A) \[f(p,q)=\frac{q}{p+q}f(p-1,q-1)\]
B) \[f(p,q)=\frac{p}{p+q}f(p-1,q-1)\]
C) \[f(p,q)=\frac{p}{p+q}f(p-1,q-1)\]
D) \[f(p,q)=-\frac{q}{p+q}f(p-1,q-1)\]
Correct Answer: B
Solution :
[b] \[f(p,q)=\int_{0}^{\pi /2}{{{\cos }^{p}}x\cos \,\,qx\,\,dx}\] \[=\left[ {{\cos }^{p}}x.\frac{\sin qx}{q} \right]_{0}^{\pi /2}+\int_{0}^{\pi /2}{\frac{p}{q}{{\cos }^{p-1}}x\sin x\sin qxdx}\]\[=0+\frac{p}{q}\int_{0}^{\pi /2}{{{\cos }^{p-1}}x[\cos (q-1)x-\cos \,\,qx\,\,\cos \,\,x]dx}\]\[[\because \cos (q-1)x=\cos qx\cos x+\sin qx\sin x\] \[\therefore \cos (q-1)x-\cos qx\] \[\cos x=\sin qx\sin x)\] \[=\frac{p}{q}f(p-1,q-1)-\frac{p}{q}f(p,q)\] \[\Rightarrow \left( 1+\frac{p}{q} \right)f(p,q)=\frac{p}{q}f(p-1,q-1)\] \[\Rightarrow f(p,q)=\frac{p}{p+q}f(p-1,q-1)\]
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