JEE Advance - Mathematics (2010 - Paper 1 Offline - No. 16)
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?
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.
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