Q.

Let f be a differentiable function defined on [0,π2] such that f(x)>0 and f(x)+∫0xf(t)1-(logef(t))2 dt=e,∀x∈[0,π2]. Then (6logef(π6))2 is equal to _____ .         [2023]


Ans.

(27)

Given, f(x)+∫0xf(t)1-(logef(t))2 dt=e             ...(i)

⇒ f(0)=e

Differentiating (i) both sides, we get

f'(x)+f(x)1-(lnf(x))2=0

Put f(x)=y

⇒dydx=-y1-(lny)2⇒∫dyy1-(lny)2=-∫dx

Put lny=t

⇒∫dt1-t2=-x+c⇒sin-1t=-x+c

⇒sin-1(lny)=-x+c

⇒sin-1(lnf(x))=-x+c;  f(0)=e

⇒π2=C⇒sin-1(ln(f(x)))=-x+π2

⇒sin-1(lnf(π6))=-π6+π2=π3

∴   lnf(π6)=sin(π3)=32

Thus,  (6logef(π6))2=(6×32)2=27