Let f : R→R be a differentiable function such that f0=0, f(π)/2=3 and f^(′)0=1. If gx=∫x(π)/2 f^(′)tcosect-cott cosect ftdt , for x∈0,(π)/2 then…

Question

Let f : R→R be a differentiable function such that f0=0, f(π)/2=3 and f^(′)0=1. If gx=∫x(π)/2 f^(′)tcosect-cott cosect ftdt , for x∈0,(π)/2 then limx→0⁡gx=

MARKS Content Team · Accepted Answer

Correct answer: 2. g(x)= ∫x(π)/2[f^(')(t)cosec t−f(t)cosec t cot t]dt =f(t)cosectx(π)/2 =f(π)/2cosec (π)/2-fx/(sin⁡x)=3-fx/(sin⁡x) ∴ limx→0⁡gx=3-limx→0⁡fx/(sin⁡x)=limx→0⁡(f^(')x)/(cos⁡x); as f^(')0=1 ⇒ limx→0⁡gx=3-1=2

Let $f : R \to R$ be a differentiable function such that $f (0) = 0, f (\frac{π}{2}) = 3$ and $f^{'} (0) = 1 .$ If $g (x) = \int_{x}^{\frac{π}{2}} [f^{'} (t) c o s e c t - \cot t c o s e c t f (t)] d t$ , for $x \in (0, \frac{π}{2}]$ then $\lim_{x \to 0} g (x) =$

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Relevant Questions

Let f : R→R be a differentiable function such that f0=0, fπ2=3 and f′0=1. If gx=∫xπ2 f′tcosect-cott cosect ftdt , for x∈0,π2 then limx→0⁡gx=

Solution

Relevant Questions

Let $f : R \to R$ be a differentiable function such that $f (0) = 0, f (\frac{π}{2}) = 3$ and $f^{'} (0) = 1 .$ If $g (x) = \int_{x}^{\frac{π}{2}} [f^{'} (t) c o s e c t - \cot t c o s e c t f (t)] d t$ , for $x \in (0, \frac{π}{2}]$ then $\lim_{x \to 0} g (x) =$