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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The \ number \ of \ integral \ solutions \ x \ of \ \log_{\left(x + \frac{7}{2}\right)} \left( \frac{x-7}{2x-3} \right)^2 \ge 0 \ is:

Choose the correct answer:

Explanation

Step 1: Logarithm ki Conditions (Domain)

Logarithm tabhi define hota hai jab:

Base > 0 aur Base $\neq$ 1:

$x + \frac{7}{2} > 0 \implies x > -3.5$

$x + \frac{7}{2} \neq 1 \implies x \neq -2.5$

Argument > 0:

$\left( \frac{x-7}{2x-3} \right)^2 > 0$

Ye hamesha true hoga siwaye jab $x = 7$ (numerator 0 ho jaye) ya $x = 1.5$ (denominator undefined ho jaye). Toh $x \neq 7$ aur $x \neq 1.5$ .

Step 2: Inequality ko Solve karna ( $\log_b a \ge 0$ )

Iske do cases bante hain base ki value ke aadhar par:

Case 1: Base > 1 ( $x + \frac{7}{2} > 1 \implies x > -2.5$ )

Jab base 1 se bada hota hai, toh inequality ka sign nahi badalta:

$\left( \frac{x-7}{2x-3} \right)^2 \ge \left( x + \frac{7}{2} \right)^0$

$\left( \frac{x-7}{2x-3} \right)^2 \ge 1 \implies \frac{(x-7)^2}{(2x-3)^2} - 1 \ge 0$

$\frac{(x-7)^2 - (2x-3)^2}{(2x-3)^2} \ge 0$

$(x-7 - 2x + 3)(x-7 + 2x - 3) \ge 0$

$(-x - 4)(3x - 10) \ge 0 \implies (x + 4)(3x - 10) \le 0$

Solution: $x \in [-4, \frac{10}{3}]$ .

Condition $x > -2.5$ ke saath intersect karne par: $x \in (-2.5, 3.33]$ .

Is range mein integers: $\{-2, -1, 0, 1, 2, 3\}$ . (Lekin $x \neq 1.5$ , jo ki integers mein nahi hai). Count = 6.

Case 2: 0 < Base < 1 ( $-3.5 < x < -2.5$ )

Jab base 1 se chhota ho, toh sign badal jata hai:

$\left( \frac{x-7}{2x-3} \right)^2 \le 1 \implies (x + 4)(3x - 10) \ge 0$

Solution: $x \le -4$ ya $x \ge \frac{10}{3}$ .

Lekin base ki condition $-3.5 < x < -2.5$ hai, jisme ye solution set fit nahi baithta. Isliye yahan se koi integer solution nahi milega.

Step 3: Final Count

Integral solutions hain: $\{-2, -1, 0, 1, 2, 3\}$ .

Total number of solutions = 6.