JEE Main 2023 — Mathematics Calculus

Given,

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

Now, solving $L.H.L.$L.H.L. at $x=\frac{\pi }{2}$x=2π​ we get,

$$\underset{x\rightarrow \frac{{\pi }^{+}}{2}}{\mathrm{lim}}{e}^{\frac{\mathrm{cot}6x}{\mathrm{cot}4x}}=\underset{x\rightarrow \frac{{\pi }^{+}}{2}}{\mathrm{lim}}{e}^{\frac{\mathrm{sin}4x\cdot \mathrm{cos}6x}{\mathrm{sin}6x\cdot \mathrm{cos}4x}}={e}^{2/3}$$x→2π+​lim​ecot4xcot6x​=x→2π+​lim​esin6x⋅cos4xsin4x⋅cos6x​=e2/3

Similarly, on simplification for $R.H.L.$R.H.L. we get

$$\underset{x\rightarrow {\frac{\pi }{2}}^{-}}{\mathrm{lim}}{(1+|\mathrm{cos}x|)}^{\frac{\lambda }{|\mathrm{cos}x|}}={e}^{\underset{x\rightarrow {\frac{\pi }{2}}^{-}}{\mathrm{lim}}\lambda }={e}^{\lambda }$$x→2π​−lim​(1+∣cosx∣)∣cosx∣λ​=ex→2π​−lim​λ=eλ

$$\because f(\frac{\pi }{2})=\mu$$∵f(2π​)=μ

For continuous function,

$$f({\frac{\pi }{2}}^{-})=f(\frac{\pi }{2})=f({\frac{\pi }{2}}^{+})$$f(2π​−)=f(2π​)=f(2π​+)

$$\Rightarrow {e}^{\frac{2}{3}}={e}^{\lambda }=\mu$$⇒e32​=eλ=μ

$$\Rightarrow \lambda =\frac{2}{3},\mu ={e}^{\frac{2}{3}}$$⇒λ=32​,μ=e32​

Now,

$$9\lambda +6{\mathrm{log}}_{e}\mu +{\mu }^{6}-{e}^{6\lambda }$$9λ+6loge​μ+μ6−e6λ

$$=9(\frac{2}{3})+6(\frac{2}{3})+{({e}^{\frac{2}{3}})}^{6}-{e}^{6(\frac{2}{3})}$$=9(32​)+6(32​)+(e32​)6−e6(32​)

$$=6+4+{e}^{4}-{e}^{4}$$=6+4+e4−e4

$$=10$$=10