Let b be a nonzero real number. Suppose f : R \to R is a differentiable function such that f(0) = 1. If the derivative f' of f satisfies the equation f'(x) = f(x) \over b^2 + x^2 for all x \in R, then which of the following statements is/are TRUE?

If b > 0, then f is an increasing function, f(x) f( - x) = 1 for all x \in R

If b > 0, then f is an increasing function

If b < 0, then f is a decreasing function

f(x) f( - x) = 1 for all x \in R

f(x) - f( - x) = 0 for all x \in R

Definite Integration PYQ — JEE Advanced 2020 · Shift 2 · Q25 with Solution

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Correct answer: A, C

We are given the differential equation

$f'(x)=\frac{f(x)}{b^2+x^2}, \qquad f(0)=1,$

where $b\neq 0$ is a real constant.

We must determine which statements are true.

Solve the differential equation

Since

$\frac{f'(x)}{f(x)}=\frac{1}{b^2+x^2},$

we separate variables and integrate:

$\int \frac{f'(x)}{f(x)}\,dx=\int \frac{1}{b^2+x^2}\,dx.$

So,

$\ln |f(x)|=\int \frac{1}{b^2+x^2}\,dx + C.$

Now,

$\int \frac{1}{b^2+x^2}\,dx = \frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right),$

because $b^2>0$ .

Thus,

$\ln |f(x)|=\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)+C.$

Using $f(0)=1$ :

$\ln 1 = 0 = \frac{1}{|b|}\tan^{-1}(0)+C \implies C=0.$

Hence,

$f(x)=\exp\left(\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right).$

Since $f(0)=1>0$ , $f(x)$ is positive for all $x$ .

Check monotonicity

From the given differential equation,

$f'(x)=\frac{f(x)}{b^2+x^2}.$

Now $b^2+x^2>0$ for all $x$ , and we already have $f(x)>0$ for all $x$ .

Therefore,

$f'(x)>0 \quad \text{for all } x\in\mathbb R.$

So $f$ is strictly increasing for every nonzero real $b$ , whether $b>0$ or $b<0$ .

Therefore:

A is true.
B is false.

Check statement C: $f(x)f(-x)=1$

Using the explicit form,

=\exp\left(-\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right).$$ Hence, $$f(x)f(-x)=\exp\left(\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right) \exp\left(-\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right)=1.$$ So **C is true**. --- 5. Check statement D: $f(x)-f(-x)=0$ This would mean $$f(x)=f(-x),$$ i.e. $f$ is even. But from statement C, $$f(x)f(-x)=1,$$ and in general a positive increasing function satisfying this need not be even. In fact, from the explicit formula, $$f(-x)=\frac{1}{f(x)}.$$ So $f(x)=f(-x)$ would imply $$f(x)^2=1 \implies f(x)=1,$$ for all $x$, which is impossible since $f'(x)=\frac{f(x)}{b^2+x^2}>0$. Hence **D is false**. --- 6. Final conclusion The true statements are: $$\boxed{A,\ C}$$ --- 7. Comparison with stored correct answer Stored correct answer: $A, C$ Our derived answer matches the stored answer.

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View written solutionFree

Correct answer: A, C

We are given the differential equation

$f'(x)=\frac{f(x)}{b^2+x^2}, \qquad f(0)=1,$

where $b\neq 0$ is a real constant.

We must determine which statements are true.

Solve the differential equation

Since

$\frac{f'(x)}{f(x)}=\frac{1}{b^2+x^2},$

we separate variables and integrate:

$\int \frac{f'(x)}{f(x)}\,dx=\int \frac{1}{b^2+x^2}\,dx.$

So,

$\ln |f(x)|=\int \frac{1}{b^2+x^2}\,dx + C.$

Now,

$\int \frac{1}{b^2+x^2}\,dx = \frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right),$

because $b^2>0$ .

Thus,

$\ln |f(x)|=\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)+C.$

Using $f(0)=1$ :

$\ln 1 = 0 = \frac{1}{|b|}\tan^{-1}(0)+C \implies C=0.$

Hence,

$f(x)=\exp\left(\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right).$

Since $f(0)=1>0$ , $f(x)$ is positive for all $x$ .

Check monotonicity

From the given differential equation,

$f'(x)=\frac{f(x)}{b^2+x^2}.$

Now $b^2+x^2>0$ for all $x$ , and we already have $f(x)>0$ for all $x$ .

Therefore,

$f'(x)>0 \quad \text{for all } x\in\mathbb R.$

So $f$ is strictly increasing for every nonzero real $b$ , whether $b>0$ or $b<0$ .

Therefore:

A is true.
B is false.

Check statement C: $f(x)f(-x)=1$

Using the explicit form,

=\exp\left(-\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right).$$ Hence, $$f(x)f(-x)=\exp\left(\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right) \exp\left(-\frac{1}{|b|}\tan^{-1}\left(\frac{x}{|b|}\right)\right)=1.$$ So **C is true**. --- 5. Check statement D: $f(x)-f(-x)=0$ This would mean $$f(x)=f(-x),$$ i.e. $f$ is even. But from statement C, $$f(x)f(-x)=1,$$ and in general a positive increasing function satisfying this need not be even. In fact, from the explicit formula, $$f(-x)=\frac{1}{f(x)}.$$ So $f(x)=f(-x)$ would imply $$f(x)^2=1 \implies f(x)=1,$$ for all $x$, which is impossible since $f'(x)=\frac{f(x)}{b^2+x^2}>0$. Hence **D is false**. --- 6. Final conclusion The true statements are: $$\boxed{A,\ C}$$ --- 7. Comparison with stored correct answer Stored correct answer: $A, C$ Our derived answer matches the stored answer.

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