If the function f(x) = \ (1 + | cos x|)^(λ)/(| cos x|), 0 < x < (π)/2 μ, x = (π)/2 e^((cot 6x)/(cot 4x)), (π)/2 < x < π. is continuous at x = (π)/2, then 9λ + 6 log e μ + μ ⁶ - e^(6λ) is equal to

Limit, Continuity and Differentiability: JEE Main 2023 Mathematics question

JEE Main 2023 · 25 Jan, Evening shift · Q4159

JEE Main 202325 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function

f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.

is continuous at

x = {\pi \over 2}

, then

9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}

is equal to

JEE Main 2023 · 25 Jan, Evening shift · Q4159

JEE Main 202325 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002

If the function

f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.

is continuous at

x = {\pi \over 2}

, then

9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}

is equal to