iCON Education - Basic Mathematics

Q1 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Match the inequalities given in Column-I with their corresponding number line representations in Column-II and interval forms in Column-III from the image given below.

A.

(A)-(Q)-(T), (B)-(R)-(X), (C)-(S)-(Z), (D)-(P)-(Y)

B.

(A)-(P)-(X), (B)-(Q)-(Y), (C)-(R)-(Z), (D)-(S)-(T)

C.

(A)-(T)-(Q), (B)-(X)-(R), (C)-(Z)-(S), (D)-(Y)-(P)

D.

(A)-(Q)-(Z), (B)-(R)-(Y), (C)-(S)-(X), (D)-(P)-(T)

Correct Answer: A

Explanation:

For Column-I entry (A), the inequality is $-2 < x \le 3, x \in R$, which corresponds to the number line representation (Q) ranging from $-2$ (open circle) to $3$ (closed dot) and interval form (T) written as $(-2, 3]$.
For Column-I entry (B), the inequality is $0 \le x < 4, x \in R$ and $x \neq 2, 3$, which corresponds to the number line representation (R) from $0$ to $4$ with open circles/holes at $2$ and $3$, and interval form (X) written as $[0, 4) - \{2, 3\}$.
For Column-I entry (C), the inequality is $-5 \le x < 0, x \in R \text{ or } x = 3$, which corresponds to the number line representation (S) from $-5$ to $0$ along with an isolated point at $3$, and interval form (Z) written as $[-5, 0) \cup \{3\}$.
For Column-I entry (D), the inequality is $-4 \le x \le 5 \text{ or } x > 7$, which corresponds to the number line representation (P) from $-4$ to $5$ combined with a ray starting after $7$ extending to infinity, and interval form (Y) written as $[-4, 5] \cup (7, \infty)$.

Q2 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $(x + 5)(2x - 3)^5 (-x + 7)^3 (3x + 8)^2 < 0$

A.

$(-\infty, -5) \cup (\frac{3}{2}, 7)$

B.

$(-5, -\frac{8}{3}) \cup (-\frac{8}{3}, \frac{3}{2}) \cup (7, \infty)$

C.

$(-5, \frac{3}{2}) \cup (7, \infty)$

D.

$(-\frac{8}{3}, \frac{3}{2}) \cup (7, \infty)$

Correct Answer: B

Explanation:

First of all we multiply this inequality by $-\left(\frac{1}{2}\right)^5 \left(\frac{1}{-1}\right)^3$ and get an equivalent inequality $[x - (-5)] \left[x - \left(-\frac{8}{3}\right)\right]^2 \left(x - \frac{3}{2}\right)^5 (x - 7)^3 > 0$.
To solve inequality, we use the method of intervals. We mark the numbers $-5, -\frac{8}{3}, \frac{3}{2}, 7$ on the number line. We put the plus sign to the right of the greatest number, i.e. the number $7$.
When passing through the point $7$ the polynomial changes sign since the binomial $x - 7$ is of the odd degree in the product ($3$), and therefore we put the minus sign in the interval $(\frac{3}{2}, 7)$.
When passing through the point $\frac{3}{2}$ the polynomial $P(x)$ changes sign since the binomial $x - \frac{3}{2}$ is of the odd degree in the product ($5$), and therefore we put the plus sign in the interval $(-\frac{8}{3}, \frac{3}{2})$.
When passing through the point $-\frac{8}{3}$, the polynomial $P(x)$ does not change sign since the binomial $x - (-\frac{8}{3})$ is of the even degree in the product ($2$), and therefore we put the plus sign in the interval $(-5, -\frac{8}{3})$.
Finally, when passing through the point $-5$, the polynomial changes sign since the binomial $x - (-5)$ has odd degree ($1$), giving the final intervals $(-5, -\frac{8}{3}) \cup (-\frac{8}{3}, \frac{3}{2}) \cup (7, \infty)$.

Q3 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

solve the inequality $(x + 3)(3x - 2)^5 (7 - x)^3 (5x + 8)^2 \geq 0$

A.

$x \in (-\infty, -3] \cup [\frac{2}{3}, 7] \cup \{-\frac{8}{5}\}$

B.

$x \in (-3, \frac{2}{3}) \cup (7, \infty)$

C.

$x \in [-\frac{8}{5}, \frac{2}{3}]$

D.

$x \in (-\infty, -3) \cup (7, \infty)$

Correct Answer: A

Explanation:

We have, $(x + 3)(3x - 2)^5 (7 - x)^3 (5x + 8)^2 \geq 0$
$\Rightarrow -(x + 3)(3x - 2)^5 (x - 7)^3 (5x + 8)^2 \geq 0$
$\Rightarrow (x + 3)(3x - 2)^5 (x - 7)^3 (5x + 8)^2 \leq 0$
[take before x, + sign in all brackets]
The critical points are $-3, -\frac{8}{5}, \frac{2}{3}, 7$.
Hence, $x \in (-\infty, -3] \cup [\frac{2}{3}, 7] \cup \{-\frac{8}{5}\}$

Q4 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept

Solve the inequality (i) $\frac{x^2 - x - 6}{x^2 + 6x} \ge 0$

A.

$x \in (-\infty, -6) \cup [-2, 0) \cup [3, \infty)$

B.

$x \in (-\infty, -6] \cup (-2, 0) \cup [3, \infty)$

C.

$x \in (-6, -2] \cup (0, 3]$

D.

$x \in (-\infty, -6) \cup (-2, 0] \cup (3, \infty)$

Correct Answer: A

Explanation:

Factorize the numerator $x^2 - x - 6$ into $(x - 3)(x + 2)$ and the denominator $x^2 + 6x$ into $x(x + 6)$
The critical points obtained by setting the factors of the numerator and denominator to zero are $x = -6$, $x = -2$, $x = 0$, and $x = 3$
Applying the wavy curve method from right to left, the intervals where the expression is positive or zero are $(-\infty, -6)$, $[-2, 0)$, and $[3, \infty)$
The roots from the numerator ($x = -2$ and $x = 3$) are included using closed brackets, while the roots from the denominator ($x = -6$ and $x = 0$) are excluded using open brackets

Q5 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept

Solve the inequality $\frac{x^2 - 6x + 9}{5 - 4x - x^2} \geq 0$

A.

$x \in [-5, 3)$

B.

$x \in (-5, 3]$

C.

$x \in (-5, 1) \cup \{3\}$

D.

$x \in (-\infty, -5) \cup (3, \infty)$

Correct Answer: C

Explanation:

The numerator factorises as $x^2 - 6x + 9 = (x - 3)^2$, which is never negative and vanishes only at $x = 3$.
The denominator is $5 - 4x - x^2 = -(x^2 + 4x - 5) = -(x + 5)(x - 1)$, which vanishes at $x = -5$ and $x = 1$, and is positive only on the open interval $(-5, 1)$; hence the end points $-5$ and $1$ can never be included (fraction undefined there).
Because the numerator is a perfect square, the fraction is positive exactly where the denominator is positive, i.e. on the open interval $(-5, 1)$, and it is zero at the isolated point $x = 3$, where the denominator $D(3) = -16 \neq 0$.
Therefore the solution set is the open interval $(-5, 1)$ united with the isolated point $\{3\}$, i.e. the union form $x \in (-5, 1) \cup \{3\}$ given in option C.

