JEE Mathematics 2023 Question with Solution

Given: $\log_2 \left( \frac{x - 7}{2x - 3} \right) \geq 0$

Find: The number of integral solutions.

Since the base is $2>1$, we use

$$\log_2 t \geq 0 \iff t \geq 1$$

with the argument defined. So we solve

$$\frac{x - 7}{2x - 3} \geq 1$$

Now,

$$\frac{x - 7}{2x - 3} - 1 \geq 0$$ $$\frac{x - 7 - (2x - 3)}{2x - 3} \geq 0$$ $$\frac{-x - 4}{2x - 3} \geq 0$$ $$\frac{x + 4}{2x - 3} \leq 0$$

Critical points are $x=-4$ and $x=\frac{3}{2}$. Using sign analysis,

$$-4 \leq x < \frac{3}{2}$$

This interval already keeps the logarithm argument positive because the quantity is at least $1$ there.

Hence the integral values are

$$-4,-3,-2,-1,0,1$$

So the number of integral solutions is $6$. Therefore, the correct option is D.

The solution states the correct option is D, although some intermediate domain discussion there is incomplete.

Given: $\log_2 \left( \frac{x - 7}{2x - 3} \right) \geq 0$

Find: Count of integral values of $x$.

Let

$$y = \frac{x - 7}{2x - 3}$$

Then

$$\log_2 y \geq 0 \Rightarrow y \geq 1$$

because the logarithm with base $2$ is increasing.

So,

$$\frac{x - 7}{2x - 3} \geq 1$$ $$\frac{x - 7}{2x - 3} - 1 \geq 0$$ $$\frac{x - 7 - 2x + 3}{2x - 3} \geq 0$$ $$\frac{-x - 4}{2x - 3} \geq 0$$ $$\frac{x + 4}{2x - 3} \leq 0$$

Now examine intervals formed by $x=-4$ and $x=\frac{3}{2}$:

• For $x<-4$, numerator is negative and denominator is negative, so the ratio is positive, which does not satisfy $\leq 0$.
• For $-4<x<\frac{3}{2}$, numerator is positive and denominator is negative, so the ratio is negative, which satisfies $\leq 0$.
• For $x>\frac{3}{2}$, numerator is positive and denominator is positive, so the ratio is positive, which does not satisfy $\leq 0$.
• At $x=-4$, the ratio is $0$, so it is included.
• At $x=\frac{3}{2}$, the expression is undefined, so it is excluded.

Therefore,

$$-4 \leq x < \frac{3}{2}$$

The integers in this interval are $-4,-3,-2,-1,0,1$, giving $6$ integral solutions.

Therefore, the correct option is D.