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TS EAMCET 9 May 2024 Shift 1 Solved Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

If the function

f(x)=\begin{cases} \frac{\cos a x-\cos 9 x}{x^2} &\text{if } x \neq 0 \\ 16 &\text{if } x=0 \end{cases}.

[TG EAPCET 9 May 2024 Shift 1]

$\;\pm 8$
$\;\pm 7$
$\;\pm 6$
$\;\pm 5$

Solution:

We have,

f(x)=\begin{cases} \frac{\cos a x-\cos 9 x}{x^2} &\text{if } x \neq 0 \\ 16 &\text{if } x=0 \end{cases}

Now,

\lim\limits_{x\to 0}f(x)=16

[ \because f(x)

is continuous at

x=0]

\lim\limits_{x\to 0}\;\frac{\cos a x-\cos 9 x}{x^2}=16

Take LHS

HS

\;\lim\limits_{x\to 0}\;\frac{\cos a x-\cos 9 x}{x^2}

\;=\lim\limits_{x\to 0}\;\frac{-a \sin a x+9 \sin 9 x}{2x}

[using L.' Hospital's rule]

\;=\lim\limits_{x\to 0}\;\frac{-a^2 \cos a x+81 \cos 9 x}{2}

\;=\;\frac{81-a^2}{2}=16 \;\; \;\text{[from, Eq. (i)]}\;

81-a^2=32 \Rightarrow a^2=49 \Rightarrow a=\pm 7

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