Algebra

Inequalities

Medium

applied

The solution of $\frac{7x+12}{x-9} < 4$ ; $\neq 9$ is:

$\{x: 9 < x < 16; x \in \mathbb{R}\}$

$\{x: -9 < x < -16; x \in \mathbb{R}\}$

$\{x: -16 < x < 9; x \in \mathbb{R}\}$

$\{x: -16 < x < -9; x \in \mathbb{R}\}$

✅ Correct Option: 3

Move all terms to one side of the inequality:

$\frac{7x+12}{x-9} - 4 < 0$

Express as a single fraction with common denominator:

$\frac{7x+12}{x-9} - \frac{4(x-9)}{x-9} < 0$

$\frac{7x+12 - 4(x-9)}{x-9} < 0$

Simplify the numerator:

$\frac{7x+12 - 4x + 36}{x-9} < 0$

$\frac{3x + 48}{x-9} < 0$

Factor out 3 from the numerator:

$\frac{3(x + 16)}{x-9} < 0$

Find the critical points where the expression equals zero or is undefined:

Numerator = 0: $x + 16 = 0$ → $x = -16$

Denominator = 0: $x - 9 = 0$ → $x = 9$ (undefined)

Test the sign in each interval:

Interval	Test value	Sign of $(x+16)$	Sign of $(x-9)$	Sign of fraction
$x < -16$	$x = -20$	$(-)$	$(-)$	$(+)$
$-16 < x < 9$	$x = 0$	$(+)$	$(-)$	$(-)$
$x > 9$	$x = 10$	$(+)$	$(+)$	$(+)$

The fraction is negative when $-16 < x < 9$ .

Note that $x = 9$ is excluded since the expression is undefined there.

The solution is $\boxed{\{x: -16 < x < 9; x \in \mathbb{R}\}}$

2026: 22 May Shift 1

Match List-I with List-II

List-I	List-II
Inequalities	Solution set
(A) If $\left\vert 3 - 5x \right\vert < 7$ , then $x \in$	(I) $\left[-\dfrac{5}{2}, \infty\right)$
(B) If $3x - 7 > 5x - 1$ , then $x \in$	(II) $(-\infty, -3)$
(C) If $\dfrac{x}{3} > \dfrac{x}{2} + 1$ , then $x \in$	(III) $\left(-\dfrac{4}{5}, 2\right)$
(D) If $5x + 4 \leq 7x + 9$ , then $x \in$	(IV) $(-\infty, -6)$

Choose the correct answer from the options given below:

2025: 13 May Shift 2

If $a > b$ and $c < 0$ , then which of the following is NOT correct?

(A) $ac < bc$

(B) $a + c < b + c$

(D) $ac > bc$

Choose the correct answer from the options given below:

2026: 31 May Shift 2

Which of the following facts are not true ?( Where $R$ is set of real numbers)

A. $|x+y| \le |x|+|y|, x,y \in R$

B. $|x-y| \ge |x|+|y|, x,y \in R$

C. $|x+y| \ge |x|+|y|, x,y \in R$

D. $|x-y| \le |x|+|y|, x,y \in R$

Choose the correct answer from the options given below:

2025: 14 May Shift 2

Which of the following inequalities are NOT correct?

(A) If $a > 1, b > 1,$ then $\log_b a + \log_a b \leq 2$

(B) For any real number $x, (9^x + 9^{1-x}) \geq 9$

(C) If $a,b,c$ are non-zero real numbers of the same sign, then $\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \leq 3$

(D) If $a,b,c$ are three distinct real numbers, then $(a + b)(b + c)(c + a) \geq 8abc$

Choose the correct answer from the options given below:

2025: 21 May Shift 2

If a, b, c are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:

2025: 13 May Shift 1

If $a, b$ and $c$ are positive real numbers, then

Match List-I with List-II

List-I	List-II
(Expression)	(The Least value of the expression)
(A) $(a + b)(b + c)(c + a)$	(I) $8abc$
(B) $(a + b + c)(ab + bc + ca)$	(II) $9a^2b^2c^2$
(C) $(a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)$	(III) $9abc$
(D) $(a + b)^2(b + c)^2(c + a)^2$	(IV) $64a^2b^2c^2$

Choose the correct answer from the options given below:

The solution of 7x+12x−9<4\frac{7x+12}{x-9} < 4x−97x+12​<4; ≠9 \neq 9=9 is:

Solution

Related questions:

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CUET Mathematics Past Year Questions (Topic-Wise)

The solution of $\frac{7x+12}{x-9} < 4$ ; $\neq 9$ is: