If \[y=a{{x}^{n+1}}+b{{x}^{-n}}\], then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\] [Karnataka CET 1993]

If \[y=a{{x}^{n+1}}+b{{x}^{-n}}\], then \[{{x}^{2}}\frac{{{d}^{2}}y}{d

1.	If \[y=a{{x}^{n+1}}+b{{x}^{-n}}\], then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\] [Karnataka CET 1993]
A.	\[n\,(n-1)y\]
B.	\[n\,(n+1)y\]
C.	ny
D.	\[{{n}^{2}}y\]
Answer» C. ny

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If \[y=a{{x}^{n+1}}+b{{x}^{-n}}\], then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=\] [Karnataka CET 1993]

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