Q.

Let f(x)=limr→x{2r2[(f(r))2–f(x)f(r)]r2–x2–r3ef(r)r} be differentiable in (–∞,0)∪(0,∞) and f(1) = 1. Then the value of ea, such that f(a) = 0, is equal to __________.          [2024]


Ans.

(2)

f(x)=limr→x{2r2((f(r))2–f(x)f(r))r2–x2–r3ef(r)/r}

Using L'Hospital rule, we have

f(x)=limr→x{2r2[2f(r)·f'(r)–f(x)·f'(r)]+[f(r)2–f(x)·f(r)]·4r2r–r3ef(r)r}

=2x2[2f(x)·f'(x)–f(x)·f'(x)]+02x–x3ef(x)x

∴ f(x)=xf(x)·f'(x)–x3·ef(x)x

Let y = f(x)

y2=xy·dydx–x3eyx ⇒ dydx=y2+x3ey/xxy

Put y=vx ⇒ dydx=v+xdvdx

⇒ v2x2+x3evxxx·vx=v+xdvdx ⇒ v2+xevv–v=xdvdx

⇒ evv=dvdx ⇒ dx=ve–vdv

⇒ ∫dx=∫v·e–vdv ⇒ x+c=–e–v(v+1)

⇒ x+c=–e–y/x(yx+1)          ... (i)

∵ y(1)=1

⇒ 1+c=–e–1(2)

⇒ c=–1–2e          ... (ii)

Now, f(a) = 0

∴ a+c=–e–0/a(0a+1)          [Using (i)]

a + c = –1

a=–1–c=–1+1+2e=2e          [Using (ii)]

∴ ea = 2.