If the function $f(x)=\left\{\begin{array}{cl}(1+|\cos x|) \frac{\lambda}{|\cos x|}, & 0<x<\frac{\pi}{2} \\ \mu \quad, & x=\frac{\pi}{2} \quad \text { is continuous at } \\ e^{\frac{\cot 6 x}{\cot 4 x}} & , \frac{\pi}{2}<x<\pi\end{array}\right.$ $\mathrm{x}=\frac{\pi}{2}$, then $9 \lambda+6 \log _{\mathrm{e}} \mu+\mu^{6}-\mathrm{e}^{6 \lambda}$ is equal to

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Continuity and Differentiability

If the function

f (x) = ⎩ ⎨ ⎧ (1 + ∣ cos x ∣) \frac{λ}{∣ c o s x ∣}, μ, e^{\frac{c o t 6 x}{c o t 4 x}} 0 < x < \frac{π}{2} x = \frac{π}{2} is continuous at, \frac{π}{2} < x < π

x = \frac{π}{2}

, then

9 λ + 6 lo g_{e} μ + μ^{6} - e^{6 λ}

is equal to

Answer: (D)

Continuity at

x = \frac{π}{2}

demands equal one-sided limits and value. From the left,

(1 + ∣ cos x ∣)^{λ /∣ c o s x ∣} \to e^{λ}

∣ cos x ∣ \to 0

. From the right, with

x = \frac{π}{2} + h

\frac{cot 6 x}{cot 4 x} = \frac{cot 6 h}{cot 4 h} \to \frac{4}{6} = \frac{2}{3}

, so the limit is

e^{2/3}

. Matching,

e^{λ} = e^{2/3} = μ

, so

λ = \frac{2}{3}

and

μ = e^{2/3}

. Then

9 λ + 6 ln μ + μ^{6} - e^{6 λ} = 6 + 4 + e^{4} - e^{4} = 10

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option