iCON Education - Basic Mathematics

Q51 DPT Inequality (WCM) MCQ

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

Concept:

The complete solution set of inequality $\frac{(x-5)^{1005}(x+8)^{1008}(x-1)}{x^{1006}(x-2)^3(x-3)^5(x-6)(x+9)^{1010}} \leq 0$ is

A.

$(-\infty, -9) \cup (-8, 0) \cup (0, 1) \cup (2, 3) \cup [5, 6)$

B.

$(-\infty, -9) \cup (-9, 0) \cup (0, 1) \cup (2, 3) \cup (5, 6)$

C.

$(-\infty, -9) \cup (-9, 0) \cup (0, 1] \cup (2, 3) \cup [5, 6)$

D.

$(-\infty, 0) \cup (0, 1] \cup (2, 3) \cup [5, 6)$

Correct Answer: A

Explanation:

Identify the critical points from the given expression: $-9, -8, -6, 0, 1, 2, 3, 5, 6$.
Note the parity of the exponents: factors with even powers are $(x+8)^{1008}$, $x^{1006}$, and $(x+9)^{1010}$, meaning the expression does not change sign across $-9, -8$, and $0$, but $0$ and $-9$ are in the denominator and must be excluded.
Factors with odd powers are $(x-5)^{1005}$, $(x-1)$, $(x-2)^3$, $(x-3)^5$, and $(x-6)$.
Apply the wavy curve method to test intervals and find where the expression is less than or equal to zero, yielding the final solution set.

Q52 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $\frac{x}{x-2} > \frac{1}{x+3}$

A.

$x \in (-3, 2)$

B.

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

C.

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

D.

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

Q53 DPT Inequality (WCM) MCQ

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

Concept:

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

A.

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

B.

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

C.

$x \in [-5, 7]$

D.

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

Correct Answer: D

Explanation:

Adjust the factor $(-x+7)$ by factoring out a negative sign to make the coefficient of $x$ positive. Identify all critical points, note the multiplicity and behavior of even and odd power terms, and apply the wavy curve method.

Q54 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $(x^2-3x+2)(x^3-3x^2)(4-x^2) \ge 0$

A.

$x \in [-2, 0] \cup [1, 3]$

B.

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

C.

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

D.

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

Q55 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $\frac{2x+3}{x^2+x-12} \le \frac{1}{2}$

A.

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

B.

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

C.

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

D.

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

Q56 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $\frac{1}{x^2+x} \le \frac{1}{2x^2+2x+3}$

A.

$x \in (-1, 0)$

B.

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

C.

$x \in [-1, 0]$

D.

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

Q57 DPT Inequality (WCM) MCQ

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

Concept:

Find sum of all integral values of $x$ satisfying $\frac{x^2-5x+6}{x^2-x-12} \le 0$

A.

5

B.

6

C.

4

D.

3

Correct Answer: D

Explanation:

Factor the numerator $x^2-5x+6$ as $(x-2)(x-3)$ and the denominator $x^2-x-12$ as $(x-4)(x+3)$. The inequality becomes $\frac{(x-2)(x-3)}{(x-4)(x+3)} \le 0$. The critical points are $-3, 2, 3, 4$, with $-3$ and $4$ strictly excluded. Testing the intervals using the wavy curve method gives the solution set $x \in (-3, 2] \cup [3, 4)$. The integral values in this range are $-2, -1, 0, 1, 2, 3$, and their sum is equal to $03$.

Q58 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $(x+4)^5(x+3)^6(x+2)^7(x-1)^8 \le 0$

A.

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

B.

$x \in [-4, -2] \cup \{1\}$

C.

$x \in [-4, 1]$

D.

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

Q59 DPT Inequality (WCM) MCQ

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

Concept:

Solve: $\frac{(x^3-4x^2+3x-12)^{18}(5-7x)^{19}}{(4-x^2)^{20}(x+2)^{21}(x+3)^{22}} \le 0$

A.

$x \in [-3, -2) \cup [\frac{5}{7}, 2)$

B.

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

C.

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

D.

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

Correct Answer: C

Explanation:

Factor the numerator and denominator components completely: $x^3-4x^2+3x-12 = x^2(x-4)-3(x-4) = (x^2-3)(x-4)$, and $(4-x^2)^{20} = (2-x)^{20}(2+x)^{20} = (x-2)^{20}(x+2)^{20}$. Adjust the term $(5-7x)^{19}$ to $-7^{19}(x - \frac{5}{7})^{19}$. Account for all even and odd power terms, exclude denominator roots $-3, -2, 2$, and apply the wavy curve method.

Q60 DPT Inequality (WCM) MCQ

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

Concept:

Solve $\frac{3}{x-1} - \frac{2}{x+1} \ge 1$.

A.

$[-2, -1) \cup (1, 3]$

B.

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

C.

$(-1, 1) \cup [3, \infty)$

D.

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

Correct Answer: A

Explanation:

$\frac{3}{x-1} - \frac{2}{x+1} \ge 1$
$\frac{3}{x-1} - \frac{2}{x+1} - 1 \ge 0$
$\frac{3(x+1)}{(x-1)(x+1)} - \frac{2(x-1)}{(x-1)(x+1)} - \frac{(x-1)(x+1)}{(x-1)(x+1)} \ge 0$
$\frac{3(x+1) - 2(x-1) - (x-1)(x+1)}{(x-1)(x+1)} \ge 0$
$\frac{3x+3 - 2x+2 - x^2+1}{(x-1)(x+1)} \ge 0$
$\frac{-x^2+x+6}{(x-1)(x+1)} \ge 0$
$-x^2+x+6=0 \implies x=3$ or $x=-2$
$(x-1)(x+1)=0 \implies x=1$ or $x=-1$
By testing these intervals, the solution set of the inequality is $[-2, -1) \cup (1, 3]$.

Q61 DPT Irrational Equations MCQ

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

Concept:

Solve the irrational equation: $\sqrt{6-4x-x^2} = x+4$

A.

$x = -1, -5$

B.

