IIT-JEE 2010 Paper 1 Offline

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MCQ (More than One Correct Answer)

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

$$f''(x)$$ exists for all $$x \in \left( {0,\infty } \right)$$

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

there exists $$\,\,\alpha > 1$$ such that $$\left| {f'\left( x \right)} \right| < \left| {f\left( x \right)} \right|$$ for all $$x \in \left( {\alpha ,\infty } \right)\,$$

there exists $$\beta > 0$$ such that $$\left| {f\left( x \right)} \right| + \left| {f'\left( x \right)} \right| \le \beta $$ for all $$x \in \left( {0,\infty } \right)$$

Explanation

$$f(x) = \ln x + \int\limits_0^x {\sqrt {1 + \sin t} \,dt} $$

$$f'(x) = {1 \over x} + \sqrt {1 + \sin x} $$

f' is not differentiable at sin x = $$-$$1

i.e. $$x = 2n\pi - {\pi \over 2},n \in N$$ in the interval (0, $$\infty$$)

$$f''(x) = - {1 \over {{x^2}}} + {{\cos x} \over {2\sqrt {1 + \sin x} }}$$

f'' does not exist for all x $$\in$$ (0, $$\infty$$)

f' exist for x > 0

we have $${1 \over x} + \sqrt {1 + \sin x} < \ln x + \int\limits_0^x {\sqrt {1 + \sin x} dx} $$

because L.H.S. is bounded and R.H.S. is not bounded so $$\exists $$ some $$\alpha$$ beyond which R.H.S. is greater than L.H.S.

i.e. $$|f'(x)| < |f(x)|$$ for all x $$\in$$ ($$\alpha$$, $$\infty$$)

$$|f| + |f'| \le \beta $$ is wrong as f is unbounded while f' is bounded.

IIT-JEE 2009

MCQ (More than One Correct Answer)

If $${I_n} = \int\limits_{ - \pi }^\pi {{{\sin nx} \over {\left( {1 + {\pi ^x}} \right)\sin x}}dx\,\,n = 0,1,2,.....,} $$ then

$${I_n} = {I_{n + 2}}$$

$$\sum\limits_{m = 1}^{10} {{I_{2m + 1}}} = 10\pi $$

$$\sum\limits_{m = 1}^{10} {{I_{2m}}} = 0$$

$${I_n} = {I_{n + 1}}$$

IIT-JEE 2009

MCQ (More than One Correct Answer)

Area of the region bounded by the curve $$y = {e^x}$$ and lines $$x=0$$ and $$y=e$$ is

$$e-1$$

$$\int\limits_1^e {\ln \left( {e + 1 - y} \right)dy} $$

$$e - \int\limits_0^1 {{e^x}dx} $$

$$\int\limits_1^e {\ln y\,dy} $$

IIT-JEE 2008

MCQ (More than One Correct Answer)

Let $$f(x)$$ be a non-constant twice differentiable function defined on $$\left( { - \infty ,\infty } \right)$$
such that $$f\left( x \right) = f\left( {1 - x} \right)$$ and $$f'\left( {{1 \over 4}} \right) = 0.$$ Then,

$$f''\left( x \right)$$ vanishes at least twice on $$\left[ {0,1} \right]$$

$$f'\left( {{1 \over 2}} \right) = 0$$

$$\int\limits_{ - 1/2}^{1/2} {f\left( {x + {1 \over 2}} \right)\sin x\,dx} = 0$$

$$\int\limits_0^{1/2} {f\left( t \right){e^{\sin \,\pi t}}dt = } \int\limits_{1/2}^1 {f\left( {1 - t} \right){e^{\sin \,\pi t}}dt} $$