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Let f : R to R be a differentiable fu...

Let ` f : R to R ` be a differentiable function such that `f (0) = 0, f(pi/2) = 3 and f'(0) = 1`.
If ` g(x) = int_(x)^(pi/2)[ f'(t) cosec t-cot t cosec t f (t)] dt" for " x in (0,pi/2]," then " lim_( x to 0) g(x) = `

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The correct Answer is:
2
g(x) = `int_(x)^(pi/2) (f'(t) ` cosec t - cosec t f (t)) dt
`:. G (x) = f(pi/2) cosec pi/2 - f(x) cosec x`
` rArr g(x) = 3 - (f(x))/(sin x) `
` underset( x to 0) lim g(x) = underset( x to 0) lim ((3 sin x - f(x))/( sin x))`
` =underset( x to 0) lim (3 cos x - f'(x))/(cos x) `
` = (3-1)/1 = 2`
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