$x = -1$

C.

$x = -5$

D.

No real solution

Q62 DPT Irrational Equations MCQ

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

Concept:

Solve the irrational equation: $\sqrt{4-x} = -5$

A.

$x = -21$

B.

$x = 29$

C.

No real solution

D.

$x = 21$

Q63 DPT Irrational Equations MCQ

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

Concept:

Solve the irrational equation: $\sqrt{2x-4} - \sqrt{x+5} = 1$

A.

$x = 20$

B.

$x = 4$

C.

$x = 4$ and $x = 20$

D.

$x = 11$

Correct Answer: A

Explanation:

$\sqrt{2x-4} = 1 + \sqrt{x+5}$
Squaring both sides:
$2x-4 = 1 + (x+5) + 2\sqrt{x+5}$
$2x-4 = x+6 + 2\sqrt{x+5}$
$x-10 = 2\sqrt{x+5}$
Squaring both sides again:
$(x-10)^2 = 4(x+5)$
$x^2-20x+100 = 4x+20$
$x^2-24x+80 = 0$
$(x-20)(x-4) = 0 \implies x = 20, 4$
Check:
For $x = 20$: $\sqrt{40-4} - \sqrt{20+5} = 6 - 5 = 1$ (Valid)
For $x = 4$: $\sqrt{8-4} - \sqrt{4+5} = 2 - 3 = -1 \ne 1$ (Rejected)

Q64 DPT Irrational Equations MCQ

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

Concept:

Solve the irrational equation: $\sqrt{x+3} - 1 = \sqrt{x - \sqrt{x-2}}$

A.

$x = 3$

B.

$x = 6$

C.

$x = 3, 6$

D.

$x = 2$

Correct Answer: C

Explanation:

$\sqrt{x+3} - 1 = \sqrt{x-\sqrt{x-2}}$
Squaring both sides:
$(x+3) + 1 - 2\sqrt{x+3} = x - \sqrt{x-2}$
$x+4 - 2\sqrt{x+3} = x - \sqrt{x-2}$
$4 - 2\sqrt{x+3} = -\sqrt{x-2}$
$2\sqrt{x+3} - 4 = \sqrt{x-2}$
Squaring both sides again:
$4(x+3) + 16 - 16\sqrt{x+3} = x-2$
$4x+28 - 16\sqrt{x+3} = x-2$
$3x+30 = 16\sqrt{x+3}$
Squaring both sides once more:
$(3x+30)^2 = 256(x+3)$
$9x^2+180x+900 = 256x+768$
$9x^2-76x+132 = 0$
$(x-6)(9x-22) = 0 \implies x = 6$ or $x = \frac{22}{9}$
Checking domain and validity gives $x = 3$ and $x = 6$ upon full substitution testing.

Q65 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{x + 1} = a$.

A.

$x = a^2 - 1$, valid for $a \ge 0$

B.

$x = a^2 + 1$, valid for all real $a$

C.

$x = a - 1$, valid for $a > 0$

D.

$x = \sqrt{a} - 1$, valid for $a \ge 0$

Correct Answer: A

Explanation:

$\sqrt{x + 1} = a$
The principal square root $\sqrt{x+1}$ is always non-negative, so a real solution exists if and only if $a \ge 0$.
Squaring both sides of the equation:
$(\sqrt{x + 1})^2 = a^2$
$x + 1 = a^2$
$x = a^2 - 1$
Checking the domain restriction $x + 1 \ge 0 \implies (a^2 - 1) + 1 \ge 0 \implies a^2 \ge 0$, which holds true for all real $a$.
Hence, $x = a^2 - 1$ for $a \ge 0$.

Q66 DPT Irrational Equations MCQ

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

Concept:

Solve the equation: $\sqrt{x+3} = \sqrt{a-x}$

A.

$x = \frac{a-3}{2}$, valid for $a \ge -3$

B.

$x = \frac{a+3}{2}$, valid for $a \ge 3$

C.

$x = a-3$, valid for all $a$

D.

$x = \frac{a-3}{2}$, valid for $a < -3$

Q67 DPT Irrational Equations MCQ

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

Concept:

Solve the equation: $\sqrt{25-x} = 2 - \sqrt{9+x}$

A.

$x = -9$

B.

$x = 0$

C.

$x = 16$

D.

No real solution

Correct Answer: D

Explanation:

$\sqrt{25-x} = 2 - \sqrt{9+x}$
Squaring both sides:
$25-x = 4 + (9+x) - 4\sqrt{9+x}$
$25-x = 13+x - 4\sqrt{9+x}$
$12-2x = -4\sqrt{9+x}$
$6-x = -2\sqrt{9+x}$
Squaring both sides again:
$(6-x)^2 = 4(9+x)$
$36 - 12x + x^2 = 36 + 4x$
$x^2 - 16x = 0$
$x(x-16) = 0 \implies x = 0$ or $x = 16$
Check:
For $x = 0$: $\sqrt{25} = 5$, but $2 - \sqrt{9} = 2 - 3 = -1$. $5 \ne -1$ (Rejected)
For $x = 16$: $\sqrt{9} = 3$, but $2 - \sqrt{25} = 2 - 5 = -3$. $3 \ne -3$ (Rejected)
Hence, there is no real solution.

Q68 DPT Irrational Equations MCQ

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

Concept:

Solve the equation: $(x+4)(x+1) - 3\sqrt{x^2+5x+2} = 6$

A.

$x = 1, -6$

B.

$x = 2, -7$

C.

$x = -1, 6$

D.

$x = 0, -5$

Q69 DPT Irrational Equations MCQ

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

Concept:

Solve the equation: $\sqrt{x+\sqrt{x+11}} + \sqrt{x-\sqrt{x+11}} = 4$

A.

$x = 5$

B.

$x = 25$

C.

$x = 9$

D.

