Lef f (0, π) rightarrow R be a function given by f ( x )= \ begin matrix ( 8/7 ) frac tan 8x tan 7x , & 0<x<π /2 \a-8, & x=π /2 \\ (1+ cot x ) b/a tanx , & π /2<x<π \\end matrix . Where a , b in Z . If f is continuous at x =π /2 , then a 2 + b 2 is equal to \!\! \!\!

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 9 April, Shift 1 — Question 26

Lef $\mathrm{f}(0, \pi) \rightarrow \mathrm{R}$ be a function given by $f\left( x \right)=\left\{ \begin{matrix}{{\left( \frac{8}{7} \right)}^{\frac{\text{tan}8x}{\text{tan}7x}}}, & 0<x<\frac{\pi }{2} \\a-8, & x=\frac{\pi }{2} \\{{(1+\left| \text{cot}x \right|)}^{\frac{b}{a}|tanx| }}, & \frac{\pi }{2}<x<\pi \\\end{matrix} \right.$ Where $\text{a},\text{b}\in \text{Z}$ .

If f is continuous at $\text{x}=\frac{\pi }{2}$ , then ${{a}^{2}}+{{b}^{2}}$ is equal to $\text{ }\!\!~\!\!\text{ }$

Answer: 81

Numerical answer — enter this value.

Step-by-step solution

LHL at $\mathrm{x}=\frac{\pi}{2}$ $\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}=\left(\frac{8}{7}\right)^{0}=1$

RHL at $\mathrm{x}=\frac{\pi}{2}$

$\lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}$

$\Rightarrow 1=\mathrm{a}-8=\mathrm{e}^{\frac{\mathrm{b}}{\mathrm{a}}}$

$\Rightarrow \mathrm{a}=9, \mathrm{~b}=0$

$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=81$

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity

Let a circle passing through (2,0) have its centre at the point ( h, k) . Let ( x c, y c ) be the point of intersection of the lines 3 x+5…

Answer: 81

Step-by-step solution

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