Let f be a real-valued function defined on the interval ( 0,\infty ) by \,f( x ) = \ln x + \int\limits_0^x \sqrt 1 + \sin t\, dt. then which of the following statement(s) is (are) true?

f'(x) exists for all x \in ( 0,\infty ) and f' is continuous on ( 0,\infty ) , but not differentiable on ( 0,\infty ), there exists \,\,\alpha \gt 1 such that | f'( x ) | \lt | f( x ) | for all x \in ( \alpha ,\infty )\,

f''(x) exists for all x \in ( 0,\infty )

f'(x) exists for all x \in ( 0,\infty ) and f' is continuous on ( 0,\infty ) , but not differentiable on ( 0,\infty )

there exists \,\,\alpha \gt 1 such that | f'( x ) | \lt | f( x ) | for all x \in ( \alpha ,\infty )\,

there exists \beta \gt 0 such that | f( x ) | + | f'( x ) | \le \beta for all x \in ( 0,\infty )

Application of Integration PYQ — JEE Advanced 2010 · Shift 1 · Q47 with Solution

View written solutionFree

Correct answer: C, B

Given function

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt, \qquad x>0.

We analyze differentiability and growth.

First derivative

Since $\ln x$ is differentiable for $x>0$ , and the integrand

g(t)=\sqrt{1+\sin t}

is continuous for all $t\in\mathbb R$ (because $1+\sin t\ge 0$ ), by the Fundamental Theorem of Calculus,

\frac{d}{dx}\left(\int_0^x \sqrt{1+\sin t}\,dt\right)=\sqrt{1+\sin x}.

Hence

f'(x)=\frac1x+\sqrt{1+\sin x}, \qquad x>0.

So $f'(x)$ exists for all $x>0$ .

Also, both $1/x$ and $\sqrt{1+\sin x}$ are continuous on $(0,\infty)$ , so $f'$ is continuous on $(0,\infty)$ .

Second derivative / differentiability of $f'$

To differentiate $\sqrt{1+\sin x}$ , formally:

\frac{d}{dx}\sqrt{1+\sin x}=\frac{\cos x}{2\sqrt{1+\sin x}}.

This exists only when $1+\sin x>0$ . But if

1+\sin x=0 \iff \sin x=-1 \iff x=\frac{(4n+3)\pi}{2}, \quad n=0,1,2,\dots

then the denominator becomes $0$ .

Indeed, near such points, $\sqrt{1+\sin x}$ behaves like $|x-a|$ -type square root singularity in derivative, so it is not differentiable there.

Therefore $f''(x)$ does not exist for all $x>0$ .

So:

A is false.
B is true: $f'$ exists everywhere on $(0,\infty)$ , is continuous there, but is not differentiable at points $x=\frac{(4n+3)\pi}{2}$ .

Check option C

We need to see whether there exists $\alpha>1$ such that

|f'(x)|<|f(x)| \quad \forall x>\alpha.

First note that for $x>0$ ,

f'(x)=\frac1x+\sqrt{1+\sin x}>0.

Also,

\sqrt{1+\sin t}\ge 0,

so for $x>1$ ,

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt >0.

Thus absolute values can be removed for large $x$ .

Now estimate growth of $f(x)$ :

Since $\sqrt{1+\sin t}$ is continuous and nonnegative, and not identically zero, its average over a period is positive. Let

m=\frac1{2\pi}\int_0^{2\pi} \sqrt{1+\sin t}\,dt >0.

Then as $x\to\infty$ ,

\int_0^x \sqrt{1+\sin t}\,dt = mx + O(1).

Hence

f(x)=\ln x + mx + O(1) \to \infty

linearly.

But

f'(x)=\frac1x+\sqrt{1+\sin x}.

Since $0\le \sqrt{1+\sin x}\le \sqrt2$ , we get

0<f'(x)\le \frac1x+\sqrt2 \le 1+\sqrt2 \quad (x\ge 1).

So $f'(x)$ is bounded, while $f(x)\to\infty$ . Therefore there exists $\alpha>1$ such that

f(x)>1+\sqrt2 \ge f'(x)

for all $x>\alpha$ .

Thus

|f'(x)|<|f(x)| \quad \forall x>\alpha.

So C is true.

Check option D

Option D claims that there exists $\beta>0$ such that

|f(x)|+|f'(x)|\le \beta \quad \forall x>0.