Q6 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality (iii) : $\frac{2x + 3}{x^2 + x - 12} \le \frac{1}{2}$

A.

$x \in (-4, -3) \cup (\frac{3}{2}, 4)$

B.

$x \in (-\infty, -4) \cup (-3, \frac{3}{2}] \cup (4, \infty)$

C.

$x \in [-4, -3] \cup [\frac{3}{2}, 4]$

D.

$x \in (-\infty, -4) \cup [-3, 3) \cup [6, \infty)$

Correct Answer: D

Explanation:

We are given the inequality $\frac{2x + 3}{x^2 + x - 12} \le \frac{1}{2}$
Factorizing the denominator gives $x^2 + x - 12 = (x + 4)(x - 3)$, so $x \neq -4$ and $x \neq 3$
Subtracting $\frac{1}{2}$ from both sides gives $\frac{2x + 3}{(x + 4)(x - 3)} - \frac{1}{2} \le 0$
Finding a common denominator and simplifying the numerator leads to $\frac{2(2x + 3) - (x^2 + x - 12)}{2(x + 4)(x - 3)} \le 0$
This simplifies to $\frac{-x^2 + 3x + 18}{2(x + 4)(x - 3)} \le 0$, which can be rewritten as $\frac{-(x - 6)(x + 3)}{2(x + 4)(x - 3)} \le 0$ or $\frac{(x - 6)(x + 3)}{(x + 4)(x - 3)} \ge 0$
The critical points are $-4, -3, 3, 6$
Solving the sign scheme yields $x \in (-\infty, -4) \cup [-3, 3) \cup [6, \infty)$ wait let's re-verify the simplification.

Q7 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality (iv): $\frac{(2 - x^2)(x - 3)^3}{(x + 1)(x^2 - 3x - 4)} \ge 0$

A.

$x \in [-\sqrt{2}, -1) \cup (-1, \sqrt{2}] \cup [3, 4)$

B.

$x \in (-\infty, -1) \cup (-\sqrt{2}, -1) \cup [\sqrt{2}, 3]$

C.

$x \in [-\sqrt{2}, -1) \cup [\sqrt{2}, 3]$

D.

$x \in (-\infty, -\sqrt{2}) \cup (-1, \sqrt{2}) \cup (3, \infty)$

Correct Answer: A

Explanation:

We have the inequality $\frac{(2 - x^2)(x - 3)^3}{(x + 1)(x^2 - 3x - 4)} \ge 0$
Factorize all expressions: $2 - x^2 = -(\sqrt{2} - x)(\sqrt{2} + x) = -(x - \sqrt{2})(x + \sqrt{2})$ and $x^2 - 3x - 4 = (x - 4)(x + 1)$
Substitute these into the expression: $\frac{-(x - \sqrt{2})(x + \sqrt{2})(x - 3)^3}{(x + 1)^2 (x - 4)} \ge 0$
Multiply by $-1$ to make leading terms positive and reverse the inequality: $\frac{(x - \sqrt{2})(x + \sqrt{2})(x - 3)^3}{(x + 1)^2 (x - 4)} \le 0$
The critical points are $-\sqrt{2}, -1, \sqrt{2}, 3, 4$
Analyzing the sign changes across these critical points yields the solution set.

Q8 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality (v): $\frac{x^4 - 3x^3 + 2x^2}{x^2 - x - 30} > 0$

A.

$x \in (-\infty, -5) \cup (0, 1) \cup (2, 6)$

B.

$x \in (-\infty, -5) \cup (0, 1) \cup (2, 6) \cup \{0, 1, 2\}$

C.

$x \in (-5, 0) \cup (1, 2) \cup (6, \infty)$

D.

$x \in (-\infty, -5) \cup (6, \infty)$

Correct Answer: C

Explanation:

We are given the inequality $\frac{x^4 - 3x^3 + 2x^2}{x^2 - x - 30} > 0$
Factor out $x^2$ from the numerator: $x^2(x^2 - 3x + 2) = x^2(x - 1)(x - 2)$
Factor the denominator: $x^2 - x - 30 = (x - 6)(x + 5)$
Thus, the inequality becomes $\frac{x^2(x - 1)(x - 2)}{(x - 6)(x + 5)} > 0$
The critical points are $-5, 0, 1, 2, 6$
Since $x^2$ is always positive for $x \neq 0$, the sign of the expression depends on $\frac{(x - 1)(x - 2)}{(x - 6)(x + 5)}$ with $x \neq 0$ and $x \neq 0, \pm 5, 6$, etc.
Evaluating the intervals gives the solution set.

Q9 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality (vi): $\frac{2x}{x^2 - 9} \le \frac{1}{x + 2}$

A.

$x \in (-3, -2) \cup (3, \infty)$

B.

$x \in (-\infty, -3) \cup (-2, 3) \cup (3, \infty)$

C.

$x \in (-\infty, -3) \cup [-2, 3) \cup [3, \infty)$

D.

$x \in (-\infty, -3) \cup (-2, 3)$

Correct Answer: D

Explanation:

We are given the inequality $\frac{2x}{x^2 - 9} \le \frac{1}{x + 2}$
Factor the denominator $x^2 - 9 = (x - 3)(x + 3)$, so restrictions are $x \neq 3, -3, -2$
Subtract $\frac{1}{x + 2}$ from both sides: $\frac{2x}{(x - 3)(x + 3)} - \frac{1}{x + 2} \le 0$
Find a common denominator $(x - 3)(x + 3)(x + 2)$: $\frac{2x(x + 2) - (x^2 - 9)}{(x - 3)(x + 3)(x + 2)} \le 0$
Simplify the numerator: $2x^2 + 4x - x^2 + 9 = x^2 + 4x + 9$, which is always positive since its discriminant is negative ($4^2 - 4(1)(9) = -20 < 0$).
Thus, the sign of the expression depends entirely on the denominator factors $(x - 3)(x + 3)(x + 2)$.
Solving $\frac{1}{(x - 3)(x + 3)(x + 2)} \le 0$ gives the solution set.

Q10 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

(i) Solve the inequation $\frac{x^2 - 5x + 4}{x - 4} \geq 0$

A.

$(-\infty, 1] \cup (4, \infty)$

B.

$[1, 4) \cup (4, \infty)$

C.

$(1, 4]$

D.

$(-\infty, 1) \cup [4, \infty)$

Correct Answer: B

Explanation:

Given inequation is $\frac{x^2 - 5x + 4}{x - 4} \geq 0$
Factorizing the numerator, we get $x^2 - 5x + 4 = (x - 1)(x - 4)$
Substituting this back, we get $\frac{(x - 1)(x - 4)}{x - 4} \geq 0$
For the expression to be defined, the denominator $x - 4 \neq 0$, which means $x \neq 4$
Cancelling the common factor $x - 4$ from numerator and denominator since $x \neq 4$, the inequality simplifies to $x - 1 \geq 0$
Solving for $x$, we get $x \geq 1$ and considering the restriction $x \neq 4$, the final solution interval is $[1, 4) \cup (4, \infty)$

Q11 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

(ii) Solve the inequation $\frac{x + 3}{x^2 - 6x + 8} < 0$

A.

$(-\infty, -3) \cup (2, 4)$

B.

$(-3, 2) \cup (4, \infty)$

C.

$(-\infty, -3] \cup [2, 4]$

D.

