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EAMCET Engineering 2010 Solved Paper

Section: Mathematics

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Question : 34 of 160

Marks: +1, -0

If

f: R \rightarrow R

f(x) = \begin{cases} \frac{1+3x^2-\cos 2x}{x^2} & \text{for } x \neq 0 \\ k & \text{for } x=0 \end{cases}

is continuous at

x=0

k

1
5
6
0

Solution:

f(x) = \begin{cases} \frac{1+3x^2-\cos 2x}{x^2} & \text{for } x \neq 0 \\ k & \text{for } x=0 \end{cases}

\;\;\text{ RHL }\;

\;f(0+h)= \lim\limits_{h \to 0} \frac{1+3(0+h)^2-\cos 2(0+h)}{(0+h)^2}

\;\; = \lim\limits_{h \to 0} \frac{1+3 h^2-\cos 2 h}{h^2}

\;\; = \lim\limits_{h \to 0} \frac{1+3 h^2- (1-2 \sin^2 h)}{h^2}

\;\; = \lim\limits_{h \to 0} \frac{1+3 h^2-1+2 \sin^2 h}{h^2}

\;\; = \lim\limits_{h \to 0} \left\{3+2 \left(\frac{\sin^2 h}{h^2}\right)\right\}

\;\; =3+2 \cdot \lim\limits_{h \to 0} \left(\frac{\sin^2}{h}\right)^2

\;\; =3+2 \cdot (1)^2 \left\{ \because \lim\limits_{x \to 0} \frac{\sin x}{x}=1\right\}

\;\; = 3+2=5

LHL

f(0-h)\;=\lim\limits_{h \to 0} \frac{1+3(0-h)^2-\cos 2(0-h)}{(0-h)^2}

\;=\lim\limits_{h \to 0} \frac{1+3 h^2-\cos 2 h}{h^2}

\;=5

Since, the function is continuous at

x=0

, then

\text{LHL}= \text{RHL}=f(0) \because f(0)=k \Rightarrow k=5

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