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ƒ(x) = cosx – ∫(x - t)f(t) x ∈[0,x] dt, then ƒ'' (x) + ƒ(x) is equal to

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ƒ(x) = cosx – ∫(x - t)f(t) x ∈[0,x] dt,  then ƒ'' (x) + ƒ(x) is equal to

(a) – cosx 

(b) – sinx 

(c)  0 

(d)  none of these

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1 Answer

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Best answer

Correct option  (a) – cosx

Explanation:

∴ f''(x) = -cos x - (f(x)1 - 0)

⇒ f''(x) + f(x) = -cosx

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