Question 1: Palette 1/111

DPT MCQ

Inequalities represent subsets of real numbers and can be expressed in different forms such as set-builder form, number line graphs, and interval notation. Open circles on a number line denote strict inequalities ($<$ or $>$) indicating that the endpoint is not included, whereas closed or solid dots denote inclusive inequalities ($\le$ or $\ge$) indicating that the endpoint is included. Set operations like union ($\cup$) combine intervals and set difference ($-$) excludes specific points or intervals.

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)

Question 2: Palette 2/111

DPT MCQ

To solve polynomial inequalities using the method of intervals, first factor the polynomial completely, identify all critical points by setting each factor to zero, plot the points on a number line, and analyze sign changes based on the multiplicity of each factor (even multiplicity causes no sign change, while odd multiplicity causes a sign change).

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

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

Question 3: Palette 3/111

DPT MCQ

The wavy curve method (or method of intervals) is used to solve polynomial inequalities of the form $P(x) \gtrless 0$. The roots of the polynomial divide the real number line into intervals. The sign of the expression changes across roots of odd multiplicity and remains the same across roots of even multiplicity, provided the coefficient of $x$ in all linear factors is positive.

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

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

Question 4: Palette 4/111

DPT MCQ

Concept

+

To solve a rational inequality of the form $\frac{P(x)}{Q(x)} \ge 0$, factorize the numerator and the denominator into linear factors to find the critical points. The expression is non-negative where the numerator and denominator have the same sign, keeping in mind that the denominator must never equal zero.

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

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

Question 5: Palette 5/111

DPT MCQ

Concept

+

Solving rational inequalities by the sign scheme (interval) method. A fraction $\frac{P(x)}{Q(x)} \geq 0$ when numerator and denominator have the same sign, or when $P(x) = 0$ with $Q(x) \neq 0$; points where $Q(x) = 0$ are always excluded. Key factorisations: $x^2 - 6x + 9 = (x - 3)^2$ (perfect square, always $\geq 0$) and $5 - 4x - x^2 = -(x + 5)(x - 1)$.

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

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

Question 6: Palette 6/111

DPT MCQ

To solve a rational inequality where a constant or fraction appears on both sides, bring all terms to one side to form a single rational expression greater than or equal to zero (or less than or equal to zero). Find a common denominator, simplify the numerator, factor all components, and use the wavy curve method while excluding values that make any denominator zero.

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

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

Question 7: Palette 7/111

DPT MCQ

Higher-order rational inequalities are solved by completely factorizing all polynomial components into linear and irreducible quadratic factors, identifying all critical points, noting the multiplicity of each root (even powers maintain sign, odd powers change sign), and finding the intervals satisfying the inequality.

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

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

Question 8: Palette 8/111

DPT MCQ

Rational inequalities involving polynomials of degree greater than two are solved by factoring out common terms, completely factoring all polynomial numerators and denominators, identifying critical points, and determining the sign intervals where the strict inequality holds.

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

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

Question 9: Palette 9/111

DPT MCQ

To solve inequalities with rational expressions on both sides, transpose all terms to one side, find a common denominator, simplify the resulting single rational expression, factor completely, and apply the wavy curve method while explicitly excluding values that make any denominator zero.

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

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

Question 10: Palette 10/111

DPT MCQ

Rational inequalities are solved by finding the critical points where the numerator or denominator equals zero, factorizing polynomial expressions, and applying the wavy curve method while taking care of the domain restrictions where the denominator is non-zero. The formula for factorizing a quadratic expression $ax^2 + bx + c$ is used to simplify the rational expression.

(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

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

Question 11: Palette 11/111

DPT MCQ

Rational inequalities containing quadratic expressions in the denominator are solved by factorizing the quadratic expression into linear factors, identifying all critical points where the numerator or denominator becomes zero, determining the sign of the rational expression in each interval using the wavy curve method, and finding the intervals where the inequality holds true.

(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

$(-3, 4)$

Question 12: Palette 12/111

DPT MCQ

To solve rational inequalities, bring all terms to one side to form a single rational expression compared to zero, find the critical points by setting the numerator and denominator to zero, determine the sign of the expression in each interval, and find the domain restrictions where the denominator equals zero.

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

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

Question 13: Palette 13/111

DPT MCQ

To solve rational inequalities involving quadratic expressions, rearrange the inequality to have zero on one side, combine the terms into a single fraction, factor the numerator and denominator, analyze the sign of the expression across different intervals, and account for any domain restrictions where the denominator becomes zero.

