The solution set of the given inequation (2x+3)/(x-1)<4 (x≠1) is: Option 1: (-∞,-1][3/2,∞) Option 2: (1,7/2) Option 3: (-∞,1)(7/2,∞) Option 4: (-∞,1][7/2,∞)

Answer: (-∞,1)(7/2,∞). Solution: Bring everything to one side: (2x+3)/(x-1)-4=(7-2x)/(x-1) 0. The critical points are 1 and 7/2, and the quotient is positive outside them. So the solution set is (-∞,1)(7/2,∞).

Applied Mathematics

Algebra

Medium

applied

The solution set of the given inequation $\frac{2x+3}{x-1}<4 (x\neq1)$ is:

$(-\infty,-1]\cup\left[\frac{3}{2},\infty\right)$

$\left(1,\frac{7}{2}\right)$

$(-\infty,1)\cup\left(\frac{7}{2},\infty\right)$

$(-\infty,1]\cup\left[\frac{7}{2},\infty\right)$

✅ Correct Option: 3

Bring everything to one side: $\frac{2x+3}{x-1}-4=\frac{7-2x}{x-1}<0$ , i.e. $\frac{2x-7}{x-1}>0$ .

The critical points are $1$ and $\frac{7}{2}$ , and the quotient is positive outside them.

So the solution set is $(-\infty,1)\cup\left(\frac{7}{2},\infty\right)$ .

2022: 25 May Shift 1

If $\begin{bmatrix} 2x + 3 & y + 1 \\ 2x - z & 3y \end{bmatrix} = \begin{bmatrix} 7 & -2 \\ -4 & -9 \end{bmatrix}$ , then matrix $\begin{bmatrix} 3z + 1 & 4 \\ -2 & 4 \end{bmatrix}$ is equal to:

2023: 7 June Shift 1

Let $P = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix}$ , $Q = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$ , $R = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$ , then the matrix S such that QS - RP = 0 will be :

2023: 11 Aug Shift 1

If $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$ , then AB is equals to :

2023: 7 June Shift 1

If A and B are events such that $P(A/B) = P(B/A)$ , then :

2023: 11 Aug Shift 1

Solution for the inequality $|3x - 7| \leq 2$ is :

2026: 14 May Shift 2

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	A matrix which is both symmetric and skew symmetric	I.	symmetric matrix
B.	All positive integral powers of a symmetric matrix	II.	unit matrix
C.	Positive odd integral powers of a skew-symmetric matrix	III.	null matrix
D.	A square matrix $A = [a_{ij}]_{n\times n}$ with $a_{ij} = \begin{cases}0, & \forall\, i \neq j \\ 1, & \forall\, i = j\end{cases}$	IV.	skew-symmetric matrix

Choose the correct answer from the options given below:

The solution set of the given inequation 2x+3x−1<4(x≠1)\frac{2x+3}{x-1}<4 (x\neq1)x−12x+3​<4(x=1) is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The solution set of the given inequation $\frac{2x+3}{x-1}<4 (x\neq1)$ is: