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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

ધારોકે એક વિધેય \(f:(0, \pi) \rightarrow {R}\) એ \(f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.\) જ્યાં \(a, b \in Z\) મુજબ આપેલ છે. જો \(x=\frac{\pi}{2}\) પર \(f\) સતત હોય, તો \(\mathrm{a}^2+\mathrm{b}^2=\) ..........

  1. A \(12\)
  2. B \(81\)
  3. C \(35\)
  4. D \(74\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(81\)

Step-by-step Solution

Detailed explanation

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