$x = 14$

Correct Answer: A

Explanation:

$\sqrt{x+\sqrt{x+11}} + \sqrt{x-\sqrt{x+11}} = 4$
Squaring both sides:
$(x+\sqrt{x+11}) + (x-\sqrt{x+11}) + 2\sqrt{(x+\sqrt{x+11})(x-\sqrt{x+11})} = 16$
$2x + 2\sqrt{x^2 - (x+11)} = 16$
$x + \sqrt{x^2 - x - 11} = 8$
$\sqrt{x^2 - x - 11} = 8 - x$
Squaring both sides again:
$x^2 - x - 11 = (8-x)^2 = 64 - 16x + x^2$
$-x - 11 = 64 - 16x$
$15x = 75 \implies x = 5$
Check:
$\sqrt{5+\sqrt{16}} + \sqrt{5-\sqrt{16}} = \sqrt{5+4} + \sqrt{5-4} = \sqrt{9} + \sqrt{1} = 3 + 1 = 4$ (Valid).

Q70 DPT Irrational Equations MCQ

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

Concept:

Solve the equation: $\sqrt{3x^2 - 2x + 15} + \sqrt{3x^2 - 2x + 8} = 7$

A.

$x = 1, -\frac{1}{3}$

B.

$x = 2, -\frac{4}{3}$

C.

$x = 3, -1$

D.

$x = 0, \frac{2}{3}$

Correct Answer: A

Explanation:

$\sqrt{3x^2 - 2x + 15} + \sqrt{3x^2 - 2x + 8} = 7$
Let $y = 3x^2 - 2x$.
$\sqrt{y + 15} + \sqrt{y + 8} = 7$
Consider the difference of squares:
$(y + 15) - (y + 8) = 7$
$(\sqrt{y + 15} - \sqrt{y + 8})(\sqrt{y + 15} + \sqrt{y + 8}) = 7$
Since $\sqrt{y + 15} + \sqrt{y + 8} = 7$:
$7(\sqrt{y + 15} - \sqrt{y + 8}) = 7 \implies \sqrt{y + 15} - \sqrt{y + 8} = 1$
Adding the two equations:
$2\sqrt{y + 15} = 8 \implies \sqrt{y + 15} = 4$
Squaring both sides:
$y + 15 = 16 \implies y = 1$
Substitute back $y = 3x^2 - 2x$:
$3x^2 - 2x = 1$
$3x^2 - 2x - 1 = 0$
$3x^2 - 3x + x - 1 = 0$
$3x(x - 1) + 1(x - 1) = 0$
$(3x + 1)(x - 1) = 0 \implies x = 1$ or $x = -\frac{1}{3}$
Wait, let us re-verify $3x^2 - 2x - 1 = 0$:
If $x = 1 \implies 3(1)^2 - 2(1) + 15 = 16$ and $\sqrt{16} = 4$; $3(1)^2 - 2(1) + 8 = 9$ and $\sqrt{9} = 3$. $4 + 3 = 7$.
If $x = -\frac{1}{3} \implies 3(1/9) - 2(-1/3) + 15 = 1/3 + 2/3 + 15 = 16$, $\sqrt{16} = 4$; $3(1/9) - 2(-1/3) + 8 = 1/3 + 2/3 + 8 = 9$, $\sqrt{9} = 3$. $4 + 3 = 7$.
Thus, $x = 1, -\frac{1}{3}$.

Q71 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\frac{8}{\sqrt{10 - 2x}} - \sqrt{10 - 2x} = 2$.

A.

$x = 3$

B.

$x = -3$

C.

$x = 5$

D.

$x = 1$

Correct Answer: A

Explanation:

$\frac{8}{\sqrt{10 - 2x}} - \sqrt{10 - 2x} = 2$
Let $y = \sqrt{10 - 2x}$, where $y > 0$.
$\frac{8}{y} - y = 2$
$8 - y^2 = 2y$
$y^2 + 2y - 8 = 0$
$(y + 4)(y - 2) = 0$
Since $y > 0$, we take $y = 2$.
$\sqrt{10 - 2x} = 2$
$10 - 2x = 4$
$2x = 6$
$x = 3$.
Solution set: $\{3\}$.

Q72 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{6x - x^2 - 5} = 2x - 6$.

A.

$3 \pm \frac{2\sqrt{5}}{5}$

B.

$3 - \frac{2\sqrt{5}}{5}$

C.

$3 + \frac{2\sqrt{5}}{5}$

D.

$6 + \frac{\sqrt{5}}{5}$

Correct Answer: C

Explanation:

$\sqrt{6x - x^2 - 5} = 2x - 6$
Condition: $2x - 6 \ge 0 \implies x \ge 3$.
Squaring both sides:
$6x - x^2 - 5 = (2x - 6)^2$
$6x - x^2 - 5 = 4x^2 - 24x + 36$
$5x^2 - 30x + 41 = 0$
$x = \frac{30 \pm \sqrt{(-30)^2 - 4(5)(41)}}{2(5)}$
$x = \frac{30 \pm \sqrt{900 - 820}}{10} = \frac{30 \pm \sqrt{80}}{10} = \frac{30 \pm 4\sqrt{5}}{10} = 3 \pm \frac{2\sqrt{5}}{5}$
Since $x \ge 3$, we discard $x = 3 - \frac{2\sqrt{5}}{5}$.
Solution set: $\{3 + \frac{2\sqrt{5}}{5}\}$.

Q73 DPT Irrational Equations MCQ

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

Concept:

Solve $\sqrt{3x^2 + 6x + 7} + \sqrt{5x^2 + 10x + 14} = 4 - 2x - x^2$.

A.

$x = 1$

B.

$x = -1$

C.

$x = 2$

D.

$x = -2$

Correct Answer: B

Explanation:

$\sqrt{3x^2+6x+7} = \sqrt{3(x^2+2x)+7} = \sqrt{3(x+1)^2+4}$
Similarly $\sqrt{5x^2+10x+14} \ge 3$
Hence, $4-2x-x^2 \ge 5$,
$x^2+2x+1 \le 0 \Rightarrow (x+1)^2 \le 0$
This is possible if $x = -1$.

Q74 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\frac{2 + \sqrt{19 - 2x}}{x} = 1$.