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

$[-1, 0]$

Question 14: Palette 14/111

DPT MCQ

To solve complex rational inequalities containing multiple fractions, group or pair terms to simplify the expression using algebraic identities, combine them into a single rational function compared to zero, find the critical points, and analyze the sign intervals while considering domain restrictions.

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

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

Question 15: Palette 15/111

DPT MCQ

To solve rational inequalities of the form $\frac{P(x)}{Q(x)} < 0$, factor both the numerator and the denominator completely into linear or irreducible quadratic factors, identify all critical points where any factor is zero, set up sign intervals, and determine where the quotient is negative.

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

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

Question 16: Palette 16/111

DPT MCQ

To solve rational inequalities containing quadratic and higher-degree terms, factor all components completely, identify the sign of irreducible quadratic expressions by checking their discriminants, determine the critical points, and analyze the sign intervals.

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

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

Question 17: Palette 17/111

DPT MCQ

To solve rational inequalities involving quadratic terms, factor both the numerator and the denominator, evaluate the signs of any irreducible quadratic factors using their discriminants, and analyze the sign intervals to find where the quotient is positive.

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

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

Question 18: Palette 18/111

DPT MCQ

Higher-degree polynomial inequalities and rational inequalities with factors raised to various powers are solved using the wavy curve method. Factors with even powers do not change sign across their critical points (except they might equal zero), while factors with odd powers result in a sign change across their critical points. Denominator factors are strictly restricted from equaling zero.

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

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

Question 19: Palette 19/111

DPT MCQ

For a rational inequality $\frac{P(x)}{Q(x)} \ge 0$, if any quadratic factor $ax^2 + bx + c$ has a negative discriminant ($D = b^2 - 4ac < 0$), its sign is constant for all real values of $x$ and depends entirely on the coefficient of $x^2$ ($a$). If $a > 0$, $ax^2 + bx + c > 0$, and if $a < 0$, $ax^2 + bx + c < 0$. These strictly positive or negative non-zero terms can be simplified to reduce the inequality to simple linear factors, which are then solved using the wavy curve method (method of intervals).

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 $2 < x \le 3$

Question 20: Palette 20/111

DPT MCQ

Rational inequalities are solved by bringing all terms to one side to form a single rational expression greater or less than zero, factorizing the numerator and denominator completely into linear terms, identifying all critical points, and determining the sign of the expression across intervals using the wavy curve method.

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

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

Question 21: Palette 21/111

DPT MCQ

To solve a polynomial inequality, find the critical points by setting the expression equal to zero, divide the number line into intervals, and determine the sign of the expression in each interval. For a product greater than or equal to zero, the solution includes the intervals where the product is positive or zero.

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

$x\in[2,3]$

Question 22: Palette 22/111

DPT MCQ

To solve rational inequalities of the form $\frac{P(x)}{Q(x)}>0$ or $<0$, find all critical points where the numerator or denominator becomes zero. Mark these points on a number line, taking care that denominator points are always open circles (excluded from the domain). Test the sign of the expression in each resulting interval.

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

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

Question 23: Palette 23/111

DPT MCQ

To solve rational inequalities like $\frac{1}{x-5}<3$, transpose all terms to one side to get a single rational expression compared to zero, find the critical points by setting the numerator and denominator equal to zero, and use the interval method (wavy curve method) to find where the inequality holds true.

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

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

Question 24: Palette 24/111

DPT MCQ

To solve rational inequalities of the form $\frac{P(x)}{Q(x)}\ge0$, find the critical points where numerator and denominator are zero, use the wavy curve method to find the solution range while strictly excluding values that make the denominator zero. Then, list all integers within this valid range and sum them up.

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

A

-14

B

+14

C

-15

D

15

Question 25: Palette 25/111

DPT MCQ

To solve a quadratic inequality like $ax^{2}+bx+c<0$, factor the quadratic expression into linear factors $(x-r_{1})(x-r_{2})<0$ where $r_{1} < r_{2}$ are the roots. The solution for a negative product lies strictly between the two roots, i.e., $x\in(r_{1}, r_{2})$.

