1+3 x^2-\cos 2 x}{x^2}, & \text { for } x \neq 0 \\k & , \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then $k$ is equal to

  • A
    1
  • Correct answer
    B
    5
  • C
    6
  • D
    0

Solution

$f(x)=\left\{\begin{array}{rr}\frac{1+3 x^2-\cos 2 x}{x^2}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.$ RHL $f(0+h)=\lim _{h \rightarrow 0} \frac{1+3(0+h)^2-\cos 2(0+h)}{(0+h)^2}$ $=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$ $=\lim _{h \rightarrow 0} \frac{1+3 h^2-\left(1-2 \sin ^2 h\right)}{h^2}$ $=\lim _{h \rightarrow 0} \frac{1+3 h^2-1+2 \sin ^2 h}{h^2}$ $=\lim _{h \rightarrow 0}\left\{3+2\left(\frac{\sin ^2 h}{h^2}\right)\right\}$ $=3+2 \cdot \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^2$ $=3+2 \cdot(1)^2 \quad\left\{\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right\}$ $=3+2=5$ LHL $f(0-h)=\lim _{h \rightarrow 0} \frac{1+3(0-h)^2-\cos 2(0-h)}{(0-h)^2}$ $=\lim _{h \rightarrow 0} \frac{1+3 h^2-\cos 2 h}{h^2}$ $=5$ Since, the function is continuous at $x=0$, then LHL $=$ RHL $=f(0) \because f(0)=k \Rightarrow k=5$

Hence, the correct answer is option: 'B'.

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