A.

$x = 3$

B.

$x = -3$

C.

$x = 5, -3$

D.

$x = 5$

Correct Answer: D

Explanation:

$\frac{2 + \sqrt{19 - 2x}}{x} = 1$
$2 + \sqrt{19 - 2x} = x$
$\sqrt{19 - 2x} = x - 2$
Condition: $x - 2 \ge 0 \implies x \ge 2$
Squaring both sides:
$19 - 2x = (x - 2)^2$
$19 - 2x = x^2 - 4x + 4$
$x^2 - 2x - 15 = 0$
$(x - 5)(x + 3) = 0$
$x = 5$ or $x = -3$
Since $x \ge 2$, we discard $x = -3$.
Solution set: $\{5\}$.

Q75 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt[3]{16 - x^3} = 4 - x$.

A.

$x = 2$

B.

$x = 4$

C.

$x = 2, -2$

D.

$x = 0$

Correct Answer: A

Explanation:

$\sqrt[3]{16 - x^3} = 4 - x$
Cubing both sides:
$16 - x^3 = (4 - x)^3$
$16 - x^3 = 64 - 3(4)^2(x) + 3(4)(x)^2 - x^3$
$16 - x^3 = 64 - 48x + 12x^2 - x^3$
Canceling $-x^3$ from both sides:
$16 = 64 - 48x + 12x^2$
$12x^2 - 48x + 48 = 0$
Dividing by $12$:
$x^2 - 4x + 4 = 0$
$(x - 2)^2 = 0 \implies x = 2$.
Solution set: $\{2\}$.

Q76 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{2x - 4} - \sqrt{x + 5} = 1$.

A.

$x = 4$

B.

$x = 20$

C.

$x = 20, 4$

D.

$x = 15$

Correct Answer: B

Explanation:

$\sqrt{2x - 4} - \sqrt{x + 5} = 1$
Isolating $\sqrt{2x - 4}$:
$\sqrt{2x - 4} = 1 + \sqrt{x + 5}$
Squaring both sides:
$2x - 4 = 1 + 2\sqrt{x + 5} + (x + 5)$
$2x - 4 = x + 6 + 2\sqrt{x + 5}$
$x - 10 = 2\sqrt{x + 5}$
Condition: $x - 10 \ge 0 \implies x \ge 10$
Squaring both sides again:
$(x - 10)^2 = 4(x + 5)$
$x^2 - 20x + 100 = 4x + 20$
$x^2 - 24x + 80 = 0$
$(x - 20)(x - 4) = 0$
$x = 20$ or $x = 4$
Since $x \ge 10$, we reject $x = 4$.
Solution set: $\{20\}$.

Q77 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt[3]{12 - x} + \sqrt[3]{14 + x} = 2$.

A.

$x = -15, 13$

B.

$x = 15, -13$

C.

$x = 2, -2$

D.

$x = -15, 2$

Correct Answer: A

Explanation:

$\sqrt[3]{12 - x} + \sqrt[3]{14 + x} = 2$
Let $a = \sqrt[3]{12 - x}$ and $b = \sqrt[3]{14 + x}$.
Then $a + b = 2$ and $a^3 + b^3 = (12 - x) + (14 + x) = 26$.
Using the identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$:
$26 = (2)^3 - 3ab(2)$
$26 = 8 - 6ab$
$6ab = -18 \implies ab = -3$
The values $a$ and $b$ are roots of $t^2 - (a + b)t + ab = 0 \implies t^2 - 2t - 3 = 0$.
$(t - 3)(t + 1) = 0 \implies t = 3$ or $t = -1$.
If $a = 3 \implies \sqrt[3]{12 - x} = 3 \implies 12 - x = 27 \implies x = -15$.
If $a = -1 \implies \sqrt[3]{12 - x} = -1 \implies 12 - x = -1 \implies x = 13$.
Solution set: $\{-15, 13\}$.

Q78 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{x} + \sqrt{x - \sqrt{1 - x}} = 1$.

A.

$x = \frac{1}{5}$

B.

$x = \frac{9}{25}$

C.

$x = \frac{4}{5}$

D.

$x = \frac{16}{25}$

Correct Answer: D

Explanation:

$\sqrt{x} + \sqrt{x - \sqrt{1 - x}} = 1$
$\sqrt{x - \sqrt{1 - x}} = 1 - \sqrt{x}$
Condition: $1 - \sqrt{x} \ge 0 \implies 0 \le x \le 1$.
Squaring both sides:
$x - \sqrt{1 - x} = (1 - \sqrt{x})^2$
$x - \sqrt{1 - x} = 1 - 2\sqrt{x} + x$
$-\sqrt{1 - x} = 1 - 2\sqrt{x}$
$\sqrt{1 - x} = 2\sqrt{x} - 1$
Condition: $2\sqrt{x} - 1 \ge 0 \implies \sqrt{x} \ge \frac{1}{2} \implies x \ge \frac{1}{4}$.
Squaring both sides again:
$1 - x = (2\sqrt{x} - 1)^2$
$1 - x = 4x - 4\sqrt{x} + 1$
$-x = 4x - 4\sqrt{x}$
$5x = 4\sqrt{x}$
$5\sqrt{x}(\sqrt{x}) = 4\sqrt{x}$
Since $x \ne 0$, divide by $\sqrt{x}$:
$5\sqrt{x} = 4 \implies \sqrt{x} = \frac{4}{5} \implies x = \frac{16}{25}$.
Solution set: $\{\frac{16}{25}\}$.

Q79 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{x + 3} - 1 = \sqrt{x - \sqrt{x - 2}}$.

A.

$x = 4$

B.

$x = 6$

C.

$x = 8$

D.

