If the function f ( x ) = { ( 1 + | cos x | ) | cos x | , 0 x 2 , x = 2 cot 6 x e cot 4 x , 2 x is continuous at x = 2 , then 9 + 6 log e + 6 - e 6 is equal to [2023]

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Solution

Q.
If the function

$f (x) =$ ${\begin{matrix} (1 + | \cos x |) \frac{λ}{| \cos x |}, & 0 < x < \frac{π}{2} \\ μ, & x = \frac{π}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{π}{2} < x < π \end{matrix}$

is continuous at $x = \frac{π}{2}$ , then $9 λ + 6 \log_{e} μ + μ^{6} - e^{6 λ}$ is equal to [2023]

1 $2 e^{4} + 8$

2 11

3 8

4 10

Ans.
(4)

Given question is incorrect. It should be $e^{(\frac{\cot 6 x}{\cot 4 x})}$ in place of $\frac{\cot 6 x}{e^{\cot 4 x}} .$

$f (x) =$ ${\begin{matrix} \frac{(1 + | \cos x |) λ}{| \cos x |}, & 0 < x < \frac{π}{2} \\ μ, & x = \frac{π}{2} \\ \frac{\cot 6 x}{e^{\cot 4 x}}, & \frac{π}{2} < x < π \end{matrix}$

Since $f (x)$ is continuous at $x = \frac{π}{2}$

$R.H.L. at x = \frac{π}{2}$

$\lim_{x \to π / 2^{+}} e^{(\frac{\cot 6 x}{\cot 4 x})} = \lim_{x \to π / 2^{+}} e^{(\frac{\cos 6 x}{\sin 6 x} \times \frac{\sin 4 x}{\cos 4 x})} = e^{2 / 3}$

$L.H.L. = \lim_{x \to π / 2} (1 + | \cos x |) \frac{λ}{| \cos x |} = e^{λ}$

Value of the function, $f (π / 2) = μ$ (given)

By the definition of continuity, $e^{2 / 3} = e^{λ} = μ$ $[∵ R.H.L = L.H.L = value of the function]$

$\Rightarrow λ = \frac{2}{3}, μ = e^{2 / 3}$

$∴$ $9 λ + 6 \log_{e} μ + μ^{6} - e^{6 λ}$

$= 9 \times \frac{2}{3} + 6 \times \frac{2}{3} + {(e^{2 / 3})}^{6} - e^{6 \times \frac{2}{3}} = 6 + 4 + e^{4} - e^{4} = 10$

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