Discussion Forum

1)

Let f:R→ R be a differentable function such that f(0)=0,  $f(\frac{\pi}{2})=3$ and f ' (0)=1

  if $g(x)=\int_{0}^{\frac{\pi}{2}} f'(t) cosec$ $ t f(t)-\cot $ $ t cosec$ $ t f(t)] dt$

  for   $x\epsilon(0,\frac{\pi}{2}]$, then $\lim_{x \rightarrow0 }g(x)=$

Answer:

Option B

Explanation:

$g(x)=\int_{x}^{\frac{\pi}{2}} f'(t) cosec$ $ t f(t)-\cot $ $ t cosec$ $ t f(t)] dt$

$g(x)= f(\frac{\pi}{2}) cosec $ $\frac{\pi}{2}-f(x) cosec$ $ x$

$\Rightarrow$   $g(x)=3-\frac{f(x)}{\sin x}$

$\lim_{x \rightarrow 0}g(x)=\lim_{x \rightarrow 0}(\frac{3\sin  x-f(x)}{\sin x})$

$=\lim_{x \rightarrow 0}\frac{3\cos x-f'(x)}{\cos x}$

$\frac{3-1}{1}=2$