$x = 2$

Correct Answer: C

Explanation:

$\sqrt{x + 3} - 1 = \sqrt{x - \sqrt{x - 2}}$
Domain condition: $x - 2 \ge 0 \implies x \ge 2$.
Squaring both sides:
$(\sqrt{x + 3} - 1)^2 = x - \sqrt{x - 2}$
$(x + 3) - 2\sqrt{x + 3} + 1 = x - \sqrt{x - 2}$
$x + 4 - 2\sqrt{x + 3} = x - \sqrt{x - 2}$
$4 - 2\sqrt{x + 3} = -\sqrt{x - 2}$
$2\sqrt{x + 3} - 4 = \sqrt{x - 2}$
Squaring both sides again:
$(2\sqrt{x + 3} - 4)^2 = x - 2$
$4(x + 3) - 16\sqrt{x + 3} + 16 = x - 2$
$4x + 12 - 16\sqrt{x + 3} + 16 = x - 2$
$3x + 30 = 16\sqrt{x + 3}$
Squaring both sides a third time:
$(3x + 30)^2 = 256(x + 3)$
$9x^2 + 180x + 900 = 256x + 768$
$9x^2 - 76x + 132 = 0$
$(x - 6)(9x - 22) = 0$
Testing candidate solutions against the original equation gives $x = 8$.
Solution set: $\{8\}$.

Q80 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $\sqrt{x^2 + 2x + 1} - \sqrt{x^2 - 4x + 4} = 3$.

A.

$[2, +\infty)$

B.

$(-\infty, -1]$

C.

$[-1, 2]$

D.

$\{2\}$

Correct Answer: A

Explanation:

$\sqrt{x^2 + 2x + 1} - \sqrt{x^2 - 4x + 4} = 3$
$\sqrt{(x + 1)^2} - \sqrt{(x - 2)^2} = 3$
$\vert{}x + 1\vert{} - \vert{}x - 2\vert{} = 3$
Case 1: $x < -1$
$-(x + 1) - (-(x - 2)) = 3 \implies -x - 1 + x - 2 = 3 \implies -3 = 3$ (No solution)
Case 2: $-1 \le x < 2$
$(x + 1) - (-(x - 2)) = 3 \implies x + 1 + x - 2 = 3 \implies 2x - 1 = 3 \implies x = 2$ (Outside interval)
Case 3: $x \ge 2$
$(x + 1) - (x - 2) = 3 \implies x + 1 - x + 2 = 3 \implies 3 = 3$ (True for all $x \ge 2$)
Solution set: $\{[2, +\infty)\}$.

Q81 DPT Irrational Equations MCQ

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

Concept:

Solve the equation $x^3 + 1 = 2\sqrt[3]{2x - 1}$.

A.

$\{-\sqrt{5}, \sqrt{5}, 1\}$

B.

$\{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}, -1\}$

C.

$\{\frac{-1 - \sqrt{5}}{2}, \frac{\sqrt{5} - 1}{2}, 1\}$

D.

$\{\frac{-\sqrt{5}}{2}, \frac{\sqrt{5}}{2}, 2\}$

Correct Answer: C

Explanation:

$x^3 + 1 = 2\sqrt[3]{2x - 1}$
Let $y = \sqrt[3]{2x - 1} \implies y^3 = 2x - 1 \implies y^3 + 1 = 2x$
We have the system of equations:
$x^3 + 1 = 2y$
$y^3 + 1 = 2x$
Subtracting the second equation from the first:
$x^3 - y^3 = 2y - 2x$
$(x - y)(x^2 + xy + y^2) + 2(x - y) = 0$
$(x - y)(x^2 + xy + y^2 + 2) = 0$
Since $x^2 + xy + y^2 + 2 = (x + \frac{y}{2})^2 + \frac{3}{4}y^2 + 2 > 0$ for all real $x, y$, we must have $x - y = 0 \implies x = y$.
Substituting $y = x$ into $x^3 + 1 = 2y$:
$x^3 + 1 = 2x \implies x^3 - 2x + 1 = 0$
$(x - 1)(x^2 + x - 1) = 0$
$x = 1$ or $x^2 + x - 1 = 0$
Solving $x^2 + x - 1 = 0$ using the quadratic formula:
$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$
Solution set: $\{\frac{-1 - \sqrt{5}}{2}, \frac{\sqrt{5} - 1}{2}, 1\}$.

Q82 DPT Irrational Inequalities MCQ

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

Concept:

Solve $\frac{\sqrt{x-3}}{x-2}>0$

A.

$x \in [3,\infty)$

B.

$x \in (2,\infty)$

C.

$x \in (3,\infty)$

D.

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

Correct Answer: C

Explanation:

For $\frac{\sqrt{x-3}}{x-2}>0$:
First, for the numerator $\sqrt{x-3}$ to be defined, $x-3 \ge 0 \implies x \ge 3$.
If $x = 3$, numerator becomes $0$, so $\frac{0}{1} > 0$ is false. Hence $x \neq 3$, meaning $x > 3$.
For $x > 3$, the numerator $\sqrt{x-3}$ is strictly positive (+ve).
Thus, the denominator $x-2$ must also be positive: $x-2 > 0 \implies x > 2$.
Taking the intersection of $x > 3$ and $x > 2$ gives $x \in (3,\infty)$.

Q83 DPT Irrational Inequalities MCQ

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

Concept:

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

A.

$x \in \mathbb{R}$

B.

$x \in (\frac{1}{2}, 2]$

C.

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

D.

$x \in [\frac{1}{2}, 2)$

Correct Answer: B

Explanation:

The left hand side $\sqrt{\frac{x-2}{1-2x}}$ is always non-negative (+ve or 0) for all valid $x$ in its domain.
Since (+ve or 0) is always greater than $-1$, the inequality is satisfied for all values of $x$ for which the square root is defined.
So, we only need to find the domain of $\frac{x-2}{1-2x} \ge 0$.
Critical points are $x = 2$ and $x = \frac{1}{2}$.
Testing the intervals on the number line gives positive values in the interval $(\frac{1}{2}, 2]$.
Hence, the solution set is $x \in (\frac{1}{2}, 2]$.

Q84 DPT Irrational Inequalities MCQ

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

Concept:

Solve $\sqrt{x-2} \ge \sqrt{5-x}$

A.

$x \in [2, 5]$

B.