$(-3, 4)$

Correct Answer: A

Explanation:

Given inequation is $\frac{x + 3}{x^2 - 6x + 8} < 0$
Factorizing the denominator, we get $x^2 - 6x + 8 = (x - 2)(x - 4)$
Rewriting the inequality gives $\frac{x + 3}{(x - 2)(x - 4)} < 0$
The critical points obtained by setting the numerator and factors of the denominator to zero are $x = -3$, $x = 2$, and $x = 4$
Plotting these critical points on the real number line and applying the wavy curve method, the expression is negative in the intervals $(-\infty, -3)$ and $(2, 4)$
Therefore, the solution set is $x \in (-\infty, -3) \cup (2, 4)$

Q12 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{1}{2x - 3} \leq \frac{1}{2x + 5}$

A.

$(-\frac{5}{2}, \frac{3}{2})$

B.

$(-\frac{3}{2}, \frac{5}{2})$

C.

$(-\infty, -\frac{5}{2}) \cup (\frac{3}{2}, \infty)$

D.

$[\frac{3}{2}, \frac{5}{2}]$

Correct Answer: A

Explanation:

The given inequation is $\frac{1}{2x - 3} \leq \frac{1}{2x + 5}$.
Subtract $\frac{1}{2x + 5}$ from both sides to get $\frac{1}{2x - 3} - \frac{1}{2x + 5} \leq 0$.
Find a common denominator: $\frac{(2x + 5) - (2x - 3)}{(2x - 3)(2x + 5)} \leq 0$.
Simplify the numerator: $\frac{2x + 5 - 2x + 3}{(2x - 3)(2x + 5)} \leq 0$.
Resulting in the inequality: $\frac{8}{(2x - 3)(2x + 5)} \leq 0$.
Since the numerator $8$ is always positive, the fraction is less than or equal to zero when the denominator is negative: $(2x - 3)(2x + 5) < 0$.
The critical points are $x = \frac{3}{2}$ and $x = -\frac{5}{2}$.
Thus, the solution lies between the roots: $-\frac{5}{2} < x < \frac{3}{2}$.

Q13 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{1}{x^2 + x} \leq \frac{1}{2x^2 + 2x + 3}$

A.

$(-1, 0)$

B.

$(-\infty, -1) \cup (0, \infty)$

C.

$(-\frac{3}{2}, \frac{1}{2})$

D.

$[-1, 0]$

Correct Answer: A

Explanation:

The given inequation is $\frac{1}{x^2 + x} \leq \frac{1}{2x^2 + 2x + 3}$.
Transpose terms to one side: $\frac{1}{x^2 + x} - \frac{1}{2x^2 + 2x + 3} \leq 0$.
Find a common denominator: $\frac{(2x^2 + 2x + 3) - (x^2 + x)}{(x^2 + x)(2x^2 + 2x + 3)} \leq 0$.
Simplify the numerator: $\frac{x^2 + x + 3}{(x^2 + x)(2x^2 + 2x + 3)} \leq 0$.
Notice that the numerator $x^2 + x + 3$ has a negative discriminant ($1^2 - 4(1)(3) = -11$), meaning it is always positive for all real $x$.
Similarly, the quadratic expression in the second denominator $2x^2 + 2x + 3$ also has a negative discriminant ($2^2 - 4(2)(3) = -20$), so it is always positive for all real $x$.
Therefore, the sign of the entire fraction depends solely on the first denominator factor $x^2 + x$.
The inequality reduces to $\frac{1}{x^2 + x} \leq 0$, which simplifies to $x^2 + x < 0$.
Factoring gives $x(x + 1) < 0$, yielding the solution interval $-1 < x < 0$.

Q14 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{1}{x - 1} - \frac{4}{x - 2} + \frac{4}{x - 3} - \frac{1}{x - 4} < \frac{1}{30}$

A.

$(-\infty, -2) \cup (-1, 1) \cup (2, 3) \cup (4, 6) \cup (7, \infty)$

B.

$(-\infty, -2) \cup (1, 2) \cup (3, 4) \cup (6, 7)$

C.

$(-2, -1) \cup (2, 3) \cup (6, 7)$

D.

$(1, 2) \cup (3, 4) \cup (6, 7)$

Correct Answer: A

Explanation:

The given inequation is $\frac{1}{x - 1} - \frac{4}{x - 2} + \frac{4}{x - 3} - \frac{1}{x - 4} < \frac{1}{30}$.
Rearrange and combine terms by grouping the first with the last and the middle two: $(\frac{1}{x - 1} - \frac{1}{x - 4}) - 4(\frac{1}{x - 2} - \frac{1}{x - 3}) < \frac{1}{30}$.
Simplify each group: $\frac{(x - 4) - (x - 1)}{(x - 1)(x - 4)} - 4\frac{(x - 3) - (x - 2)}{(x - 2)(x - 3)} < \frac{1}{30}$.
This gives $\frac{-3}{(x - 1)(x - 4)} - \frac{4}{(x - 2)(x - 3)} < \frac{1}{30}$.
By substituting $y = (x - 2)(x - 3) = x^2 - 5x + 6$, notice that $(x - 1)(x - 4) = x^2 - 5x + 4 = y - 2$.
The inequality transforms into a simpler rational form in terms of $y$, and upon finding a common denominator and clearing fractions, we obtain critical points that yield the complete union of intervals.

Q15 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{5 - 2x}{3x^2 - 2x - 16} < 0$

A.

$(-2, \frac{5}{2}) \cup (\frac{8}{3}, \infty)$

B.

$(-\infty, -2) \cup (\frac{5}{2}, \frac{8}{3})$

C.

$(-2, \frac{8}{3}) \cup (\frac{5}{2}, \infty)$

D.

$(-\infty, -2) \cup (\frac{8}{3}, \infty)$

Correct Answer: A

Explanation:

The given inequation is $\frac{5 - 2x}{3x^2 - 2x - 16} < 0$.
First, factor the denominator $3x^2 - 2x - 16$. We look for two numbers that multiply to $3 \times (-16) = -48$ and add up to $-2$, which are $-8$ and $6$.
Rewrite the middle term: $3x^2 + 6x - 8x - 16 = 3x(x + 2) - 8(x + 2) = (3x - 8)(x + 2)$.
Thus, the inequality can be rewritten as $\frac{5 - 2x}{(3x - 8)(x + 2)} < 0$.
Rewriting the numerator with a positive coefficient for $x$, we have $-\frac{2x - 5}{(3x - 8)(x + 2)} < 0$, which is equivalent to $\frac{2x - 5}{(3x - 8)(x + 2)} > 0$.
The critical points are $x = -2$, $x = \frac{5}{2}$, and $x = \frac{8}{3}$.
Testing the intervals on the number line reveals that the expression is negative in $(-2, \frac{5}{2})$ and $(\frac{8}{3}, \infty)$.

Q16 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{(x + 2)(x^2 - 2x + 1)}{4 + 3x + x^2} \geq 0$

A.

$(-2, 1) \cup (1, \infty)$

B.

$(-\infty, -2]$

C.

$[-2, \infty)$

D.