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

A

$x\in(5,7)$

B

$x\in[5,7)$

C

$x\in(5,7]$

D

$x\in[5,7]$

Question 26: Palette 26/111

DPT MCQ

To solve rational inequalities involving quadratic expressions, factor both the numerator and the denominator into linear factors, find all critical points where factors equal zero, and use the sign scheme (wavy curve method) while keeping in mind that points from the denominator must be strictly excluded.

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

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

Question 27: Palette 27/111

DPT MCQ

To solve rational inequalities like $\frac{P(x)}{Q(x)}\le 2$, bring all terms to one side to compare with zero, combine them into a single rational expression, find the critical points from the numerator and denominator, and determine the sign scheme using the wavy curve method while excluding values that make the denominator zero.

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

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

Question 28: Palette 28/111

DPT MCQ

To solve rational inequalities of the form $\frac{P(x)}{Q(x)}\le0$, factor the numerator and denominator into linear factors, find all critical points where factors equal zero, use the wavy curve method to identify the intervals where the expression is less than or equal to zero while strictly excluding denominator roots, and then find the largest integer in the solution set.

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

A

5

B

3

C

4

D

6

Question 29: Palette 29/111

DPT MCQ

Rational inequalities are solved by bringing all terms to one side to form a single rational expression, finding the critical points where the numerator and denominator are zero, and using the wavy curve method to determine the sign of the expression in various intervals. The denominator must never be zero. Formula: If $\frac{P(x)}{Q(x)}\ge0$, find roots of $P(x)=0$ and $Q(x)=0$, plot them on a number line, and check intervals.

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

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

Question 30: Palette 30/111

DPT MCQ

Rational inequalities involving quadratic expressions are solved by factorizing both the numerator and the denominator into linear factors, identifying the critical points, and using the wavy curve method to determine the sign of the expression in each interval while ensuring the denominator does not equal zero. Formula: If $\frac{(x-a)(x-b)}{(x-c)(x-d)}\ge0$, find roots of all factors, place them on the number line in increasing order, and assign alternating signs starting from the right for odd multiplicities.

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

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

Question 31: Palette 31/111

DPT MCQ

Quadratic inequalities of the form $ax^2 + bx + c < 0$ are solved by finding the roots of the corresponding quadratic equation $ax^2 + bx + c = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. If $a > 0$, the expression is negative between the two roots.

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

$x\in(-1,2)$

Question 32: Palette 32/111

DPT MCQ

Rational inequalities with powers are solved by factoring the numerator and denominator. Even powers are always non-negative and do not change the sign of the inequality, but they introduce restrictions where the denominator becomes zero. Odd powers retain the sign of the underlying linear factors.

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

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

Question 33: Palette 33/111

DPT MCQ

Higher-order polynomial inequalities are solved using the wavy curve method (Wilsons method). Factors with even exponents are non-negative for all real numbers and their roots are included if the inequality is non-strict ($\ge 0$). Factors with odd exponents dictate the sign change across critical points.

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

$3.0$

Question 34: Palette 34/111

DPT MCQ

Rational inequalities are solved by checking the sign of each factor. Quadratic expressions with a negative discriminant and a positive leading coefficient are always positive for all real numbers and can be factored out from the inequality.

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

$x\le1$

Question 35: Palette 35/111

DPT MCQ

Rational inequalities containing factors raised to high powers are simplified by examining the parity of the exponents. Even exponents are non-negative for all real numbers and their roots must be checked separately for inclusion or exclusion, while odd exponents determine the sign changes across critical points.

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

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

Question 36: Palette 36/111

DPT MCQ

Rational inequalities are solved by factoring both the numerator and the denominator completely into linear and irreducible quadratic factors, identifying all critical points and domain restrictions, and applying the wavy curve method while accounting for even-power factors and point-wise equality.

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

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

Question 37: Palette 37/111

DPT MCQ

Rational inequalities with high-power factors are solved by identifying critical points from numerator and denominator, considering the parities of exponents, and determining the sign intervals via the wavy curve method while accounting for domain restrictions and point-wise equality.

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

Question 38: Palette 38/111

DPT MCQ

Double inequalities of the form $f(x) < g(x) < h(x)$ are solved by splitting them into two separate inequalities, $f(x) < g(x)$ and $g(x) < h(x)$, finding the common solution set of $x$ satisfying both, and then counting the integers within that range.

