ધારો કે {x} એ x નો અપૂર્ણાંક ભાગ દર્શાવે છે અ | vedclass

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(A) જમણી બાજુની લક્ષ $(R)$ માટે: $\lim _{x \rightarrow 0^{+}} f(x) = \lim _{h \rightarrow 0^{+}} f(h) = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) જમણી બાજુની લક્ષ $(R)$ માટે: $\lim _{x ightarrow 0^{+}} f(x) = \lim _{h ightarrow 0^{+}} f(h) = \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2) \sin ^{-1}(1-h)}{h(1-h^2)}$ $= \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2)}{h} \cdot \frac{\sin ^{-1}(1)}{1} = \frac{\pi}{2} \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2)}{h}$. ધારો કે $\cos ^{-1}(1-h^2) = 	heta$,તો $\cos 	heta = 1-h^2$,તેથી $h = \sqrt{1-\cos 	heta} = \sqrt{2} \sin(	heta/2)$. જ્યારે $h ightarrow 0^{+}$,ત્યારે $	heta ightarrow 0^{+}$. $R = \frac{\pi}{2} \lim _{	heta ightarrow 0^{+}} \frac{	heta}{\sqrt{2} \sin(	heta/2)} = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2} \cdot (1/2)} = \frac{\pi}{\sqrt{2}}$. તેથી,$R^2 = \frac{\pi^2}{2}$. ડાબી બાજુની લક્ષ $(L)$ માટે: $\lim _{x ightarrow 0^{-}} f(x) = \lim _{h ightarrow 0^{+}} f(-h) = \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(1-(1-h)^2) \sin ^{-1}(1-(1-h))}{(1-h)-(1-h)^3} = \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(2h-h^2) \sin ^{-1}(h)}{(1-h)(1-(1-h)^2)} = \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(2h-h^2) \sin ^{-1}(h)}{(1-h)(2h-h^2)} = \lim _{h ightarrow 0^{+}} \frac{\cos ^{-1}(0) \cdot h}{1 \cdot 2h} = \frac{\pi/2}{2} = \frac{\pi}{4}$. તેથી,$L^2 = \frac{\pi^2}{16}$. અંતે,$\frac{32}{\pi^2} (L^2 + R^2) = \frac{32}{\pi^2} \left(\frac{\pi^2}{16} + \frac{\pi^2}{2}ight) = \frac{32}{\pi^2} \left(\frac{\pi^2 + 8\pi^2}{16}ight) = 2 \cdot 9 = 18$.

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