Let f:(0, \pi) arrow R be a function given by f(x)= beginarraycc(\frac87)^\frac\tan 8 x\tan 7 x, & 0\lt x\lt \frac\pi2 a-8, & x=\frac\pi2 (1+|\cot x|)^\fracba|\tan x|, & \frac\pi2 \lt x \lt \pi\endarray. where a, b \in Z . If f is continuous at x=\frac\pi2 , then a^2+b^2 is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q53 with Solution

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Correct answer: 81

For continuity at $x=\frac{\pi}{2}$ , we need $\lim_{x\to (\pi/2)^-} f(x)=f\left(\frac{\pi}{2}\right)=\lim_{x\to (\pi/2)^+} f(x).$ Given $f\left(\frac{\pi}{2}\right)=a-8.$
Compute the left-hand limit: $\lim_{x\to (\pi/2)^-}\left(\frac87\right)^{\frac{\tan 8x}{\tan 7x}}.$ Let $x=\frac\pi2-h$ with $h\to 0^+$ . Then $8x=4\pi-8h \implies \tan 8x=\tan(-8h)\sim -8h,$

\implies \tan 7x=\tan\left(\frac\pi2-7h\right)=\cot 7h\sim \frac{1}{7h}.$$ Hence $$\frac{\tan 8x}{\tan 7x}\sim \frac{-8h}{1/(7h)}=-56h^2\to 0.$$ Therefore $$\lim_{x\to (\pi/2)^-} f(x)=\left(\frac87\right)^0=1.$$ So continuity gives $$a-8=1 \implies a=9.$$ 3. Compute the right-hand limit: $$\lim_{x\to (\pi/2)^+}(1+|\cot x|)^{\frac ba |\tan x|}.$$ Let $x=\frac\pi2+h$ with $h\to 0^+$. Then $$|\cot x|=|\cot(\tfrac\pi2+h)|=|- an h|\sim h,$$ $$|\tan x|=|\tan(\tfrac\pi2+h)|=| -\cot h|\sim \frac1h.$$ So the expression becomes of the standard form $$(1+h)^{\frac ba \cdot \frac1h} \to e^{b/a}.$$ Thus $$\lim_{x\to (\pi/2)^+} f(x)=e^{b/a}.$$ Continuity requires $$e^{b/a}=a-8=1.$$ Hence $$\frac ba=0 \implies b=0.$$ Since $a=9$, we get $$a^2+b^2=9^2+0^2=81.$$ 4. Final answer: $$\boxed{81}$$

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Correct answer: 81

For continuity at $x=\frac{\pi}{2}$ , we need $\lim_{x\to (\pi/2)^-} f(x)=f\left(\frac{\pi}{2}\right)=\lim_{x\to (\pi/2)^+} f(x).$ Given $f\left(\frac{\pi}{2}\right)=a-8.$
Compute the left-hand limit: $\lim_{x\to (\pi/2)^-}\left(\frac87\right)^{\frac{\tan 8x}{\tan 7x}}.$ Let $x=\frac\pi2-h$ with $h\to 0^+$ . Then $8x=4\pi-8h \implies \tan 8x=\tan(-8h)\sim -8h,$

\implies \tan 7x=\tan\left(\frac\pi2-7h\right)=\cot 7h\sim \frac{1}{7h}.$$ Hence $$\frac{\tan 8x}{\tan 7x}\sim \frac{-8h}{1/(7h)}=-56h^2\to 0.$$ Therefore $$\lim_{x\to (\pi/2)^-} f(x)=\left(\frac87\right)^0=1.$$ So continuity gives $$a-8=1 \implies a=9.$$ 3. Compute the right-hand limit: $$\lim_{x\to (\pi/2)^+}(1+|\cot x|)^{\frac ba |\tan x|}.$$ Let $x=\frac\pi2+h$ with $h\to 0^+$. Then $$|\cot x|=|\cot(\tfrac\pi2+h)|=|- an h|\sim h,$$ $$|\tan x|=|\tan(\tfrac\pi2+h)|=| -\cot h|\sim \frac1h.$$ So the expression becomes of the standard form $$(1+h)^{\frac ba \cdot \frac1h} \to e^{b/a}.$$ Thus $$\lim_{x\to (\pi/2)^+} f(x)=e^{b/a}.$$ Continuity requires $$e^{b/a}=a-8=1.$$ Hence $$\frac ba=0 \implies b=0.$$ Since $a=9$, we get $$a^2+b^2=9^2+0^2=81.$$ 4. Final answer: $$\boxed{81}$$