Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) e 0 for all x \in R. If | \beginmatrix f(x) & f'(x) f'(x) & f''(x) \endmatrix | = 0, for all x \in R, then the value of f(1) lies in the interval :

Definite Integration PYQ — JEE Main 2021 · 24 Feb · Shift 2 · Q28 with Solution

View written solutionFree

Correct answer: D

Interpret the determinant condition

Given

\begin{vmatrix} f(x) & f'(x)\\ f'(x) & f''(x) \end{vmatrix}=0

for all $x\in\mathbb R$ .

So,

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

Hence,

f(x)f''(x)=\big(f'(x)\big)^2. \qquad (1)

Also, $f'(x)\ne 0$ for all $x\in\mathbb R$ .

Convert into a simpler differential equation

From (1), divide by $f(x)^2$ (we will justify that $f(x)\neq 0$ where needed):

\frac{f''(x)}{f(x)}=\left(\frac{f'(x)}{f(x)}\right)^2.

Now observe that

\left(\frac{f'(x)}{f(x)}\right)'= rac{f(x)f''(x)-\big(f'(x)\big)^2}{f(x)^2}=0.

Therefore,

\frac{f'(x)}{f(x)}=c

for some constant $c$ .

So,

f'(x)=cf(x).

This is a first-order linear differential equation whose solution is

f(x)=Ae^{cx}.

Use the initial conditions

Given $f(0)=1$ , so

A=1.

Thus,

f(x)=e^{cx}.

Then

f'(x)=ce^{cx}.

Using $f'(0)=2$ ,

c=2.

Hence,

f(x)=e^{2x}.

Find $f(1)$

f(1)=e^2.

Now,

e^2\approx 7.389.

Therefore,

f(1)\in (6,9).

So the correct option is D.

Verification of nonzero condition

For $f(x)=e^{2x}$ ,

f'(x)=2e^{2x}\neq 0

for all $x\in\mathbb R$ , so the condition is satisfied.

Compare with stored correct answer

Derived answer: D : $(6,9)$ .

Stored correct answer: D.

They match.

More from Definite Integration

View written solutionFree

Correct answer: D

Interpret the determinant condition

Given

\begin{vmatrix} f(x) & f'(x)\\ f'(x) & f''(x) \end{vmatrix}=0

for all $x\in\mathbb R$ .

So,

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

Hence,

f(x)f''(x)=\big(f'(x)\big)^2. \qquad (1)

Also, $f'(x)\ne 0$ for all $x\in\mathbb R$ .

Convert into a simpler differential equation

From (1), divide by $f(x)^2$ (we will justify that $f(x)\neq 0$ where needed):

\frac{f''(x)}{f(x)}=\left(\frac{f'(x)}{f(x)}\right)^2.

Now observe that

\left(\frac{f'(x)}{f(x)}\right)'= rac{f(x)f''(x)-\big(f'(x)\big)^2}{f(x)^2}=0.

Therefore,

\frac{f'(x)}{f(x)}=c

for some constant $c$ .

So,

f'(x)=cf(x).

This is a first-order linear differential equation whose solution is

f(x)=Ae^{cx}.

Use the initial conditions

Given $f(0)=1$ , so

A=1.

Thus,

f(x)=e^{cx}.

Then

f'(x)=ce^{cx}.

Using $f'(0)=2$ ,

c=2.

Hence,

f(x)=e^{2x}.

Find $f(1)$

f(1)=e^2.

Now,

e^2\approx 7.389.

Therefore,

f(1)\in (6,9).

So the correct option is D.

Verification of nonzero condition

For $f(x)=e^{2x}$ ,

f'(x)=2e^{2x}\neq 0

for all $x\in\mathbb R$ , so the condition is satisfied.

Compare with stored correct answer

Derived answer: D : $(6,9)$ .

Stored correct answer: D.

They match.

Definite Integration question

Definite Integration question

More from Definite Integration

Definite Integration question

Definite Integration question

More from Definite Integration