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  • Let f : [2, 4]  be a differentiable function such that <img alt=

Let f : [2, 4]\rightarrow \mathbb{R}  be a differentiable function such that \left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1, x \in[2,4]
with f(2)=\frac{1}{2} and f(4) =\frac{1}{4}  Consider the following two statements :

(A): f(x) \leq 1$, for all $x \in[2,4] (B)f(x) \geq \frac{1}{8}$, for all $x \in[2,4]

 

Option: 1

Only statement (B) is true
 


Option: 2

Only statement (A) is true
 


Option: 3

Neither statement (A) nor statement (B) is true
 


Option: 4

Both the statements (A) and (B) are true


Answers (1)

best_answer

\begin{aligned} & x \ell \operatorname{nxf} f^{\prime}(x)+\ln x f(x)+f(x) \geq 1, x \in[2,4] \\ & \text { And } f(2)=\frac{1}{2}, f(4)=\frac{1}{4} \\ & \text { Now } x \ln x, \frac{d y}{d x}+(\ln +1) y \geq 1 \\ & \frac{d}{d x}(y \cdot x \ln x) \geq 1 \\ & \frac{d}{d x}(f(x) \cdot x \ln x) \geq 1 \\ & \Rightarrow \frac{d}{d x}(x \ln x f(x)-x) \geq 0, x \in[2,4] \\ & \Rightarrow \text { The function } g(x)=x \ln x f(x)-x \text { is increasing in } \\ & {[2,4]} \\ & \text { And } g(2)=2 \ln 2 f(2)-2=\ln 2-2 \\ & g(2)=4 \ln 4 f(4)-4=\ln 4-4 \\ & =2(\ln 2-2) \\ & \text { Now } g(2) \leq g(x) \leq g(4) \end{aligned}

\begin{aligned} & \operatorname{Ln} 2-2 \leq x \ln x(x)-x \leq 2(\ln 2-2) \\ & \frac{\ln 2-2}{x \ln x}+\frac{1}{\ln x} \leq f(x) \leq \frac{2(\ln 2-2)}{x \ln x}+\frac{1}{\ln x} \end{aligned} Now for \mathrm{x} \in[2,4] \begin{aligned} & \frac{2(\ell \ln 2-2)}{\mathrm{x} \ln x}+\frac{1}{\ell \mathrm{nx}}<\frac{2(\ln 2-2)}{2 \ln 2}+\frac{1}{\ln 2}=1-\frac{1}{\ln 2}<1 \\ & \Rightarrow \mathrm{f}(\mathrm{x}) \leq 1 \text { for } \mathrm{x} \in[2,4] \\ & \frac{\ln 2-2}{\mathrm{x} \ln \mathrm{x}}+\frac{1}{\ln x} \geq \frac{\ln 2-2}{4 \ln 4}+\frac{1}{\ln 4}=\frac{1}{8}+\frac{1}{2 \ln 2}>\frac{1}{8} \\ & \Rightarrow \mathrm{f}(\mathrm{x}) \geq \frac{1}{8} \text { for } \mathrm{x} \in[2,4] \end{aligned} Hence both A and B are true.

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Ritika Harsh

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