QJEEM23Mab16b: JEE Main 2023 Mathematics - Class 12 JEEM23Mab16b Solution: Let . Then 18 ∫ 1 2 f ( x ) d x is equal to :

Question

Verified Solution • 2025–26 Session moderate

Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$ . Then $18 \int_{1}^{2} f (x) d x$ is equal to :

Accepted Answer

We have,

5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x&gt;0

..........(i)

On replacing

x

by

\frac{1}{x}

in (i), we get

5 f\left(\frac{1}{x}\right)+4 f(x)=x+3

..........(ii)

Now, using Eq. (i)

\times 5-

(ii)

\times 4

, we get

\begin{aligned}&amp; 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\&amp;\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\&amp;\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii)\end{aligned}

\begin{aligned}\therefore \quad &amp; 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\= &amp; 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x\end{aligned}

\begin{aligned}&amp; =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\&amp; =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\&amp; =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\&amp; =10 \log _e 2-6\end{aligned}

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