JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 5)
If $$f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$$, then the least value of $$f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$$ is :
2
4
0
8
Explanation
Given that $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}$
Let us consider a similar function of $(x)$,
$\therefore f(x)=\frac{A x+B}{C x-A}$
$\text { Now, } $
$$ \begin{aligned} &f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\right)+B}{C\left(\frac{A x+B}{C x-A}\right)-A} \\\\ & =\frac{A^2 x+A B+B C x-A B}{A C x+B C-A C x+A^2} \\\\ & =\frac{x\left(A^2+B C\right)}{\left(B C+A^2\right)}=x \end{aligned} $$
$$ \begin{aligned} & \therefore \quad f(f(x))=x \\\\ & \text { Similarly, } f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\\\ & \text { Apply, AM } \geq \text { GM } \\\\ & \left(x+\frac{4}{x}\right) \geq 2 \sqrt{x \cdot \frac{4}{x}}=4(x>0) \\\\ & \Rightarrow f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \geq 4 \end{aligned} $$
Hence, the least value is 4 .
Let us consider a similar function of $(x)$,
$\therefore f(x)=\frac{A x+B}{C x-A}$
$\text { Now, } $
$$ \begin{aligned} &f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\right)+B}{C\left(\frac{A x+B}{C x-A}\right)-A} \\\\ & =\frac{A^2 x+A B+B C x-A B}{A C x+B C-A C x+A^2} \\\\ & =\frac{x\left(A^2+B C\right)}{\left(B C+A^2\right)}=x \end{aligned} $$
$$ \begin{aligned} & \therefore \quad f(f(x))=x \\\\ & \text { Similarly, } f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\\\ & \text { Apply, AM } \geq \text { GM } \\\\ & \left(x+\frac{4}{x}\right) \geq 2 \sqrt{x \cdot \frac{4}{x}}=4(x>0) \\\\ & \Rightarrow f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \geq 4 \end{aligned} $$
Hence, the least value is 4 .
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