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If f(x)= (pcosx+qsinx)(x 2 +ax+b) given ∫f(x)dx from lower limit -π/2 to upper limit π/2 vanishes for all p,q ϵ R, then the value of π 2 +4b+a = ….......... ?

If  f(x)= (pcosx+qsinx)(x2+ax+b)
given  ∫f(x)dx from lower limit -π/2  to upper limit π/2 vanishes for all p,q ϵ R, then 
the value of π2+4b+a = ….......... ?
 

Grade:12

1 Answers

Arju mukherjee
13 Points
2 years ago
f(x)=(pcosx + qsinx)(x2+\alphax+\beta) [Given]
\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}} f(x) dx =\int_{0}^{\frac{\pi }{2}}[ f(x)+f(-x)] dx
\Rightarrow \int_{0}^{\frac{\pi }{2}} [(p\cos x + q\sin x)(x^{2}+\alpha x+\beta ) +(p\cos x - q\sin x )(x^{2}-\alpha x+\beta )] dx
\Rightarrow 2p\int_{0}^{\frac{\pi }{2}} (x^{2}+\beta )\cos x dx+2\alpha q\int_{0}^{\frac{\pi }{2}} x\sin x dx
\Rightarrow 2\left [p(x^{2}+\beta )\sin x + \left ( q\alpha - 2p \right )\left ( \sin x-x\cos x \right ) \right ]_{0}^{\frac{\pi }{2}}
\Rightarrow 2\left [ p(\frac{\pi ^{2}}{4}+\beta ) + q\alpha - 2p \right ]
\Rightarrow 2\left [ \frac{(\pi ^{2}+4\beta )}{4} + q\alpha - 2p \right ]
\Rightarrow 2\left [\left (\frac{(\pi ^{2}+4\beta )}{4}-2 \right )p + q\alpha \right ]
As it is vanishes for all value of p and q, co-efficient of p and q must be equal to 0.
Hence,
\therefore 2\left (\frac{(\pi ^{2}+4\beta )}{4}-2 \right )=0 \Rightarrow (\pi ^{2}+4\beta )=8
\therefore 2\alpha =0\Rightarrow \alpha =0
Hence,
(\pi ^{2}+4\beta +\alpha )=8
 
 

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