Let be a function given by where . If is continuous at , then is equal to

Question

Let $f : (0, π) \to R$ be a function given by $f (x) =$ ${\begin{matrix} \sin^{2} x, & 0 < x < \frac{π}{2} \\ a - 8, & x = \frac{π}{2} \\ {(1 + | \cot x |)}^{\frac{b}{a} | \tan x |}, & \frac{π}{2} < x < π \end{matrix}$

where $a, b \in Z$ . If $f$ is continuous at $x = \frac{π}{2}$ , then $a^{2} + b^{2}$ is equal to _____

Smart Achievers · Accepted Answer

(81)

$f$ is continuous at $x = \frac{π}{2} \Rightarrow \lim_{x \to π / 2^{-}} f (x) = \lim_{x \to π / 2^{+}} f (x) = f (\frac{π}{2})$

$\Rightarrow \lim_{x \to π / 2^{-}} {(\frac{8}{7})}^{\frac{\tan 8 x}{\tan 7 x}} = \lim_{x \to π / 2^{+}} {(1 + | \cot x |)}^{\frac{b}{a} | \tan x |} = a - 8$

$\Rightarrow {(\frac{8}{7})}^{0} = e^{\lim_{x \to π / 2^{+}} {(1 + \cot x)}^{- 1} \frac{b}{a} \tan x} = a - 8 \Rightarrow 1 = e^{\lim_{x \to π / 2^{+}} \frac{b}{a}} = a - 8$ $\Rightarrow a - 8 = 1,$ $e^{b / a} = 1$

$\Rightarrow a = 9, b = 0,$ $∴$ $a^{2} + b^{2} = 81 + 0 = 81$

Solution

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Solution

Q. Let f:(0,π)→R be a function given by f(x)={sin2x,0<x<π2a-8,x=π2(1+|cotx|)ba|tan x|,π2<x<π where a,b∈Z. If f is continuous at x=π2, then a2+b2 is equal to _____

Ans. (81) f is continuous at x=π2⇒limx→π/2-f(x)=limx→π/2+f(x)=f(π2) ⇒limx→π/2-(87)tan8xtan7x=limx→π/2+(1+|cotx|)ba|tanx|=a-8 ⇒(87)0=elimx→π/2+(1+cotx)-1batanx=a-8⇒1=elimx→π/2+ba=a-8⇒a-8=1, eb/a=1 ⇒a=9, b=0, ∴ a2+b2=81+0=81

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