Solution of \( \displaystyle\frac{x^2-4x+7}{x^2-7x+12}\le\displaystyle\frac{2}{3} \) is/are:
(A) \( x\in[-3,1] \)
(B) \( x\in[3,4] \)
(C) \( x\in(1,3] \)
(D) \( x\in(3,4) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
Since the denominator can be negative, bring everything to one side instead of cross-multiplying:
\( \displaystyle \frac{3(x^2-4x+7)-2(x^2-7x+12)}{3(x^2-7x+12)}\le 0 \)
\( \displaystyle \frac{x^2+2x-3}{3(x^2-7x+12)}\le 0 \)
\( \displaystyle \frac{(x+3)(x-1)}{(x-3)(x-4)}\le 0 \)
The critical points are −3, 1, 3, 4. Checking the sign on each interval, the expression is ≤ 0 on \( [-3,1] \) and on \( (3,4) \); x = 3 and x = 4 are excluded, since the denominator is 0 there.
So the solution set is \( [-3,1]\cup(3,4) \).
Checking the statements
(A) \( x\in[-3,1] \) — Correct. This is part of the solution set.
(B) \( x\in[3,4] \) — Incorrect. x = 3 and x = 4 make the denominator 0.
(C) \( x\in(1,3] \) — Incorrect. At x = 2 the expression is \( \frac{5}{2}\gt 0 \), and x = 3 is not allowed.
(D) \( x\in(3,4) \) — Correct. This is part of the solution set.