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

A

1

B

3

C

4

D

5

Question 39: Palette 39/111

DPT MCQ

Polynomial inequalities of degree four are solved using the wavy curve method by identifying all critical points. The number of integer solutions contained within the resulting intervals depends on the position of the variable parameter $n$, allowing us to solve for $n$ given a specific count of integer solutions.

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

$8$

Question 40: Palette 40/111

DPT MCQ

Rational inequalities are solved by bringing all terms to one side to form a single fraction greater than or equal to zero, factoring both the numerator and denominator completely, and applying the wavy curve method based on the critical points.

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

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

Question 41: Palette 41/111

DPT MCQ

Rational inequalities are solved by bringing all terms to one side to form a single rational expression of the form $\frac{P(x)}{Q(x)} > 0$ or $< 0$, finding the critical points where the numerator or denominator equals zero, and testing intervals on the real number line while ensuring the denominator is non-zero.

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

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

Question 42: Palette 42/111

DPT MCQ

Rational inequalities are solved by finding a common denominator to combine all fractions into a single rational expression of the form $\frac{P(x)}{Q(x)} > 0$, determining the critical points from the numerator and denominator, and applying the wavy curve method over intervals while excluding values that make the denominator zero.

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

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

Question 43: Palette 43/111

DPT MCQ

Rational inequalities are solved by bringing all terms to one side, finding a common denominator, determining the critical points, and testing intervals. Prime numbers are integers strictly greater than $1$ that have no positive divisors other than $1$ and themselves.

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

A

1

B

2

C

3

D

4

Question 44: Palette 44/111

DPT MCQ

Rational inequalities are solved by finding the roots of the numerator and denominator, factorizing all expressions completely, finding the valid sign intervals for each inequality using the wavy curve method, and then taking the intersection of the solution sets to determine the set $S$.

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

$(-10,-4)$

Question 45: Palette 45/111

DPT MCQ

The wavy curve method is used to solve polynomial and rational inequalities by finding the critical points where the expression equals zero, arranging them on a number line, and determining the sign of the expression in each interval.

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

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

Question 46: Palette 46/111

DPT MCQ

The wavy curve method for rational inequalities involves finding critical points from both the numerator and the denominator, ensuring denominator roots are excluded, and analyzing the sign changes across intervals while noting the multiplicity of the factors.

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

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

Question 47: Palette 47/111

DPT MCQ

Factor all quadratic and higher-degree polynomial expressions in the numerator and denominator completely into linear factors, find all critical points, account for even multiplicity factors and denominator restrictions, and apply the wavy curve method.

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

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

Question 48: Palette 48/111

DPT MCQ

To find the simultaneous solution of multiple inequalities, we solve each inequality independently to obtain their respective solution sets and then find their intersection. For rational inequalities, we use the wavy curve method by finding critical points from the numerator and denominator. For quadratic inequalities, we find the roots of the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ and determine the interval where the expression is non-positive.

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

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

Question 49: Palette 49/111

DPT MCQ

To solve rational inequalities using the wavy curve method, factor all quadratic polynomials in the numerator and denominator completely into linear factors, find the critical points, handle any quadratic factors with negative discriminants that have constant positive signs, rewrite factors with negative leading coefficients to make the coefficient of $x$ positive, and then evaluate the greatest common divisor (GCD) of the resulting integers.

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

Question 50: Palette 50/111

DPT MCQ

To solve complex rational inequalities, we can pair fractions strategically using algebra (or partial fraction/grouping techniques) to simplify the expression into a single rational function. Then, we find all critical points from the numerator and denominator, exclude values where the denominator becomes zero, and use the wavy curve method to determine the intervals satisfying the inequality.

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

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

Question 51: Palette 51/111

DPT MCQ

To solve high-degree rational inequalities using the wavy curve method, identify all critical points from the numerator and denominator, analyze the parity of exponents (even exponents do not change the sign of the expression across the critical point, while odd exponents do), exclude values that make the denominator zero, and determine the intervals where the expression satisfies the inequality.

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

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

Question 52: Palette 52/111

DPT MCQ

To solve a rational inequality, transfer all terms to one side to form a single rational expression with a common denominator, find the critical points by setting the numerator and denominator to zero, exclude values that make the denominator zero, and use the wavy curve method to determine the sign in each interval.

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

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

Question 53: Palette 53/111

DPT MCQ

When solving polynomial inequalities using the wavy curve method, factors with even powers do not change the sign of the expression across their critical points, whereas factors with odd powers do. We also adjust any factors where the coefficient of $x$ is negative to make it positive, noting the corresponding change in the inequality sign.