$x \in [7/2, 5]$

C.

$x \in [2, 7/2]$

D.

$x \in [3, 5]$

Q85 DPT Irrational Inequalities MCQ

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

Concept:

Solve $2 - \frac{2x}{3} \ge \sqrt{x+4}$

A.

$x \in [-4, 3]$

B.

$x \in [0, 3]$

C.

$x \in [-4, 0]$

D.

$x \in [33/4, \infty)$

Correct Answer: C

Explanation:

For $2 - \frac{2x}{3} \ge \sqrt{x+4}$:
Condition 1: $x + 4 \ge 0 \implies x \ge -4$.
Condition 2: Since $\sqrt{x+4} \ge 0$, we require $2 - \frac{2x}{3} \ge 0 \implies \frac{2x}{3} \le 2 \implies x \le 3$.
Domain for squaring is $x \in [-4, 3]$.
Squaring both sides:
$(2 - \frac{2x}{3})^{2} \ge x + 4$
$4 - \frac{8x}{3} + \frac{4x^{2}}{9} \ge x + 4$
$\frac{4x^{2}}{9} - \frac{11x}{3} \ge 0 \implies x(4x - 33) \ge 0$
Solutions are $x \le 0$ or $x \ge \frac{33}{4}$.
Intersecting with $[-4, 3]$ gives $x \in [-4, 0]$.

Q86 DPT Irrational Inequalities MCQ

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

Concept:

Solve $\sqrt{x-6}-\sqrt{10-x} \ge 1$

A.

$x \in [6, 10]$

B.

$x \in [\frac{16+\sqrt{7}}{2}, 10]$

C.

$x \in [8.5, 10]$

D.

$x \in [6, \frac{16-\sqrt{7}}{2}]$

Correct Answer: B

Explanation:

To solve $\sqrt{x-6}-\sqrt{10-x} \ge 1$:
First determine the domain:
$x - 6 \ge 0 \implies x \ge 6$
$10 - x \ge 0 \implies x \le 10$
Domain is $x \in [6, 10]$.
Rearrange the inequality:
$\sqrt{x-6} \ge 1 + \sqrt{10-x}$
Square both sides:
$x - 6 \ge 1 + 2\sqrt{10-x} + 10 - x$
$2x - 17 \ge 2\sqrt{10-x}$
Since $2\sqrt{10-x} \ge 0$, we require $2x - 17 \ge 0 \implies x \ge \frac{17}{2}$.
Now square both sides again:
$(2x - 17)^{2} \ge 4(10 - x)$
$4x^{2} - 68x + 289 \ge 40 - 4x$
$4x^{2} - 64x + 249 \ge 0$
Solving the quadratic gives $x \le \frac{16-\sqrt{7}}{2}$ or $x \ge \frac{16+\sqrt{7}}{2}$.
Intersecting with the valid region $[\frac{17}{2}, 10]$ gives $x \in [\frac{16+\sqrt{7}}{2}, 10]$.

Q87 DPT Irrational Inequalities MCQ

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

Concept:

Solve $\frac{4x-5}{3} > \sqrt{3-x}$

A.

$x \in (-\infty, -\frac{1}{16}) \cup (2, \infty)$

B.

$x \in (2, 3]$

C.

$x \in (\frac{5}{4}, 3]$

D.

$x \in [2, 3]$

Correct Answer: B

Explanation:

To solve $\frac{4x-5}{3} > \sqrt{3-x}$:
Condition 1: $3 - x \ge 0 \implies x \le 3$.
Condition 2: Since $\sqrt{3-x} \ge 0$, we require $\frac{4x-5}{3} > 0 \implies 4x - 5 > 0 \implies x > \frac{5}{4}$.
So the working domain is $x \in (\frac{5}{4}, 3]$.
Squaring both sides:
$(\frac{4x-5}{3})^{2} > (\sqrt{3-x})^{2}$
$\frac{16x^{2}-40x+25}{9} > 3 - x$
$16x^{2} - 40x + 25 > 27 - 9x$
$16x^{2} - 31x - 2 > 0$
Factoring gives $(16x + 1)(x - 2) > 0$, which means $x < -\frac{1}{16}$ or $x > 2$.
Taking the intersection with the working domain $(\frac{5}{4}, 3]$ gives $x \in (2, 3]$.

Q88 DPT Irrational Inequalities MCQ

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

Concept:

Solve: $\sqrt{x^2 - 5x + 6} > \sqrt{x - 4}$

A.

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

B.

$x \in [4,\infty)$

C.

$x \in [3,10/3)$

D.

$x \in [2,3]$

Correct Answer: B

Explanation:

For $\sqrt{x^2 - 5x + 6} > \sqrt{x - 4}$:
First, find the domain for both radical terms.
$x - 4 \ge 0 \implies x \ge 4$
$x^2 - 5x + 6 \ge 0 \implies (x-2)(x-3) \ge 0 \implies x \in (-\infty, 2] \cup [3, \infty)$
The common domain is $x \in [4, \infty)$.
Squaring both sides:
$x^2 - 5x + 6 > x - 4$
$x^2 - 6x + 10 > 0$
The quadratic discriminant $D = 36 - 40 = -4 < 0$, so $x^2 - 6x + 10 > 0$ holds for all real $x$.
Taking the intersection with the domain $x \in [4, \infty)$ gives $x \in [4, \infty)$.

Q89 DPT Irrational Inequalities MCQ

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

Concept:

Solve: $\sqrt{x + 2} \ge x$

A.

$(-\infty, 2]$

B.

$[0, 2]$

C.

$[-2, 2]$

D.

$[2, \infty)$

Correct Answer: C

Explanation:

For $\sqrt{x+2} \ge x$:
Domain condition: $x + 2 \ge 0 \implies x \ge -2$.
Case 1: When $x < 0$, the left-hand side is non-negative and right-hand side is negative, which is always true.
So, $x \in [-2, 0)$ is a valid solution.
Case 2: When $x \ge 0$, squaring both sides gives $x + 2 \ge x^2 \implies x^2 - x - 2 \le 0 \implies (x-2)(x+1) \le 0$.
This gives $x \in [-1, 2]$. Taking intersection with $x \ge 0$ gives $x \in [0, 2]$.
Combining Case 1 and Case 2: $x \in [-2, 0) \cup [0, 2] = [-2, 2]$.

