कथन (A): int_{frac{pi}{6}}^{frac{pi}{3}} frac | vedclass

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(A) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$गुणधर्म $\int_{a}^

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$A$ सत्य है,$R$ सत्य है और $R$,$A$ की सही व्याख्या है

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(A) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करते हुए,जहाँ $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}} dx}{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}}+(\cos(\frac{\pi}{2}-x))^{\sqrt{2}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\cos x)^{\sqrt{2}} dx}{(\cos x)^{\sqrt{2}}+(\sin x)^{\sqrt{2}}} \dots (2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$I = \frac{\pi}{12}$. अतः,कथन $(A)$ सत्य है।कारण $(R)$ के लिए,गुणधर्म $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$ का उपयोग करते हुए:यहाँ $a = \frac{\pi}{6}$ और $b = \frac{\pi}{3}$ है,इसलिए $\frac{b-a}{2} = \frac{\frac{\pi}{3} - \frac{\pi}{6}}{2} = \frac{\frac{\pi}{6}}{2} = \frac{\pi}{12}$.अतः,कारण $(R)$ सत्य है और यह $(A)$ की सही व्याख्या करता है।

कथन $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$
कारण $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$

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कथन $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$कारण $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$