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

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

Question 54: Palette 54/111

DPT MCQ

Factor all polynomial expressions completely into linear and irreducible quadratic factors, find the critical points, handle terms with even powers or repeated roots carefully, and use the wavy curve method to find the solution set.

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

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

Question 55: Palette 55/111

DPT MCQ

To solve a rational inequality involving a constant on one side, move all terms to the left-hand side to obtain a rational expression less than or equal to zero, find a common denominator, factor the numerator and denominator completely, and apply the wavy curve method while excluding denominator roots.

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

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

Question 56: Palette 56/111

DPT MCQ

For rational inequalities where both denominators can be analyzed for sign or quadratic positivity, substitute $y = x^2+x$ to simplify the expression into a rational inequality in terms of $y$, solve for $y$, and then substitute back to find the corresponding intervals for $x$.

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

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

Question 57: Palette 57/111

DPT MCQ

To solve rational inequalities using the wavy curve method, factor both the numerator and denominator completely into linear factors, find all critical points, exclude values that make the denominator zero, and determine the sign of the expression across the intervals. Then, identify all integers belonging to the solution set and compute their sum.

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

Question 58: Palette 58/111

DPT MCQ

When solving polynomial inequalities using the wavy curve method, factors with even exponents do not alter the sign of the expression across their critical points (though points where the expression equals zero are included if the inequality is non-strict), while factors with odd exponents cause a sign change.

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

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

Question 59: Palette 59/111

DPT MCQ

To solve high-power rational inequalities using the wavy curve method, simplify each polynomial factor completely, note the powers (even powers yield non-negative values and do not change signs, while odd powers change signs), adjust factors with negative leading coefficients to make the coefficient of $x$ positive, and exclude values that make the denominator zero.

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

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

Question 60: Palette 60/111

DPT MCQ

To solve rational inequalities of the form $\frac{A(x)}{B(x)} \ge C$, first rewrite the inequality so that one side is $0$ by subtracting $C$ from both sides. Combine the terms into a single rational expression over a common denominator to obtain $\frac{P(x)}{Q(x)} \ge 0$. Determine the critical points where the numerator equals zero ($P(x) = 0$) or the denominator equals zero ($Q(x) = 0$). Plot these critical numbers on a number line to divide it into test intervals. Test a value in each interval using the wavy curve method (sign scheme) to identify where the expression satisfies the inequality condition, remembering that points making the denominator zero are excluded.

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

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

Question 61: Palette 61/111

DPT MCQ

An irrational equation of the form $\sqrt{f(x)} = g(x)$ requires $g(x) \ge 0$ and $f(x) \ge 0$. Squaring both sides gives $f(x) = (g(x))^2$. Potential solutions must be checked in the original equation to eliminate extraneous roots.

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

Question 62: Palette 62/111

DPT MCQ

The principal square root of a real number is always non-negative, meaning $\sqrt{f(x)} \ge 0$ for all real values of $x$ where defined. An equation of the form $\sqrt{f(x)} = c$ with $c < 0$ has no real solutions.

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

A

$x = -21$

B

$x = 29$

C

No real solution

D

$x = 21$

Question 63: Palette 63/111

DPT MCQ

To solve equations with multiple radical terms, isolate one radical term, square both sides to reduce the number of radicals, isolate the remaining radical term, square again, and check potential roots in the original equation.

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

A

$x = 20$

B

$x = 4$

C

$x = 4$ and $x = 20$

D

$x = 11$

Question 64: Palette 64/111

DPT MCQ

For equations with nested radicals, clear outer radicals first by squaring both sides after verifying domain restrictions ($x \ge 2$), then isolate remaining radicals and square again to solve the resulting polynomial.

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

A

$x = 3$

B

$x = 6$

C

$x = 3, 6$

D

$x = 2$

Question 65: Palette 65/111

DPT MCQ

To solve an radical equation of the form $\sqrt{f(x)} = g(x)$, ensure the non-negativity constraints for real solutions: the principal square root $\sqrt{f(x)} \ge 0$, which implies $g(x) \ge 0$, and the radicand $f(x) \ge 0$. Square both sides of the equation to eliminate the radical sign: $f(x) = (g(x))^2$.

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 $a \ge 0$

Question 66: Palette 66/111

DPT MCQ

An irrational equation of the form $\sqrt{f(x)} = \sqrt{g(x)}$ is valid only when $f(x) \ge 0$ and $g(x) \ge 0$. Squaring both sides yields $f(x) = g(x)$, which gives $x = \frac{a-3}{2}$, valid provided $a \ge -3$.

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 $a < -3$

Question 67: Palette 67/111

DPT MCQ

For equations involving two square root radicals $\sqrt{f(x)} = c - \sqrt{g(x)}$, square both sides to isolate the remaining radical. After simplifying and squaring again, check solutions against the original equation to eliminate extraneous roots.

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

A

$x = -9$

B

$x = 0$

C

$x = 16$

D

No real solution

Question 68: Palette 68/111

DPT MCQ

Equations containing expressions of the form $P(x) + c\sqrt{P(x)+d} = k$ can be solved using substitution $y = \sqrt{P(x)+d}$, where $y \ge 0$. This transforms the equation into a standard quadratic in terms of $y$.

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

$x = 0, -5$

Question 69: Palette 69/111

DPT MCQ

To solve symmetric nested radical equations of the form $\sqrt{A+B} + \sqrt{A-B} = C$, square both sides using $(a+b)^2 = a^2 + 2ab + b^2$ to simplify the expression, then solve for the variable while checking domain conditions.

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

A

$x = 5$

B

$x = 25$

C

$x = 9$

D

$x = 14$

Question 70: Palette 70/111

DPT MCQ

For equations involving symmetric algebraic structures under radicals, substitution of repeating expressions such as $y = 3x^2 - 2x$ simplifies the equation to a single-variable radical relation. Alternatively, using the algebraic identity $(A-B)(A+B) = A^2 - B^2$ yields a system of linear equations in radical terms.

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

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

Question 71: Palette 71/111

DPT MCQ

For algebraic equations involving repeating radical terms of the form $\frac{a}{f(x)} - f(x) = b$, apply a substitution $y = f(x)$ with the constraint $y > 0$ to convert the irrational equation into a standard quadratic equation $y^2 + by - a = 0$. Solve for valid positive values of $y$, and back-substitute to find $x$.

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

A

$x = 3$

B

$x = -3$

C

$x = 5$

D

$x = 1$

Question 72: Palette 72/111

DPT MCQ

For radical equations of the form $\sqrt{f(x)} = g(x)$, the non-negativity constraint of the principal square root requires $g(x) \ge 0$, which sets the domain restriction. Squaring both sides yields the polynomial equation $f(x) = (g(x))^2$. After solving the resulting quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, extraneous roots that do not satisfy $g(x) \ge 0$ must be discarded.

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

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

Question 73: Palette 73/111

DPT MCQ

Bounding method (Inequality analysis of expressions) and completing the square formula $(a+b)^2 = a^2 + 2ab + b^2$.
If $f(x) \ge k$ and $g(x) \le k$, then the equation $f(x) = g(x)$ holds true if and only if $f(x) = k$ and $g(x) = k$ simultaneously.

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

$x = -2$

Question 74: Palette 74/111

DPT MCQ

For radical equations of the form $\frac{a + \sqrt{f(x)}}{x} = 1$, clear the denominator to get $\sqrt{f(x)} = x - a$. Apply the non-negativity condition for the principal square root ($x - a \ge 0$) and ensure the domain constraint ($f(x) \ge 0$). Square both sides to convert the irrational equation into a standard quadratic equation, solve for $x$, and discard any extraneous roots.

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

A

$x = 3$

B

$x = -3$

C

$x = 5, -3$

D

$x = 5$

Question 75: Palette 75/111

DPT MCQ

For cubic radical equations of the form $\sqrt[3]{f(x)} = g(x)$, eliminate the cube root by cubing both sides of the equation: $f(x) = (g(x))^3$. Use the binomial expansion formula $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ to simplify the equation, collect like terms, and solve the resulting polynomial. Unlike square roots, cube roots are defined for all real numbers and do not produce extraneous sign-related roots.

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

A

$x = 2$

B

$x = 4$

C

$x = 2, -2$

D

$x = 0$

Question 76: Palette 76/111

DPT MCQ

For equations with two square root terms $\sqrt{f(x)} - \sqrt{g(x)} = c$, isolate one radical term to get $\sqrt{f(x)} = c + \sqrt{g(x)}$, then square both sides to eliminate one square root. Isolate the remaining radical term, apply non-negativity conditions to avoid extraneous solutions, and square both sides a second time to reduce the equation to a standard quadratic form.