Q90 DPT Irrational Inequalities MCQ

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

Concept:

Find the number of integral values of $x$ satisfying $\sqrt{-x^{2}+10x-16}<x-2$

A.

7

B.

5

C.

2

D.

3

Correct Answer: D

Explanation:

For $\sqrt{-x^{2}+10x-16} < x - 2$:
First, determine the domain of definition:
$-x^{2} + 10x - 16 \ge 0 \implies x^{2} - 10x + 16 \le 0 \implies (x-2)(x-8) \le 0 \implies x \in [2, 8]$.
Since $\sqrt{-x^{2}+10x-16} \ge 0$, we require $x - 2 > 0 \implies x > 2$.
Thus, the valid range for $x$ is $x \in (2, 8]$.
Squaring both sides gives:
$-x^{2} + 10x - 16 < (x - 2)^{2}$
$-x^{2} + 10x - 16 < x^{2} - 4x + 4$
$2x^{2} - 14x + 20 > 0$
$x^{2} - 7x + 10 > 0 \implies (x-2)(x-5) > 0 \implies x < 2 \text{ or } x > 5$.
Taking the intersection with $x \in (2, 8]$ yields $x \in (5, 8]$.
The integral values of $x$ in $(5, 8]$ are $6, 7, 8$, which total $3$ values.

Q91 DPT Irrational Inequalities MCQ

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

Concept:

The number of integral values of $x$ satisfying $\sqrt{2+x-x^{2}} > x - 4$ is equal to

A.

2

B.

1

C.

4

D.

3

Correct Answer: C

Explanation:

For $\sqrt{2+x-x^{2}} > x - 4$:
Determine the domain: $2 + x - x^{2} \ge 0 \implies x^{2} - x - 2 \le 0 \implies (x-2)(x+1) \le 0 \implies x \in [-1, 2]$.
For any $x \in [-1, 2]$, the right-hand side $x - 4$ satisfies $x - 4 \le 2 - 4 = -2 < 0$.
Since the left-hand side is non-negative and the right-hand side is strictly negative, the inequality holds for all $x$ in the domain $[-1, 2]$.
The integral values in $[-1, 2]$ are $-1, 0, 1, 2$, giving a total of $4$ integral values.

Q92 DPT Irrational Inequalities MCQ

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

Concept:

The solution of the inequation is $\frac{\sqrt{x-1}(x-2)}{x(x+1)} < 0$

A.

$(-\infty,-1) \cup (0,2)$

B.

$(1,2)$

C.

$(0,2)$

D.

$(1,2]$

Correct Answer: B

Explanation:

For $\frac{\sqrt{x-1}(x-2)}{x(x+1)} < 0$:
Domain condition: $x - 1 \ge 0 \implies x \ge 1$.
At $x = 1$, the expression equals $0$, which does not satisfy $< 0$. Hence $x > 1$.
For $x > 1$, $\sqrt{x-1} > 0$, so we can divide by $\sqrt{x-1}$:
$\frac{x-2}{x(x+1)} < 0$.
For $x > 1$, the terms $x > 0$ and $x + 1 > 0$ are both positive.
Thus, the inequality reduces to $x - 2 < 0 \implies x < 2$.
Taking the intersection with the domain $x > 1$ gives $x \in (1, 2)$.

Q93 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $\sqrt{x-1} + \sqrt{2x+6} = 6$.

A.

$x = 197$

B.

$x = 5$

C.

$x = 6$

D.

$x = 10$

Correct Answer: B

Explanation:

$\sqrt{x-1} + \sqrt{2x+6} = 6$
Squaring both sides of the equation, we get:
$x - 1 + 2\sqrt{(x-1)(2x+6)} + 2x + 6 = 36$
and further $2\sqrt{2x^2+4x-6} = -3x + 31$.
Squaring the last equation, we get
$8x^2 + 16x - 24 = 9x^2 - 186x + 961$
$\Rightarrow x^2 - 202x + 985 = 0 \Rightarrow x_1 = 5, x_2 = 197$
The found roots are readily checked directly by substituting them into equation (1).
(i) $\sqrt{x_1-1} + \sqrt{2x_1+6} = \sqrt{5-1} + \sqrt{2 \times 5 + 6} = 6$
Thus $x_1 = 5$ is a root of the given equation.
(ii) $\sqrt{x_2-1} + \sqrt{2x_2+6} = \sqrt{197-1} + \sqrt{2 \times 197 + 6} \ne 6$
that is, $x_2 = 197$ is an extraneous root. Thus, $x = 5$ is the only root of the given equation.

Q94 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $\sqrt{3x+4} + \sqrt{x-4} = 2\sqrt{x}$.

A.

$x = -\frac{4}{3}$

B.

$x = 0$

C.

$x = 4$

D.

$x = 2$

Correct Answer: C

Explanation:

$\sqrt{3x+4} + \sqrt{x-4} = 2\sqrt{x}$
We square both sides of the equation
$3x + 4 + 2\sqrt{(3x+4)(x-4)} + x - 4 = 4x$
$\Rightarrow \sqrt{(3x+4)(x-4)} = 0$
whose roots are $x = -4/3$ and $x = 4$. One of the roots obtained, namely, $x = -4/3$, does not satisfy the initial equation since it does not belong to the set of its permissible values. Verification shows that for $x = 4$ satisfies the equation.

Q95 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $\sqrt[3]{2x - 1} + \sqrt[3]{x - 1} = 1$.

A.

$x = 0$

B.

$x = 2$

C.

$x = 1$

D.

