JEE Advanced · Mathematics · 23. C&D
Let \(f\) be a real-valued function defined on the interval \((0, \infty)\), by \(f(x)=\ln x+\int_0^x \sqrt{1+\sin t} d t\). Then which of the following statement(s) is (are) true ?
- A
\(f^{\prime \prime}(x)\) exists for all \(x \in(0, \infty)\)
- B
\(f^{\prime}(x)\) exists for all \(x \in(0, \infty)\) and \(f^{\prime}\) is continuous on \((0, \infty)\), but not differentiable on \((0, \infty)\)
- C
there exists \(\alpha>1\) such that \(\left|f^{\prime}(x)\right| < |f(x)|\) for all \(x \in(\alpha, \infty)\) - D
there exists \(\beta>0\) such that \(|f(x)|+\left|f^{\prime}(x)\right| \leq \beta \quad\) from all \(x \in(0, \infty)\)
Answer & Solution
Correct Answer
(C)
there exists \(\alpha>1\) such that \(\left|f^{\prime}(x)\right| < |f(x)|\) for all \(x \in(\alpha, \infty)\)
Step-by-step Solution
Detailed explanation
Here, \(f^{\prime}(x)=\frac{1}{x}+\sqrt{1+\sin x}, x>0\) but \(f(x)\) is not differentiable in \((0, \infty)\) as \(\sin x\) may be \(-1\) and then \(f^{\prime \prime}(x)=-\frac{1}{x^2}+\frac{\cos x}{2 \sqrt{1+\sin x}}\) will not exists.
\(\Rightarrow f^{\prime}(x)\) is continuous for all \(x \in(0, \infty)\) but \(f^{\prime}(x)\) is not differentiable on \((0, \infty)\).
\(\therefore\) Option (b) is true.
Also, \(\quad f^{\prime}(x) \leq 3\), if \(x>1\) and \(\quad f(x)>3\), if \(x>e^3\)
\(\therefore\) Let \(\alpha=e^3\)
\(\Rightarrow\) Option (c) is true.
(d) is not possible as \(f(x) \rightarrow \infty\) when \(x \rightarrow \infty\).
Hence, (b, c) is the correct option.
\(\Rightarrow f^{\prime}(x)\) is continuous for all \(x \in(0, \infty)\) but \(f^{\prime}(x)\) is not differentiable on \((0, \infty)\).
\(\therefore\) Option (b) is true.
Also, \(\quad f^{\prime}(x) \leq 3\), if \(x>1\) and \(\quad f(x)>3\), if \(x>e^3\)
\(\therefore\) Let \(\alpha=e^3\)
\(\Rightarrow\) Option (c) is true.
(d) is not possible as \(f(x) \rightarrow \infty\) when \(x \rightarrow \infty\).
Hence, (b, c) is the correct option.
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