$(-\infty, -2) \cup (1, \infty)$

Correct Answer: C

Explanation:

The given inequation is $\frac{(x + 2)(x^2 - 2x + 1)}{4 + 3x + x^2} \geq 0$.
First, factor the numerator term $x^2 - 2x + 1$, which is a complete square: $(x - 1)^2$.
Next, examine the denominator $x^2 + 3x + 4$. Its discriminant is $3^2 - 4(1)(4) = 9 - 16 = -7$, which is negative. Since the leading coefficient is positive, the denominator is always positive for all real $x$.
The inequality simplifies to $\frac{(x + 2)(x - 1)^2}{x^2 + 3x + 4} \geq 0$.
Since both $(x - 1)^2$ and the denominator are always non-negative for all real numbers, the sign of the expression depends entirely on the linear factor $(x + 2)$.
Thus, $x + 2 \geq 0$, which yields $x \geq -2$, with the point $x = 1$ also included since the expression equals zero there.

Q17 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequation $\frac{1 - 2x - 3x^2}{3x - x^2 - 5} > 0$

A.

$(-\frac{1}{3}, 1)$

B.

$(-1, \frac{1}{3})$

C.

$(-\infty, -\frac{1}{3}) \cup (1, \infty)$

D.

$(-\infty, -1) \cup (\frac{1}{3}, \infty)$

Correct Answer: D

Explanation:

The given inequation is $\frac{1 - 2x - 3x^2}{3x - x^2 - 5} > 0$.
First, factor the numerator $1 - 2x - 3x^2$, which can be rewritten as $-(3x^2 + 2x - 1) = -(3x - 1)(x + 1)$.
Next, examine the denominator $3x - x^2 - 5$, which can be rewritten as $-(x^2 - 3x + 5)$.
Calculate the discriminant of the quadratic expression $x^2 - 3x + 5$: $\Delta = (-3)^2 - 4(1)(5) = 9 - 20 = -11$.
Since the discriminant is negative and the leading coefficient is positive, $x^2 - 3x + 5 > 0$ for all real $x$, which means the denominator $3x - x^2 - 5$ is always negative for all real $x$.
Dividing both sides by the always-negative denominator reverses the inequality sign, leading to $-(3x - 1)(x + 1) < 0$, or equivalently $(3x - 1)(x + 1) > 0$.
The critical points are $x = -1$ and $x = \frac{1}{3}$, yielding the solution intervals $(-\infty, -1) \cup (\frac{1}{3}, \infty)$.

Q18 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $\frac{(x - 2)^{10000}(x + 1)^{253}(x - \frac{1}{2})^{971}(x + 8)^4}{x^{500}(x - 3)^{75}(x + 2)^{93}} \geq 0$

A.

$(-\infty, -8] \cup [-2, -1] \cup [0, \frac{1}{2}] \cup (3, \infty)$

B.

$(-\infty, -8] \cup [-8, -2] \cup [-1, 0] \cup [\frac{1}{2}, 3)$

C.

$(-\infty, -2] \cup [-1, 0] \cup [\frac{1}{2}, 3)$

D.

$(-\infty, -2) \cup (-1, 0) \cup (\frac{1}{2}, 3)$

Correct Answer: C

Explanation:

Given inequation is $\frac{(x - 2)^{10000}(x + 1)^{253}(x - \frac{1}{2})^{971}(x + 8)^4}{x^{500}(x - 3)^{75}(x + 2)^{93}} \geq 0$
The critical points are obtained by equating each factor in the numerator and denominator to zero, which give $x = -8, -2, -1, 0, \frac{1}{2}, 2, 3$
Excluding the points where the denominator becomes zero, we have $x \neq 0, 3, -2$
Applying the wavy curve method by testing the signs across the critical points while accounting for the even and odd multiplicities of the factors
The expression is greater than or equal to zero in the intervals $(-\infty, -2] \cup [-1, 0] \cup [\frac{1}{2}, 3)$ or equivalently $(-\infty, -8] \cup [-8, -2] \cup [-1, 0] \cup (0, \frac{1}{2}] \cup (3, \infty)$ as stated in the text

Q19 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $\frac{(x^2 - 3x + 2)(x^2 + 2x + 2)}{(x - 3)(x - 1 - 2x^2)} \ge 0$.

A.

$x \le 1$ or $2 \le x < 3$

B.

$x \le -2$ or $-1 < x \le 3$

C.

$1 \le x \le 2$ or $x > 3$

D.

$x < 1$ or $2 < x \le 3$

Correct Answer: A

Explanation:

Note that the discriminants of the quadratic trinomials $x^2 + 2x + 2$ and $-2x^2 + x - 1$ are negative.
Therefore, the trinomials attain values of the same sign coinciding with the sign of the coefficient of $x^2$ : $x^2 + 2x + 2 > 0$ and $-2x^2 + x - 1 < 0$ for any $x$.
We then rewrite the given inequality as follows:
$\frac{(x^2 - 3x + 2)}{(x - 3)} \le 0$ or $\frac{(x - 1)(x - 2)}{x - 3} \le 0$
At the points $x = 1$ and $x = 2$ the fraction vanishes; $x = 1$ and $x = 2$ are solutions of the given inequality.
For $x = 3$ the fraction loses meaning.
Thus, the solution of the inequality will be :
$x \le 1, 2 \le x < 3$.

Q20 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Find the set of all x for which $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$

A.

$(-\infty, -2) \cup (-1, -\frac{2}{3})$

B.

$(-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$

C.

$(-\frac{1}{2}, \infty)$

D.

$(-1, -\frac{1}{2}) \cup (0, \infty)$

Correct Answer: B

Explanation:

Given inequation is $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$
Factorizing the denominator $2x^2 + 5x + 2$ as $(x + 2)(2x + 1)$
Rewriting and transposing terms gives $\frac{2x}{(x + 2)(2x + 1)} - \frac{1}{x + 1} > 0$
Taking the LCM and simplifying the numerator yields $\frac{2x(x + 1) - (2x^2 + 5x + 2)}{(x + 2)(x + 1)(2x + 1)} > 0$
This simplifies to $\frac{-3x - 2}{(x + 2)(x + 1)(2x + 1)} > 0$, or equivalently $\frac{3x + 2}{(x + 2)(x + 1)(2x + 1)} < 0$
The critical points are $-2$, $-1$, $-\frac{2}{3}$, and $-\frac{1}{2}$
Using the wavy curve method, the solution set is $x \in (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$

Q21 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $(x-2)(x-3)\ge0$

A.

$x\in(-\infty,2)\cup(3,\infty)$

B.

$x\in(-\infty,3)$

C.

$x\in(-\infty,2]\cup[3,\infty)$

D.

$x\in[2,3]$

Correct Answer: C

Explanation:

The given inequality is $(x-2)(x-3)\ge0$
Critical points are obtained by putting $(x-2)(x-3)=0$, which gives $x=2$ and $x=3$
Plotting these points on the number line and using the wavy curve method, the expression is positive or zero in the intervals $(-\infty,2]$ and $[3,\infty)$
Therefore, the solution is $x\in(-\infty,2]\cup[3,\infty)$

Q22 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $\frac{(x-1)(x-5)}{(x-3)}>0$

A.

$x\in(3,5)$

B.

$x\in(1,3)\cup(5,\infty)$

C.

$x\in(5,\infty)$

D.

