JEE 2023 — Mathematics PYQ

Logarithm ${\log }_{a}b$ tabhi define hota hai jab:

1. Argument positive ho: b > 0

   x - 2 > 0 \implies \mathbf{x > 2}

2. Base positive ho: a > 0

   x + 1 > 0 \implies x > -1

3. Base $1$ ke barabar na ho: $a\ne 1$

   $$x+1\ne 1⟹x\ne 0$$

In teeno conditions ka intersection lene par hamein milta hai: x > 2.

Step 2: Denominator ki conditions

Yahan do cheezein hain:

1. Logarithm inside denominator: $e^{2{\log }_{e}x}$ mein ${\log }_{e}x$ tabhi define hoga jab x > 0.

2. Denominator zero nahi hona chahiye:

   Pehle denominator ko simplify karte hain:

   $$e^{2{\log }_{e}x}-(2x+3)=e^{{\log }_e{x}^2}-(2x+3)=x^2-2x-3$$

   Ab, $x^2-2x-3\ne 0$ hona chahiye:

   $$(x-3)(x+1)\ne 0$$
   $$\mathbf{x}\ne 3\text{ aur }\mathbf{x}\ne -1$$

Step 3: Sabhi conditions ka Intersection

Ab hamare paas ye conditions hain:

• Numerator se: $x\in (2,\infty )$

• Denominator ke log se: $x\in (0,\infty )$

• Denominator $\ne 0$ se: $x\ne 3,-1$

Jab hum in sabka common part (intersection) nikalte hain:

Yaani:

Correct Option:

Sahi jawab option (3) $(2,\infty )-\{3\}$ hai.