A) \[y'\left( \frac{\pi }{4} \right)+y'\left( \frac{-\pi }{4} \right)=-\sqrt{2}\]
B) \[y'\left( \frac{\pi }{4} \right)-y'\left( \frac{-\pi }{4} \right)=\pi -\sqrt{2}\]
C) \[y\left( \frac{\pi }{4} \right)-y\left( -\frac{\pi }{4} \right)=\sqrt{2}\]
D) \[y\left( \frac{\pi }{4} \right)+y\left( -\frac{\pi }{4} \right)=\frac{{{\pi }^{2}}}{2}+2\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}+y(tan\,x)=2x+{{x}^{2}}\tan x\] \[I.F={{e}^{\int_{{}}^{{}}{\tan xdx}}}={{e}^{\ln .\sec x}}=\sec x\] \[\therefore \]\[y.\sec x=\int_{{}}^{{}}{\left( 2x+{{x}^{2}}\tan x \right)}\sec x.dx\] \[=\int_{{}}^{{}}{2x\sec xdx}+\int_{{}}^{{}}{{{x}^{2}}}\left( \sec x.\tan x \right)dx\] \[y\sec x={{x}^{2}}\sec x+\lambda \] \[\Rightarrow y={{x}^{2}}+\lambda \cos x\] \[y(0)=0+\lambda =1\] \[\Rightarrow \lambda =1\] \[y={{x}^{2}}+\cos x\] \[y\left( \frac{\pi }{4} \right)=\frac{{{\pi }^{2}}}{16}+\frac{1}{\sqrt{2}}\] \[y\left( -\frac{\pi }{4} \right)=\frac{{{\pi }^{2}}}{16}+\frac{1}{\sqrt{2}}\] \[y'(x)=2x-sinx\] \[y'\left( \frac{\pi }{4} \right)=\frac{\pi }{2}-\frac{1}{\sqrt{2}}\] \[y'\left( \frac{-\pi }{4} \right)=\frac{-\pi }{2}+\frac{1}{\sqrt{2}}\] \[y'\left( \frac{\pi }{4} \right)-y'\left( \frac{-\pi }{4} \right)=\pi -\sqrt{2}\]
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