Let a differentiable function f satisfy f(x)+ _3^x f(t)tdt= x+1,\ x 3. Then 12f(8) is equal to

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Correct answer: C. (c) Solution Differentiate w.r.t x f (x )+ f (x )x= 12 x+1 I.f =e^ 1x dx=e^ lnx=x xf (x )= x2 x+1 dx x+1=t^2 = t^2-1 `2t2 tdt xf (x )= t^33-t+c xf (x )= (x+1)^ 323- x+1+c Also putting x=3 in given equation f (x )+0= 4 f (x )=2 ⇒C=8- 83= 163 f (x )= (x+1)^ 323+ x+1+ 163x f (8 )= 9-3+ 1638= 3424 ⇒12f (8x )=17

$Let a differentiable function f satisfy f (x) + \int_{3}^{x} \frac{f ( t )}{t} d t = x + 1, x \geq 3 . Then 12 f (8) is equal to$

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