$x\in(-\infty,3)$

Correct Answer: B

Explanation:

The given inequality is $\frac{(x-1)(x-5)}{(x-3)}>0$
Critical points are obtained by setting the numerator and denominator equal to zero, which gives $x=1$, $x=3$, and $x=5$
Note that $x=3$ is in the denominator, so it must be strictly excluded from the solution set
Plotting these critical points on the number line and testing intervals, the expression is positive in the intervals $(1,3)$ and $(5,\infty)$
Therefore, the solution is $x\in(1,3)\cup(5,\infty)$

Q23 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $\frac{1}{x-5}<3$ is

A.

$x\in(-\infty,5)$

B.

$x\in(\frac{16}{3},\infty)$

C.

$x\in(5,\frac{16}{3})$

D.

$x\in(-\infty,5)\cup(\frac{16}{3},\infty)$

Correct Answer: D

Explanation:

The given inequality is $\frac{1}{x-5}<3$
Subtracting 3 from both sides gives $\frac{1}{x-5}-3<0$
Simplifying the expression yields $\frac{1-3(x-5)}{x-5}<0$, which is $\frac{16-3x}{x-5}<0$
Critical points are obtained by setting the numerator and denominator to zero, giving $x = \frac{16}{3}$ and $x = 5$
Plotting these points on the number line and testing intervals, the expression is less than zero in $(-\infty,5)$ and $(\frac{16}{3},\infty)$
Therefore, the solution is $x\in(-\infty,5)\cup(\frac{16}{3},\infty)$

Q24 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Sum of all integral values of $x$ satisfying: $\frac{x+5}{1-x}\ge0$

A.

-14

B.

+14

C.

-15

D.

15

Correct Answer: C

Explanation:

The given inequality is $\frac{x+5}{1-x}\ge0$
Multiplying numerator and denominator by $-1$ to make the coefficient of $x$ positive gives $\frac{x+5}{x-1}\le0$
Critical points are obtained by setting numerator and denominator to zero, yielding $x = -5$ and $x = 1$
The expression is less than or equal to zero in the closed-open interval $[-5, 1)$
The integral values satisfying this inequality are $-5, -4, -3, -2, -1, 0$
Summing these integral values gives $(-5)+(-4)+(-3)+(-2)+(-1)+0 = -15$

Q25 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $(x^{2}-12x+35)<0$

A.

$x\in(5,7)$

B.

$x\in[5,7)$

C.

$x\in(5,7]$

D.

$x\in[5,7]$

Correct Answer: A

Explanation:

The given inequality is $(x^{2}-12x+35)<0$
Factoring the quadratic expression gives $(x-5)(x-7)<0$
The roots of the corresponding equation are $x=5$ and $x=7$
Since the inequality is strictly less than zero, the solution is the open interval between the roots
Therefore, the solution is $x\in(5,7)$

Q26 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Values of $x$ satisfying: $\frac{x^{2}+x-2}{x^{2}-x-12}\ge0$

A.

$x\in(-\infty,-3]\cup[-2,1]\cup[4,\infty)$

B.

$x\in(-\infty,-3)\cup[-2,1]\cup(4,\infty)$

C.

$x\in(-3,4)$

D.

$x\in(-3,-2]\cup(4,\infty)$

Correct Answer: B

Explanation:

The given inequality is $\frac{x^{2}+x-2}{x^{2}-x-12}\ge0$
Factoring the numerator and denominator gives $\frac{(x+2)(x-1)}{(x-4)(x+3)}\ge0$
Critical points are obtained by setting each factor to zero: $x = -2, 1, 4, -3$
The values $x = 4$ and $x = -3$ make the denominator zero, so they are strictly excluded (open intervals)
The values $x = -2$ and $x = 1$ make the numerator zero, so they are included (closed brackets)
Using the wavy curve method, the expression is non-negative in the intervals $(-\infty,-3)$, $[-2,1]$, and $(4,\infty)$

Q27 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of $\frac{x}{x+3}\le2$

A.

$x\in[-6,\infty)$

B.

$x\in(-\infty,-6]$

C.

$x\in(-\infty,-6]\cup(-3,\infty)$

D.

$x\in[-6,-3)$

Correct Answer: C

Explanation:

The given inequality is $\frac{x}{x+3}\le2$
Subtracting 2 from both sides gives $\frac{x}{x+3}-2\le0$
Simplifying the expression yields $\frac{x-2(x+3)}{x+3}\le0$, which becomes $\frac{-x-6}{x+3}\le0$
Multiplying by $-1$ reverses the inequality sign to give $\frac{x+6}{x+3}\ge0$
Critical points are obtained by setting the numerator and denominator to zero, yielding $x = -6$ and $x = -3$
The value $x = -3$ makes the denominator zero, so it is strictly excluded (open bracket), whereas $x = -6$ is included (closed bracket)
Using the wavy curve method, the expression is non-negative in $(-\infty,-6]$ and $(-3,\infty)$
Therefore, the solution is $x\in(-\infty,-6]\cup(-3,\infty)$

Q28 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Largest integral value of $x$ satisfying: $\frac{x^{2}-9}{(x-1)(x-6)}\le0$

A.

5

B.

3

C.

4

D.

6

Correct Answer: A

Explanation:

The given inequality is $\frac{x^{2}-9}{(x-1)(x-6)}\le0$
Factoring the numerator gives $\frac{(x-3)(x+3)}{(x-1)(x-6)}\le0$
Critical points are obtained by setting each factor to zero: $x = -3, 3, 1, 6$
The values $x = 1$ and $x = 6$ make the denominator zero, so they are strictly excluded (open intervals)
The values $x = -3$ and $x = 3$ make the numerator zero, so they are included (closed brackets)
Using the wavy curve method, the expression is non-positive in the intervals $[-3,1)$ and $[3,6)$
The integers satisfying this inequality are $-3, -2, -1, 0, 3, 4, 5$
The largest integral value among these is $5$

Q29 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $\frac{2x-3}{3x-5}\ge3$ is

A.

$x\in[\frac{5}{3},\frac{12}{7}]$

B.

$x\in(\frac{5}{3},\frac{12}{7})$

C.

$x\in[\frac{5}{3},\frac{12}{7})$

D.

$x\in(\frac{5}{3},\frac{12}{7}]$

Correct Answer: D

Explanation:

The given inequality is $\frac{2x-3}{3x-5}\ge3$. Subtracting 3 from both sides gives $\frac{2x-3}{3x-5}-3\ge0$. Taking the LCM results in $\frac{2x-3-3(3x-5)}{3x-5}\ge0$, which simplifies to $\frac{-7x+12}{3x-5}\ge0$. Multiplying by $-1$ reverses the inequality sign to get $\frac{7x-12}{3x-5}\le0$. The critical points are $x=\frac{5}{3}$ and $x=\frac{12}{7}$. Testing the intervals gives the solution as $x\in(\frac{5}{3},\frac{12}{7}]$.

Q30 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality $\frac{x^{2}-x-6}{x^{2}+6x}\ge0$ is

A.

$x\in(-\infty,-6)\cup[2,0)\cup[3,\infty)$

B.

$x\in(-\infty,-6)\cup[-2,0)\cup[3,\infty)$

C.

$x\in(-\infty,-5)\cup[-2,0)\cup[3,\infty)$

D.

$x\in(-\infty,-6)\cup[-2,0)\cup[2,\infty)$

Correct Answer: B

Explanation:

The given inequality is $\frac{x^{2}-x-6}{x^{2}+6x}\ge0$. Factorizing the numerator gives $(x-3)(x+2)$ and factorizing the denominator gives $x(x+6)$, so the inequality becomes $\frac{(x-3)(x+2)}{x(x+6)}\ge0$. The critical points are $-6, -2, 0, 3$. Testing the intervals yields the solution as $x\in(-\infty,-6)\cup[-2,0)\cup[3,\infty)$.

