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Question: Let f be twice differentiable function such that \(\frac{n^{a}\sin^{2}(n!)}{n + 1}\)and \(\lim_{n \r...

Let f be twice differentiable function such that nasin⁡2(n!)n+1\frac{n^{a}\sin^{2}(n!)}{n + 1}and lim⁡n→∞(0.2)log⁡5(1/4+1/8+1/16+...nterms)\lim_{n \rightarrow \infty}(0.2)^{\log_{\sqrt{5}}(1/4 + 1/8 + 1/16 + ...nterms)}. If lim⁡x→3\lim_{x \rightarrow 3}then lim⁡x→0\lim_{x \rightarrow 0} is equal to

A

22

B

11

C

0

D

None of these

Answer

11

Explanation

Solution

Differentiating the given relation h(x)=[f(x)]2+[g(x)]2h ( x ) = [ f ( x ) ] ^ { 2 } + [ g ( x ) ] ^ { 2 } w.r.t x, we get h′(x)=2f(x)f′(x)+2g(x)g′(x)h ^ { \prime } ( x ) = 2 f ( x ) f ^ { \prime } ( x ) + 2 g ( x ) g ^ { \prime } ( x ) .......(i)

But we are given and f′(x)=g(x)f ^ { \prime } ( x ) = g ( x ) so that

f′(x)=g′(x)f ^ { \prime } ( x ) = g ^ { \prime } ( x )

Then (1) may be re-written as

h′(x)=−2f′′(x)f′(x)+2f′(x)f′′(x)=0h ^ { \prime } ( x ) = - 2 f ^ { \prime \prime } ( x ) f ^ { \prime } ( x ) + 2 f ^ { \prime } ( x ) f ^ { \prime \prime } ( x ) = 0 Thus h′(x)=0h ^ { \prime } ( x ) = 0

Whence by integrating, we get h(x)=h ( x ) = constant = c (say). Hence h(x)=ch ( x ) = c for all x.

In particular, h(5)=ch ( 5 ) = c But we are given h(5)=11h ( 5 ) = 11

It follows that c=11c = 11 and we have h(x)=11h ( x ) = 11 for all x. Therefore, h(10) = 11.