माना फलन y=f(x) अंतराल (-5,5) तीन बार अवकलनीय | vedclass

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Question

$y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x)$

$ \left.\frac{d y}{d x}ight)_{(1, f(1))}=f^{\prime}(1)=	an \frac{\pi}{6}=\frac{1}{\sqrt{3}} \

Vedclass · Accepted Answer

$26$

Vedclass · Answer

$y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x)$

$ \left.\frac{d y}{d x}ight)_{(1, f(1))}=f^{\prime}(1)=	an \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} $

$ \left.\frac{d y}{d x}ight)_{(3, f(3))}=f^{\prime}(3)=	an \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 $

$ 27 \int_1^3\left(\left(f^{\prime}(t)ight)^2+1ight) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} $

$ I=\int_1^3\left(\left(f^{\prime}(t)ight)^2+1ight) f^{\prime \prime}(t) d t $

$ f(t)=z \Rightarrow f^{\prime}(t) d t=d z $

$ z=f^{\prime}(3)=1$

$z=f^{\prime}(1)=\frac{1}{\sqrt{3}} $

$I=\int_{1 / \sqrt{3}}^1\left(z^2+1ight) d z=\left(\frac{z^3}{3}+zight)_{1 / \sqrt{3}}^1$

$ =\left(\frac{1}{3}+1ight)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}ight)$

$ =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} $

$ \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}ight)=36-10 \sqrt{3} $

$ \alpha=36, \beta=-10$

$ \alpha+\beta=36-10=26$

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