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JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2025 · Paper 2 · Mathematics · Application of Derivatives

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be...

Let

\mathbb{R}

denote the set of all real numbers. Let

f: \mathbb{R} \rightarrow \mathbb{R}

be defined by

f(x)=

\begin{cases}\frac{6 x+\sin x}{2 x+\sin x} & \text{ if } x \neq 0 \\ \frac{7}{3} & \text{ if } x=0\end{cases}$$ Then which of the following statements is (are) TRUE? Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6 x+ x)/(2 x+ x) & if x ≠ 0

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6 x+ x)/(2 x+ x) & if x ≠ 0

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6 x+ x)/(2 x+ x) & if x ≠ 0

Options

AThe point $x=0$ is a point of local maxima of $f$
BThe point $x=0$ is a point of local minima of $f$
CNumber of points of local maxima of $f$ in the interval $[\pi, 6 \pi]$ is 3
DNumber of points of local minima of $f$ in the interval $[2 \pi, 4 \pi]$ is 1

Correct answer

B. The point $x=0$ is a point of local minima of $f$

Step-by-step solution

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6 x+ x)/(2 x+ x) & if x ≠ 0

since

3>\frac{7}{3} \Rightarrow 3>f(0)

\Rightarrow x=0

is local minima ....option (B) is correct Now,

\begin{aligned} & f(x)=\frac{6 x+\sin x}{2 x+\sin x}=1+\frac{4 x}{2 x+\sin x} \\ & f^{\prime}(x)=\frac{4[(2 x+\sin x) \cdot 1-x(2+\cos x)]}{(2 x+\sin x)^2}=\frac{4(\sin x-x \cos x)}{(2 x+\sin x)^2} \\ & =\frac{4 \cos x(\tan x-x)}{(2 x+\sin x)^2} \end{aligned}

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be defined by f(x)= cases (6 x+ x)/(2 x+ x) & if x ≠ 0

Options (C) \& (D) are also correct.

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