Q31 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solution of inequality: $2x^{2}-2x-1<0$

A.

$x\in(\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2})$

B.

$x\in[\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}]$

C.

$x\in(-\infty,\frac{1-\sqrt{3}}{2})\cup(\frac{1+\sqrt{3}}{2},\infty)$

D.

$x\in(-1,2)$

Correct Answer: A

Explanation:

The given inequality is $2x^{2}-2x-1<0$.
First, find the roots of the equation $2x^{2}-2x-1=0$.
Using the quadratic formula, $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)}$.
This simplifies to $x = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}$.
Since the coefficient of $x^2$ is positive and the inequality is strictly less than zero, the solution lies between the two roots.
Thus, the solution set is $(\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2})$.

Q32 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Complete set of values of $x$: $\frac{(x^{2}-11x+24)^{101}}{(x^{2}-10x+24)^{100}}\le0$

A.

$x\in[3,8]$

B.

$x\in[3,4)\cup(6,8]$

C.

$x\in[3,8]-\{4,6\}$

D.

$x\in(-\infty,3]\cup[8,\infty)$

Correct Answer: C

Explanation:

The given inequality is $\frac{(x^{2}-11x+24)^{101}}{(x^{2}-10x+24)^{100}}\le0$.
Factor the quadratic expressions: $x^2 - 11x + 24 = (x-3)(x-8)$ and $x^2 - 10x + 24 = (x-4)(x-6)$.
Rewrite the inequality as $\frac{((x-3)(x-8))^{101}}{((x-4)(x-6))^{100}}\le0$.
Since the denominator has an even power ($100$), it is always positive for all real numbers except at $x = 4$ and $x = 6$ where it is zero.
Therefore, the sign of the expression depends entirely on the numerator's odd power, which simplifies to $(x-3)(x-8) \le 0$.
This gives $x \in [3,8]$, while excluding the points where the denominator is zero ($x \neq 4, 6$).
Thus, the final solution set is $x\in[3,8]-\{4,6\}$.

Q33 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

The solution set for $x(x+2)^{2}(x-1)^{5}(2x-3)(x-3)^{4}\ge0$ is given by $x\in[a,b]\cup[c,\infty]\cup\{-2\}$ then value of $a+b+c$ is equal to

A.

$1.5$

B.

$2.0$

C.

$2.5$

D.

$3.0$

Correct Answer: C

Explanation:

The given inequality is $x(x+2)^{2}(x-1)^{5}(2x-3)(x-3)^{4}\ge0$.
Find the critical points by equating each factor to zero: $x = 0$, $x = -2$ (even multiplicity), $x = 1$ (odd multiplicity), $x = 3/2$ (odd multiplicity), and $x = 3$ (even multiplicity).
The even power factors $(x+2)^2$ and $(x-3)^4$ are always positive except at $x = -2$ and $x = 3$, where the expression is zero and thus satisfied.
Analyzing the intervals for the remaining terms $x(x-1)(2x-3) \ge 0$ combined with the critical points gives the solution set.
Matching the given form $x\in[a,b]\cup[c,\infty]\cup\{-2\}$, we identify the intervals and values of $a$, $b$, and $c$, leading to the required sum $a+b+c = 2.5$.

Q34 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Complete set of values of $x$ satisfying: $\frac{(x-1)(x^{2}-x+12)}{(3+x^{2})}\le0$

A.

$x\ge1$

B.

$x\in(-\infty,-3]\cup(-\sqrt{3},1]\cup(\sqrt{3},4]$

C.

$x\in(-\infty,-3]\cup[1,4]$

D.

$x\le1$

Correct Answer: D

Explanation:

The given inequality is $\frac{(x-1)(x^{2}-x+12)}{(3+x^{2})}\le0$.
Consider the quadratic factor in the numerator, $x^2 - x + 12$. Its discriminant is $b^2 - 4ac = (-1)^2 - 4(1)(12) = 1 - 48 = -47 < 0$.
Since the discriminant is negative and the leading coefficient is positive ($1 > 0$), the quadratic expression $x^2 - x + 12$ is strictly positive for all real numbers $x$.
Similarly, the denominator $3 + x^2$ is always positive for all real numbers $x$.
Dividing both sides by these always-positive expressions leaves us with $x - 1 \le 0$, which simplifies to $x \le 1$.

Q35 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Complete set of values of $x$ satisfying: $\frac{(x-1)^{15}x^{28}}{(x+1)^{2021}(x+2)^{2020}}\ge0$

A.

$x\in(-\infty,-1)\cup[1,\infty)$

B.

$x\in(-1,1]-\{0\}$

C.

$x\in(-\infty,-2)\cup(0,\infty)$

D.

$x\in(-\infty,-2)\cup(-2,-1)\cup[1,\infty)\cup\{0\}$

Correct Answer: D

Explanation:

The given inequality is $\frac{(x-1)^{15}x^{28}}{(x+1)^{2021}(x+2)^{2020}}\ge0$.
Analyze the powers of each factor: $(x-1)^{15}$ and $(x+1)^{2021}$ have odd powers, so their signs depend directly on their linear factors.
The terms $x^{28}$ and $(x+2)^{2020}$ have even powers, meaning they are non-negative everywhere. Specifically, $x = 0$ makes the expression zero (which is included since the inequality is $\ge 0$), whereas $x = -2$ makes the denominator zero and must be excluded.
Using the wavy curve method for the remaining terms and accounting for the points of zero and undefined values, the solution set is found.
Thus, the complete set of values of $x$ is $x\in(-\infty,-2)\cup(-2,-1)\cup[1,\infty)\cup\{0\}$.

Q36 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

The complete solution set of the inequality $\frac{x^{4}-3x^{3}+2x^{2}}{x^{2}-x-30}\ge0$ is:

A.

$(-\infty,-5)\cup(1,2)\cup(6,\infty)\cup\{0\}$

B.

$(-\infty,-5)\cup[1,2]\cup(6,\infty)\cup\{0\}$

C.

$(-\infty,-5)\cup[1,2]\cup[6,\infty)\cup\{0\}$

D.

$(-\infty,-5]\cup[1,2]\cup[6,\infty)$

Correct Answer: B

Explanation:

The given inequality is $\frac{x^{4}-3x^{3}+2x^{2}}{x^{2}-x-30}\ge0$.
Factor the numerator: $x^4 - 3x^3 + 2x^2 = x^2(x^2 - 3x + 2) = x^2(x-1)(x-2)$.
Factor the denominator: $x^2 - x - 30 = (x-6)(x+5)$.
Rewrite the inequality as $\frac{x^2(x-1)(x-2)}{(x-6)(x+5)}\ge0$.
The critical points are $x = -5$, $x = 0$ (with even multiplicity $2$), $x = 1$, $x = 2$, and $x = 6$.
The factor $x^2$ is always non-negative and equals zero at $x = 0$, which satisfies the inequality.
The denominator factors impose restrictions $x \neq -5$ and $x \neq 6$.
Using the wavy curve method for the remaining factors, we obtain the solution set $(-\infty,-5)\cup[1,2]\cup(6,\infty)\cup\{0\}$.

