IIT-JEE 2010 · Paper 1 · Mathematics Q40

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MathematicsMultiple correct+3Question 40 of 84

Let f be a real-valued function defined on the interval $(0,\infty)$ by $f(x)=\ln x+\displaystyle\int_0^x\sqrt{1+\sin t}\,dt$ . Then which of the following statement(s) is (are) true?

f''(x)

exists for all

x\in(0,\infty)

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>1

such that

|f'(x)|<|f(x)|

for all

x\in(\alpha,\infty)

there exists

\beta>0

such that

|f(x)|+|f'(x)|\le\beta

for all

x\in(0,\infty)

Answer and worked solution

Official answer: B, C

First derivative. $f'(x)=\dfrac1x+\sqrt{1+\sin x}$ , which exists and is continuous on $(0,\infty)$ .

Second derivative. $\sqrt{1+\sin x}=\left|\sin\frac x2+\cos\frac x2\right|$ has corners where $1+\sin x=0$ , i.e. $x=\frac{3\pi}2+2k\pi$ . So $f'$ is not differentiable there: (A) false, (B) true.

Growth. $\sqrt{1+\sin t}\ge0$ has a positive average ( $\frac{2\sqrt2}\pi$ ), so $f(x)\to\infty$ roughly linearly, while $|f'(x)|\le\frac1x+\sqrt2$ stays bounded for $x>1$ . Hence for large $x$ , $|f'(x)|<|f(x)|$ : (C) true, and $|f|+|f'|$ is unbounded: (D) false.

(B), (C)

Topics:Differentiability Fundamental theorem of calculus

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IIT-JEE 2010 · Paper 1, Mathematics question 40