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

A

$x = 4$

B

$x = 20$

C

$x = 20, 4$

D

$x = 15$

Question 77: Palette 77/111

DPT MCQ

For equations containing the sum of two cube roots $\sqrt[3]{A} + \sqrt[3]{B} = C$, use algebraic substitution $a = \sqrt[3]{A}$ and $b = \sqrt[3]{B}$. This gives the linear system $a + b = C$ and $a^3 + b^3 = A + B$. Apply the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ to find the product $ab$, then form a quadratic equation to solve for $a$ and $b$, and finally solve for $x$.

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

$x = -15, 2$

Question 78: Palette 78/111

DPT MCQ

For nested radical equations of the form $\sqrt{x} + \sqrt{x - \sqrt{1 - x}} = 1$, isolate the nested radical term to obtain $\sqrt{x - \sqrt{1 - x}} = 1 - \sqrt{x}$. Apply domain and non-negativity constraints, square both sides sequentially to eliminate the radicals step-by-step, reduce the resulting equation to a polynomial form, and solve for $x$.

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

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

Question 79: Palette 79/111

DPT MCQ

For radical equations of the form $\sqrt{x + a} - 1 = \sqrt{x - \sqrt{x - b}}$, square both sides to eliminate the outermost square root on the right side. Simplify the resulting equation to isolate the remaining inner square root term, apply non-negativity constraints, and square again to convert the expression into a standard polynomial equation to solve for $x$.

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

A

$x = 4$

B

$x = 6$

C

$x = 8$

D

$x = 2$

Question 80: Palette 80/111

DPT MCQ

For radical equations involving expressions of the form $\sqrt{A^2}$, apply the principal square root identity $\sqrt{A^2} = \vert{}A\vert{}$. This converts the equation into an absolute value equation $\vert{}f(x)\vert{} - \vert{}g(x)\vert{} = c$. Divide the real number line into critical intervals based on the zeros of the absolute value terms, remove the absolute value bars by determining their signs in each interval, and find the corresponding interval solutions.

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

A

$[2, +\infty)$

B

$(-\infty, -1]$

C

$[-1, 2]$

D

$\{2\}$

Question 81: Palette 81/111

DPT MCQ

For symmetrical cubic-radical equations of the form $x^3 + 1 = 2\sqrt[3]{2x - 1}$, introduce a variable substitution $y = \sqrt[3]{2x - 1}$. Cubing both sides yields $y^3 + 1 = 2x$. Subtracting this equation from $x^3 + 1 = 2y$ forms a system of symmetric equations $(x^3 - y^3) = 2(y - x)$, which factorizes to show $x = y$. Substituting $y = x$ back into the cubic equation allows solution by standard polynomial factoring.

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

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

Question 82: Palette 82/111

DPT MCQ

For an expression involving a principal square root and rational functions:
1. The domain requires the expression under the square root to be non-negative: $f(x) \ge 0$.
2. The denominator cannot be zero: $g(x) \neq 0$.
3. Since $\sqrt{f(x)} > 0$ strictly, $f(x)$ must be strictly positive ($f(x) > 0$), making the numerator positive. For the fraction to be positive, the denominator must also be positive: $g(x) > 0$.

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

A

$x \in [3,\infty)$

B

$x \in (2,\infty)$

C

$x \in (3,\infty)$

D

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

Question 83: Palette 83/111

DPT MCQ

1. The principal square root of any real expression is non-negative whenever it is defined: $\sqrt{f(x)} \ge 0$.
2. Since any non-negative quantity is always strictly greater than any negative number ($-1 < 0$), the inequality holds true for all $x$ in the domain of the expression.
3. The domain requirement is $\frac{f(x)}{g(x)} \ge 0$ with $g(x) \neq 0$.