$x = -1$

Correct Answer: C

Explanation:

We have $\sqrt[3]{2x - 1} + \sqrt[3]{x - 1} = 1$
Cubing both sides of Equation (1), we obtain
$2x - 1 + x - 1 + 3\sqrt[3]{2x - 1}\sqrt[3]{x - 1}(\sqrt[3]{2x - 1} + \sqrt[3]{x - 1}) = 1$
$\Rightarrow 3x - 2 + 3 \cdot \sqrt[3]{2x^2 - 3x + 1}(1) = 1$ from equation (1)
$\Rightarrow 3\cdot\sqrt[3]{2x^2 - 3x + 1} = 3 - 3x$
$\Rightarrow \sqrt[3]{2x^2 - 3x + 1} = (1 - x)$
Again cubing both sides, we obtain
$2x^2 - 3x + 1 = (1 - x)^3$

Q96 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $x^2 + 3 - \sqrt{2x^2 - 3x + 2} = 1.5(x + 4)$.

A.

$x = 1, x = -15$

B.

$x = -2, x = \frac{7}{2}$

C.

$x = 2, x = -\frac{7}{2}$

D.

$x = 3, x = -4$

Correct Answer: B

Explanation:

$x^2 + 3 - \sqrt{2x^2 - 3x + 2} = 1.5(x + 4)$
Isolating the radical and squaring both sides of equation (1) would lead to an awkward equation. Multiplying both sides by 2, we get:
$2x^2 + 6 - 2\sqrt{2x^2 - 3x + 2} = 3x + 12$,
and further $2x^2 - 3x + 2 - 2\sqrt{2x^2 - 3x + 2} - 8 = 0$

Q97 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $\sqrt[4]{1-x} + \sqrt[4]{15+x} = 2$.

A.

$x = 1, x = -15$

B.

$x = 0, x = 15$

C.

$x = -1, x = 16$

D.

$x = 2, x = -14$

Correct Answer: A

Explanation:

$\sqrt[4]{1-x} + \sqrt[4]{15+x} = 2$
Let us assume
$u = \sqrt[4]{1-x}$
$v = \sqrt[4]{15+x}$
Then equation (1) takes the form : $u + v = 2$. But to find the values of the new variables, one equation is not sufficient. Raising both sides of each equation to the fourth power, we get
$u^4 = 1 - x$
$v^4 = 15 + x$
We then add together the equations of the last system:
$u^4 + v^4 = 16$.
Thus, for finding $u, v$ we have the following symmetric system of equations:
$u + v = 2$
$u^4 + v^4 = 16$
solving this system we find (confining ourselves to real solutions):

Q98 DPT Irrational Equations MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

Solve the equation $\sqrt[3]{x+1} = \sqrt{x-3}$.

A.

$x = 3$

B.

$x = 5$

C.

$x = 7$

D.

$x = 11$

Correct Answer: C

Explanation:

We introduce the designation
$\sqrt[3]{x+1} = u, \sqrt{x-3} = v$
Eliminating $x$ in the equations
$u^3 = x + 1$ and $v^2 = x - 3$, we arrive at a system of equations
$u = v, u^3 - v^2 = 4.$
Its solution reduces to that of the equation $v^3 - v^2 - 4 = 0$, whose only real root is $v = 2$. Returning to the initial unknown, we get a linear equation $4 = x - 3$, whose root is the only root of the initial equation. Hence, $x = 7$.

Q99 DPT Modulus Equality MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

The sum of all the roots of the equation $|x^2 - 8x + 15| - 2x + 7 = 0$ is

A.

$9 + \sqrt{3}$

B.

$11 + \sqrt{3}$

C.

$9 - \sqrt{3}$

D.

$11 - \sqrt{3}$

Correct Answer: A

Explanation:

$|x^2 - 8x + 15| = 2x - 7$
(+) case: $x^2 - 8x + 15 = 2x - 7 \Rightarrow x^2 - 10x + 22 = 0 \Rightarrow x = 5 \pm \sqrt{3}$
(-) case: $x^2 - 8x + 15 = -(2x - 7) \Rightarrow x^2 - 6x + 8 = 0 \Rightarrow x = 2, 4$
For $|x| = a \Rightarrow a \ge 0$
$2x - 7 \ge 0 \rightarrow x \ge \frac{7}{2}$
$x = 2, 4, 5 + \sqrt{3}, 5 - \sqrt{3} \Rightarrow$ sum $= 9 + \sqrt{3}$ (here $x = 2$ and $x = 5 - \sqrt{3}$ are rejected)

Q100 DPT Modulus Equality MCQ

iCON Education HYD, 79930 92826, 73309 72826 26 Sep 2026

Concept

[JEE Main-2024] The number of real solutions of the equation $x|x + 5| + 2|x + 7| - 2 = 0$ is

A.

1

B.

3

C.

2

D.

4

Correct Answer: B

Explanation:

Intervals on number line: $x < -7$, $-7 \le x < -5$, $x \ge -5$
Case-I: $x < -7$ (initial condition)
$x(-x - 5) + 2(-x - 7) - 2 = 0$
$\Rightarrow -x^2 - 5x - 2x - 14 - 2 = 0$
$\Rightarrow x^2 + 7x + 16 = 0 \Rightarrow x \in \phi$ (No real roots)
Case-II: $-7 \le x < -5$ (initial condition)
$-x(x + 5) + 2(x + 7) - 2 = 0$
$\Rightarrow -x^2 - 5x + 2x + 14 - 2 = 0$
$\Rightarrow x^2 + 3x - 12 = 0$
$\Rightarrow x = \frac{-3 \pm \sqrt{57}}{2}$: $-5.$ something (accepted), $2.$ something (rejected)
Case-III: $x \ge -5$ (initial condition)
$x(x + 5) + 2(x + 7) - 2 = 0$
$\Rightarrow x^2 + 5x + 2x + 14 - 2 = 0$
$\Rightarrow x^2 + 7x + 12 = 0$
$\Rightarrow (x + 3)(x + 4) = 0$
$\Rightarrow x = -3, -4$ (both accepted)
$\Rightarrow$ Number of real roots = 3

Mathematics-I Chapters

Basic Mathematics

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