Q37 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Number of positive integral values of $x$ satisfying the inequality $\frac{(x-4)^{2017}\cdot(x+8)^{2016}(x+1)}{x^{2016}(x-2)^{3}\cdot(x+3)^{5}\cdot(x-6)(x+9)^{2018}}\le0$ is

A.

0

B.

1

C.

2

D.

3

Correct Answer: D

Explanation:

The given inequality is $\frac{(x-4)^{2017}\cdot(x+8)^{2016}(x+1)}{x^{2016}(x-2)^{3}\cdot(x+3)^{5}\cdot(x-6)(x+9)^{2018}}\le0$.
Analyze the parity of each factor's exponent. Even powers like $(x+8)^{2016}$, $x^{2016}$, and $(x+9)^{2018}$ are non-negative for all real numbers and affect the expression only through zeros or undefined points.
Odd powers like $(x-4)^{2017}$, $(x+1)$, $(x-2)^3$, $(x+3)^5$, and $(x-6)$ dictate the sign changes across their respective critical points.
Determine the intervals where the expression is less than or equal to zero.
Check for positive integer values of $x$ lying within the valid solution intervals. The positive integral values satisfying the inequality are counted to be 3.

Q38 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

The number of the integral solutions of $x^{2}+9<(x+3)^{2}<8x+25$ is

A.

1

B.

3

C.

4

D.

5

Correct Answer: D

Explanation:

The given double inequality is $x^{2}+9<(x+3)^{2}<8x+25$.
Split into two inequalities:
First part: $x^2 + 9 < (x+3)^2 \implies x^2 + 9 < x^2 + 6x + 9 \implies 6x > 0 \implies x > 0$.
Second part: $(x+3)^2 < 8x + 25 \implies x^2 + 6x + 9 < 8x + 25 \implies x^2 - 2x - 16 < 0$.
Solve the quadratic inequality $x^2 - 2x - 16 < 0$ by finding the roots $x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-16)}}{2} = 1 \pm \sqrt{17}$.
Since $\sqrt{17} \approx 4.123$, the roots are approximately $-3.123$ and $5.123$, giving $x \in (1 - \sqrt{17}, 1 + \sqrt{17})$.
Intersecting this with $x > 0$ yields the solution range $x \in (0, 1 + \sqrt{17}) \approx (0, 5.123)$.
The integers in this range are $1, 2, 3, 4,$ and $5$, giving a total of $5$ integral solutions.

Q39 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

If $n>0$ and exactly $15$ integers satisfy $(x+6)(x-4)(x-5)(2x-n)\le0$, then sum of digits of the least possible value of $n$ is

A.

$5$

B.

$6$

C.

$7$

D.

$8$

Correct Answer: C

Explanation:

The given inequality is $(x+6)(x-4)(x-5)(2x-n)\le0$ where $n>0$.
The critical points are $-6$, $4$, $5$, and $\frac{n}{2}$.
Depending on the value of $\frac{n}{2}$, the intervals where the expression is less than or equal to zero change.
We are given that exactly $15$ integers satisfy this inequality.
By testing the placement of $\frac{n}{2}$ relative to the other critical points to capture $15$ integers, we find the least possible value for $n$.
The least possible value of $n$ corresponds to the sum of its digits being equal to $7$.

Q40 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Find the set of all $x$ for which $\frac{2x}{(2x^{2}+5x+2)}>\frac{1}{(x+1)}$

A.

$x\in(2,-1)\cup(\frac{2}{3},-\frac{1}{2})$

B.

$x\in(-2,-1)\cup(-\frac{2}{3},-\frac{1}{2})$

C.

$x\in(-2,1)\cup(-\frac{2}{3},\frac{1}{2})$

D.

$x\in(2,1)\cup(\frac{2}{3},\frac{1}{2})$

Correct Answer: B

Explanation:

The given inequality is $\frac{2x}{(2x^{2}+5x+2)}>\frac{1}{(x+1)}$.
First, factor the denominator of the first term: $2x^2 + 5x + 2 = (2x + 1)(x + 2)$.
Rearrange terms by moving everything to the left side: $\frac{2x}{(2x+1)(x+2)} - \frac{1}{x+1} > 0$.
Find a common denominator: $\frac{2x(x+1) - (2x+1)(x+2)}{(2x+1)(x+2)(x+1)} > 0$.
Simplify the numerator: $2x^2 + 2x - (2x^2 + 5x + 2) = -3x - 2$.
Rewrite the inequality as $\frac{-3x - 2}{(2x+1)(x+2)(x+1)} > 0$.
Multiply by $-1$ and reverse the inequality sign: $\frac{3x + 2}{(2x+1)(x+2)(x+1)} < 0$.
The critical points are $x = -2$, $x = -1$, $x = -\frac{2}{3}$, and $x = -\frac{1}{2}$.
Applying the wavy curve method for strictly less than zero yields the solution set $x\in(-2,-1)\cup(-\frac{2}{3},-\frac{1}{2})$.

Q41 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve $\frac{1}{x-2}+\frac{1}{x-1}>\frac{1}{x}$

A.

$x\in(-\sqrt{2},0)\cup(1,\sqrt{2})\cup(2,\infty)$

B.

$x\in(-\infty,-\sqrt{2})\cup(0,1)\cup(\sqrt{2},2)$

C.

$x\in(-\infty,0)\cup(1,2)\cup(3,\infty)$

D.

$x\in(-\sqrt{2},1)\cup(2,\infty)$

Correct Answer: A

Explanation:

The given inequality is $\frac{1}{x-2}+\frac{1}{x-1}>\frac{1}{x}$. Transposing all terms to the left side gives $\frac{1}{x-2}+\frac{1}{x-1}-\frac{1}{x}>0$. Taking the common denominator $x(x-1)(x-2)$, the numerator simplifies to $x^2-2$. Thus, the inequality becomes $\frac{x^2-2}{x(x-1)(x-2)}>0$. The critical points are $x = -\sqrt{2}, 0, 1, \sqrt{2}, 2$. Testing the sign in each interval yields the final solution set.

Q42 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve $\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$

A.

$x\in(-\infty,-3)\cup(-2,1)\cup(3,\infty)$

B.

$x\in(-\infty,-2)\cup(-1,3)\cup(4,\infty)$

C.

$x\in(-2,-1)\cup(3,4)$

D.

$x\in(-\infty,-1)\cup(2,4)\cup(5,\infty)$

Correct Answer: B

Explanation:

The given inequality is $\frac{20}{(x-3)(x-4)}+\frac{10}{x-4}+1>0$. Combining terms with the common denominator $(x-3)(x-4)$, the expression simplifies to $\frac{20+10(x-3)+(x-3)(x-4)}{(x-3)(x-4)}>0$, which reduces to $\frac{x^2+3x-2}{(x-3)(x-4)}>0$. Factoring the numerator gives $\frac{(x+2)(x+1)}{(x-3)(x-4)}>0$. The critical points are $-2, -1, 3, 4$, and testing the intervals yields the solution set.

Q43 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

The number of prime numbers satisfying the inequality $\frac{x^2-1}{2x+5}<3$ is

A.

1

B.

2

C.

3

D.

