JEE 2023 — Mathematics PYQ

Step 1: Integral ko break karna

Chunki $|{\log }_{e}x|$ function $x=1$ par apna sign badalta hai (x < 1 ke liye negative aur x > 1 ke liye positive), hum integral ko do hisson mein todte hain:

Step 2: Definite Integral solve karna

Humein pata hai ki $\int {\log }_{e}xdx=x{\log }_{e}x-x$.

• Pehle part ke liye:

  $${\int }_{1/3}^1(-{\log }_{e}x)dx=-[x{\log }_{e}x-x{]}_{1/3}^1$$
  $$=-\left((1{\log }_{e}1-1)-\left(\frac{1}{3}{\log }_e\frac{1}{3}-\frac{1}{3}\right)\right)$$
  $$=-\left(-1-\left(-\frac{1}{3}{\log }_{e}3-\frac{1}{3}\right)\right)=-\left(-1+\frac{1}{3}{\log }_{e}3+\frac{1}{3}\right)$$
  $$=-\left(-\frac{2}{3}+\frac{1}{3}{\log }_{e}3\right)=\frac{2}{3}-\frac{1}{3}{\log }_{e}3$$

• Dusre part ke liye:

  $${\int }_1^3({\log }_{e}x)dx=[x{\log }_{e}x-x{]}_1^3$$
  $$=(3{\log }_{e}3-3)-(1{\log }_{e}1-1)=3{\log }_{e}3-3+1=3{\log }_{e}3-2$$

Step 3: Dono ko jodna

Step 4: Diye gaye form mein convert karna

Comparing with $\frac{m}{n}{\log }_e\left(\frac{n^2}{e}\right)$:

(Yahan $4$ aur $3$ coprime hain, yani $\text{gcd}(4,3)=1$)

Step 5: Final value nikalna

Humein $m^2+n^2-5$ ki value chahiye:

Answer:

Iska sahi uttar $20$ hai.