JEE Main 2024 9 April Morning Shift
Exam JEE Main
Year 2024
Date 9 April
Shift Morning Shift
Subject Mathematics
Chapter Continuity and Differentiability JEE Main 2024 · 9 April · Morning Shift · Mathematics · Continuity and Differentiability
Let f:(0, π) arrow R be a function given by f(x)= (\ array cc...
Let
f : ( 0 , π ) → R f:(0, \pi) \rightarrow \mathbf{R} f : ( 0 , π ) → R be a function given by
f ( x ) = f(x)= f ( x ) =
(\left\{
( 8 7 ) tan 8 x tan 7 x , 0 < x < π 2 a − 8 , x = π 2 ( 1 + ∣ cot x ∣ ) b ∣ tan x ∣ , π 2 < x < π \begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+|\cot x|)^{\frac{\mathrm{b}}{}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array} ( 7 8 ) t a n 7 x t a n 8 x , a − 8 , ( 1 + ∣ cot x ∣ ) b ∣ t a n x ∣ , 0 < x < 2 π x = 2 π 2 π < x < π \right.)
where
a , b ∈ Z \mathrm{a}, \mathrm{b} \in \mathbf{Z} a , b ∈ Z . If
f f f is continuous at
x = π 2 x=\frac{\pi}{2} x = 2 π , then
a 2 + b 2 \mathrm{a}^2+\mathrm{b}^2 a 2 + b 2 is equal to
Step-by-step solution
LHL at
x = π 2 lim x → π 2 ( 8 7 ) tan 8 x tan 7 x = ( 8 7 ) 0 = 1 \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 x = 2 π lim x → 2 π ( 7 8 ) t a n 7 x t a n 8 x = ( 7 8 ) 0 = 1 RHL at
x = π 2 \mathrm{x}=\frac{\pi}{2} x = 2 π lim x → π 2 ( 1 + ∣ cot x ∣ ) b lan tan x = e lim x − π 2 ∣ cot x ∣ b a ∣ tan x ∣ = e b a \begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{\operatorname{lan}} \tan x} \\
& =e^{\lim _{x-\frac{\pi}{2}}|\cot x| \frac{b}{a}|\tan x|}=e^{\frac{b}{a}}
\end{aligned} x → 2 π lim ( 1 + ∣ cot x ∣ ) lan b t a n x = e l i m x − 2 π ∣ c o t x ∣ a b ∣ t a n x ∣ = e a b ⇒ 1 = a − 8 = e b a ⇒ a = 9 , b = 0 ⇒ a 2 + b 2 = 81 \begin{aligned} & \Rightarrow 1=a-8=e^{\frac{b}{a}} \\ & \Rightarrow a=9, b=0 \\ & \Rightarrow a^2+b^2=81\end{aligned} ⇒ 1 = a − 8 = e a b ⇒ a = 9 , b = 0 ⇒ a 2 + b 2 = 81 $$
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