But from above,

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt \to \infty \quad \text{as } x\to\infty.

Hence $|f(x)|+|f'(x)|\to\infty$ , so it cannot be bounded on $(0,\infty)$ .

Therefore D is false.

Final conclusion

The true statements are:

\boxed{\text{B, C}}

Comparison with stored answer

Stored correct answer: C, B.

This matches exactly (order irrelevant).

More from Application of Integration

View written solutionFree

Correct answer: C, B

Given function

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt, \qquad x>0.

We analyze differentiability and growth.

First derivative

Since $\ln x$ is differentiable for $x>0$ , and the integrand

g(t)=\sqrt{1+\sin t}

is continuous for all $t\in\mathbb R$ (because $1+\sin t\ge 0$ ), by the Fundamental Theorem of Calculus,

\frac{d}{dx}\left(\int_0^x \sqrt{1+\sin t}\,dt\right)=\sqrt{1+\sin x}.

Hence

f'(x)=\frac1x+\sqrt{1+\sin x}, \qquad x>0.

So $f'(x)$ exists for all $x>0$ .

Also, both $1/x$ and $\sqrt{1+\sin x}$ are continuous on $(0,\infty)$ , so $f'$ is continuous on $(0,\infty)$ .

Second derivative / differentiability of $f'$

To differentiate $\sqrt{1+\sin x}$ , formally:

\frac{d}{dx}\sqrt{1+\sin x}=\frac{\cos x}{2\sqrt{1+\sin x}}.

This exists only when $1+\sin x>0$ . But if

1+\sin x=0 \iff \sin x=-1 \iff x=\frac{(4n+3)\pi}{2}, \quad n=0,1,2,\dots

then the denominator becomes $0$ .

Indeed, near such points, $\sqrt{1+\sin x}$ behaves like $|x-a|$ -type square root singularity in derivative, so it is not differentiable there.

Therefore $f''(x)$ does not exist for all $x>0$ .

So:

A is false.
B is true: $f'$ exists everywhere on $(0,\infty)$ , is continuous there, but is not differentiable at points $x=\frac{(4n+3)\pi}{2}$ .

Check option C

We need to see whether there exists $\alpha>1$ such that

|f'(x)|<|f(x)| \quad \forall x>\alpha.

First note that for $x>0$ ,

f'(x)=\frac1x+\sqrt{1+\sin x}>0.

Also,

\sqrt{1+\sin t}\ge 0,

so for $x>1$ ,

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt >0.

Thus absolute values can be removed for large $x$ .

Now estimate growth of $f(x)$ :

Since $\sqrt{1+\sin t}$ is continuous and nonnegative, and not identically zero, its average over a period is positive. Let

m=\frac1{2\pi}\int_0^{2\pi} \sqrt{1+\sin t}\,dt >0.

Then as $x\to\infty$ ,

\int_0^x \sqrt{1+\sin t}\,dt = mx + O(1).

Hence

f(x)=\ln x + mx + O(1) \to \infty

linearly.

But

f'(x)=\frac1x+\sqrt{1+\sin x}.

Since $0\le \sqrt{1+\sin x}\le \sqrt2$ , we get

0<f'(x)\le \frac1x+\sqrt2 \le 1+\sqrt2 \quad (x\ge 1).

So $f'(x)$ is bounded, while $f(x)\to\infty$ . Therefore there exists $\alpha>1$ such that

f(x)>1+\sqrt2 \ge f'(x)

for all $x>\alpha$ .

Thus

|f'(x)|<|f(x)| \quad \forall x>\alpha.

So C is true.

Check option D

Option D claims that there exists $\beta>0$ such that

|f(x)|+|f'(x)|\le \beta \quad \forall x>0.

But from above,

f(x)=\ln x+\int_0^x \sqrt{1+\sin t}\,dt \to \infty \quad \text{as } x\to\infty.

Hence $|f(x)|+|f'(x)|\to\infty$ , so it cannot be bounded on $(0,\infty)$ .

Therefore D is false.

Final conclusion

The true statements are:

\boxed{\text{B, C}}

Comparison with stored answer

Stored correct answer: C, B.

This matches exactly (order irrelevant).

Application of Integration question

Application of Integration question

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Application of Integration question

Application of Integration question

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