Let f:( 0,\infty ) \to ( 0,\infty ) be a differentiable function such that f(1) = e and \mathop \lim \limits_t \to x t^2f^2(x) - x^2f^2(t) \over t - x = 0 . If f(x) = 1, then x is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2020 · 4 Sep · Shift 2 · Q29 with Solution

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Correct answer: A

Interpret the given limit

We are given

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

for all $x>0$ .

Here, $x$ is fixed and $t\to x$ .

Let us simplify this limit.

Since $f(x)$ is constant with respect to $t$ ,

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

can be treated directly.

Differentiate using the limit form

Define

\phi(t)=t^2f^2(x)-x^2f^2(t).

Then the given limit is

\lim_{t\to x}\frac{\phi(t)-\phi(x)}{t-x},

because

\phi(x)=x^2f^2(x)-x^2f^2(x)=0.

Hence,

\phi'(x)=0.

Now compute $\phi'(t)$ :

\phi'(t)=2t f^2(x)-x^2\cdot 2f(t)f'(t).

So at $t=x$ ,

\phi'(x)=2x f^2(x)-2x^2 f(x)f'(x)=0.

Divide by $2x f(x)$ (valid since $x>0$ and $f(x)>0$ ):

f(x)-x f'(x)=0.

Thus,

x f'(x)=f(x).

Equivalently,

\frac{f'(x)}{f(x)}=\frac1x.

Solve the differential equation

Integrate both sides:

\int \frac{f'(x)}{f(x)}\,dx=\int \frac1x\,dx.

This gives

\ln f(x)=\ln x + C.

Therefore,

f(x)=Cx

for some positive constant $C$ .

Use the condition $f(1)=e$

Since

f(1)=C\cdot 1=e,

we get

C=e.

Hence,

f(x)=ex.

Find $x$ when $f(x)=1$

Given $f(x)=1$ ,

ex=1 \implies x=\frac1e.

Check options

A: $\frac1e$ ✅
B: $e$ ❌
C: $\frac1{2e}$ ❌
D: $2e$ ❌

Therefore, the correct answer is

\boxed{\frac1e}.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: A

Interpret the given limit

We are given

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

for all $x>0$ .

Here, $x$ is fixed and $t\to x$ .

Let us simplify this limit.

Since $f(x)$ is constant with respect to $t$ ,

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

can be treated directly.

Differentiate using the limit form

Define

\phi(t)=t^2f^2(x)-x^2f^2(t).

Then the given limit is

\lim_{t\to x}\frac{\phi(t)-\phi(x)}{t-x},

because

\phi(x)=x^2f^2(x)-x^2f^2(x)=0.

Hence,

\phi'(x)=0.

Now compute $\phi'(t)$ :

\phi'(t)=2t f^2(x)-x^2\cdot 2f(t)f'(t).

So at $t=x$ ,

\phi'(x)=2x f^2(x)-2x^2 f(x)f'(x)=0.

Divide by $2x f(x)$ (valid since $x>0$ and $f(x)>0$ ):

f(x)-x f'(x)=0.

Thus,

x f'(x)=f(x).

Equivalently,

\frac{f'(x)}{f(x)}=\frac1x.

Solve the differential equation

Integrate both sides:

\int \frac{f'(x)}{f(x)}\,dx=\int \frac1x\,dx.

This gives

\ln f(x)=\ln x + C.

Therefore,

f(x)=Cx

for some positive constant $C$ .

Use the condition $f(1)=e$

Since

f(1)=C\cdot 1=e,

we get

C=e.

Hence,

f(x)=ex.

Find $x$ when $f(x)=1$

Given $f(x)=1$ ,

ex=1 \implies x=\frac1e.

Check options

A: $\frac1e$ ✅
B: $e$ ❌
C: $\frac1{2e}$ ❌
D: $2e$ ❌

Therefore, the correct answer is

\boxed{\frac1e}.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability