Let $\mathrm{f}:(0, \infty) \rightarrow(0, \infty)$ be a differentiable function such that $\mathrm{f}(1)=\mathrm{e}$ and $\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$. If $f(x)=1$, then $x$ is equal to:

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Continuity and Differentiability

Let

f : (0, \infty) \to (0, \infty)

be a differentiable function such that

f (1) = e

and

lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0

. If

f (x) = 1

, then

x

is equal to:

Answer: (B)

The given limit is the

t

-derivative of

t^{2} f^{2} (x) - x^{2} f^{2} (t)

evaluated at

t = x

and set to zero, which gives

2 x f^{2} (x) - x^{2} \cdot 2 f (x) f^{'} (x) = 0

, i.e.

f (x) - x f^{'} (x) = 0

. Thus

\frac{f ^{'} ( x )}{f ( x )} = \frac{1}{x}

, and integrating gives

ln f (x) = ln x + C

; the condition

f (1) = e

fixes

C = 1

, so

ln f (x) = ln x + 1

. Setting

f (x) = 1

gives

0 = ln x + 1

, so

x = \frac{1}{e}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option