4

Correct Answer: D

Explanation:

The given inequality is $\frac{x^2-1}{2x+5}<3$. Subtracting $3$ from both sides gives $\frac{x^2-1}{2x+5}-3<0$, which simplifies to $\frac{x^2-6x-16}{2x+5}<0$, or $\frac{(x-8)(x+2)}{2x+5}<0$. The solution set for $x$ is $(-\infty, -\frac{5}{2}) \cup (-2, 8)$. The integers in the valid range are $-1, 0, 1, 2, 3, 4, 5, 6, 7$ (excluding values less than $-\frac{5}{2}$ and the interval $[-\frac{5}{2}, -2]$). Among these, the prime numbers are $2, 3, 5, 7$, making a total of $4$ prime numbers.

Q44 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

If $S$ is the set of all real $x$ such that $\frac{x^2(5-x)(1-2x)}{(5x+1)(x+2)}$ is negative and $\frac{3x+1}{6x^3+x^2-x}$ is positive, then $S$ contains

A.

$(1,4)$

B.

$(5,11)$

C.

$(-\frac{3}{2},-\frac{1}{2})$

D.

$(-10,-4)$

Correct Answer: A,C

Explanation:

The first inequality is $\frac{x^2(5-x)(1-2x)}{(5x+1)(x+2)}<0$, which yields the solution set $x \in (-2, -\frac{1}{5}) \cup (\frac{1}{2}, 5)$ after considering $x \neq 0$ (though $x=0$ makes the expression zero, not negative). The second inequality is $\frac{3x+1}{x(2x+1)(3x-1)}>0$, which gives the solution set $x \in (-\infty, -\frac{1}{2}) \cup (-\frac{1}{3}, 0) \cup (0, \frac{1}{3}) \cup (\frac{1}{3}, \infty)$. Taking the intersection of both solution sets gives $S = (-2, -\frac{1}{2}) \cup (-\frac{1}{3}, -\frac{1}{5}) \cup (\frac{1}{2}, 5)$. Both intervals $(1,4)$ and $(-\frac{3}{2}, -\frac{1}{2})$ are fully contained within $S$, making both options A and C correct.

Q45 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $(x+1)(x-3)(x+4)(5-x) < 0$.

A.

$x \in (-\infty, -4) \cup (-1, 3) \cup (5, \infty)$

B.

$x \in (-4, -1) \cup (3, 5)$

C.

$x \in [-4, -1] \cup [3, 5]$

D.

$x \in (-\infty, -\infty)$

Q46 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $\frac{(x+2)(4-x)^2}{(x+1)(x+3)(1-x)^3} \leq 0$.

A.

$x \in (-\infty, -3) \cup (-2, -1) \cup (1, 4]$

B.

$x \in (-3, -2) \cup (-1, 1) \cup [4, \infty)$

C.

$x \in (-\infty, -3) \cup [-2, -1) \cup (1, 4]$

D.

$x \in (-3, -2] \cup (-1, 1) \cup (4, \infty)$

Q47 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve the inequality $\frac{(x-1)^2(x^2-4)(x^2-5x+6)}{(x^3+6x^2)} \geq 0$.

A.

$x \in (-6, -2) \cup \{1, 2\} \cup [3, \infty)$

B.

$x \in (-\infty, -6) \cup [-2, 1] \cup [3, \infty)$

C.

$x \in (-6, -2] \cup (1, 2) \cup (3, \infty)$

D.

$x \in [-6, -2] \cup \{1, 2\} \cup (3, \infty)$

Q48 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

The complete solution of $\frac{x^2-1}{x+3} \geq 0$ and $x^2-5x+2 \leq 0$ is:

A.

$x \in \left[\frac{5-\sqrt{17}}{2}, \frac{5+\sqrt{17}}{2}\right]$

B.

$x \in \left[1, \frac{5+\sqrt{17}}{2}\right]$

C.

$x \in (-3, -1]$

D.

$x \in (-3, -1] \cup [1, \infty)$

Correct Answer: B

Explanation:

The given system requires finding values of $x$ satisfying both $\frac{x^2-1}{x+3} \geq 0$ and $x^2-5x+2 \leq 0$ simultaneously.
First, factor the numerator of the rational inequality as $\frac{(x-1)(x+1)}{x+3} \geq 0$, which gives the solution set $(-3, -1] \cup [1, \infty)$ after considering critical points $-3, -1, 1$ and excluding the denominator root $-3$.
Second, solve the quadratic inequality $x^2-5x+2 \leq 0$ by finding its roots $x = \frac{5 \pm \sqrt{17}}{2}$, which yields the interval $\left[\frac{5-\sqrt{17}}{2}, \frac{5+\sqrt{17}}{2}\right]$.
Taking the intersection of these two solution sets yields the final solution range.

Q49 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

If $f(x) = \frac{(x^2-x+6)(x-5)}{(x+2)(3x-8-x^2)}$ and solution set of $f(x) \leqslant 0$ is $x \in (-\infty, -a) \cup [b, \infty)$ where $a, b \in N$, then value of $GCD(2a^4 + 4, b^2 - 1)$ is ________.

A.

2

B.

4

C.

6

D.

8

Correct Answer: B

Explanation:

First, simplify the quadratic factors in the given rational function $f(x) = \frac{(x^2-x+6)(x-5)}{(x+2)(3x-8-x^2)}$.
The term $x^2-x+6$ has a negative discriminant ($\Delta = (-1)^2 - 4(1)(6) = -23$), meaning it is always positive for all real $x$ and can be omitted from sign analysis.
The denominator factor $3x-8-x^2$ can be rewritten as $-(x^2-3x+8)$ or factored into $-(x-8)(x+5)$ after adjusting terms, or more precisely rearranged as $-(x^2-3x+8)$ where $x^2-3x+8$ is always positive, leaving a negative sign outside.
Simplifying the expression and solving the inequality $f(x) \leqslant 0$ yields the solution set $x \in (-\infty, -2) \cup [5, \infty)$, giving $a = 2$ and $b = 5$.
Finally, calculate $GCD(2a^4 + 4, b^2 - 1)$ by substituting $a = 2$ and $b = 5$.

Q50 DPT Inequality (WCM) MCQ

iCON Education HYD, 79930 92826, 73309 72826 13 Sep 2026

Concept:

Solve: $\frac{1}{x-1} - \frac{4}{x-2} + \frac{4}{x-3} - \frac{1}{x-4} < \frac{1}{30}$

A.

$x \in (-\infty, -2) \cup (-1, 1) \cup (2, 3) \cup (4, 6) \cup (7, \infty)$

B.

$x \in (-\infty, -2] \cup [-1, 1] \cup [2, 3] \cup [4, 6] \cup [7, \infty)$

C.

$x \in (-2, -1) \cup (1, 2) \cup (3, 4) \cup (6, 7)$

D.

$x \in (-\infty, -\infty)$

Correct Answer: A

Explanation:

Group the terms pairwise to simplify the expression: pair $\frac{1}{x-1}$ with $-\frac{1}{x-4}$ and $-\frac{4}{x-2}$ with $\frac{4}{x-3}$.
Simplify each pair to get a common denominator and combine them.
Subtract $\frac{1}{30}$ from both sides, put everything over a common denominator, and factor the numerator completely.
Find the critical points and apply the wavy curve method to obtain the solution set $x \in (-\infty, -2) \cup (-1, 1) \cup (2, 3) \cup (4, 6) \cup (7, \infty)$.

Mathematics-I Chapters

Basic Mathematics

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