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

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

Question 84: Palette 84/111

DPT MCQ

For an irrational inequality of the form $\sqrt{f(x)} \ge \sqrt{g(x)}$:
1. $g(x) \ge 0$
2. $f(x) \ge 0$
3. Squaring both sides yields $f(x) \ge g(x)$

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

$x \in [3, 5]$

Question 85: Palette 85/111

DPT MCQ

For an inequality of the form $A(x) \ge \sqrt{B(x)}$:
1. $B(x) \ge 0$ (expression under square root must be non-negative)
2. $A(x) \ge 0$ (since $\sqrt{B(x)} \ge 0$, the left-hand side must also be non-negative)
3. Squaring both sides gives $(A(x))^{2} \ge B(x)$

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

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

Question 86: Palette 86/111

DPT MCQ

For an irrational inequality containing two radical terms $\sqrt{f(x)} - \sqrt{g(x)} \ge c$:
1. Establish the domain such that both expressions under the square roots are non-negative: $f(x) \ge 0$ and $g(x) \ge 0$.
2. Rearrange the inequality to isolate one radical term on each side so that both sides remain non-negative: $\sqrt{f(x)} \ge c + \sqrt{g(x)}$.
3. Square both sides to eliminate the outer radicals, simplify, and ensure non-negativity before squaring again if necessary.

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

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

Question 87: Palette 87/111

DPT MCQ

For an irrational inequality of the form $f(x) > \sqrt{g(x)}$:
1. $g(x) \ge 0$ (expression under square root must be non-negative to define the real domain)
2. $f(x) > 0$ (since $\sqrt{g(x)} \ge 0$, the left-hand side must be strictly positive)
3. Squaring both sides yields $(f(x))^{2} > g(x)$

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

$x \in [2, 3]$

Question 88: Palette 88/111

DPT MCQ

For an irrational inequality of the form $\sqrt{f(x)} > \sqrt{g(x)}$:
1. Both expressions under the square roots must be non-negative: $g(x) \ge 0$ and $f(x) \ge 0$.
2. Squaring both sides gives $f(x) > g(x)$.

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

$x \in [2,3]$

Question 89: Palette 89/111

DPT MCQ

For an irrational inequality of the form $\sqrt{f(x)} \ge g(x)$:
1. $f(x) \ge 0$ (expression under square root must be non-negative).
2. Case 1: If $g(x) < 0$, the inequality $\text{non-negative} \ge \text{negative}$ is always true for all valid $x$ in the domain.
3. Case 2: If $g(x) \ge 0$, both sides are non-negative, so we square both sides: $f(x) \ge (g(x))^2$.

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

A

$(-\infty, 2]$

B

$[0, 2]$

C

$[-2, 2]$

D

$[2, \infty)$

Question 90: Palette 90/111

DPT MCQ

For an irrational inequality of the form $\sqrt{f(x)} < g(x)$:
1. $f(x) \ge 0$ (expression under square root must be non-negative)
2. $g(x) > 0$ (since $\sqrt{f(x)} \ge 0$, the right-hand side must be strictly positive)
3. Squaring both sides yields $f(x) < (g(x))^{2}$

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

A

7

B

5

C

2

D

3

Question 91: Palette 91/111

DPT MCQ

For an irrational inequality of the form $\sqrt{f(x)} > g(x)$:
1. The domain requires $f(x) \ge 0$.
2. Case 1: If $g(x) < 0$, the inequality $\text{non-negative} > \text{negative}$ is always true for all valid $x$ in the domain.
3. Case 2: If $g(x) \ge 0$, both sides are non-negative, so squaring both sides gives $f(x) > (g(x))^{2}$.

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

Question 92: Palette 92/111

DPT MCQ

For a rational inequality containing a principal square root term $\frac{\sqrt{f(x)} \cdot g(x)}{h(x)} < 0$:
1. $f(x) \ge 0$ defines the real domain.
2. If $f(x) = 0$, the numerator becomes $0$, making the quotient equal to $0$, which does not satisfy strict inequality $< 0$. Thus, $f(x) > 0$.
3. Since $\sqrt{f(x)} > 0$, it is strictly positive and can be omitted when determining the sign of the inequality, leaving $\frac{g(x)}{h(x)} < 0$ subject to $f(x) > 0$.

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

$(1,2]$

Question 93: Palette 93/111

DPT MCQ

Method of elimination of radicals by repeated squaring (Irrational Equations).
Formulas involved: $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$.
Any extraneous roots introduced during the squaring process must be eliminated by substituting the found values back into the original equation.

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

A

$x = 197$

B

$x = 5$

C

$x = 6$

D

$x = 10$

Question 94: Palette 94/111

DPT MCQ

Method of elimination of radicals by squaring both sides (Irrational Equations).
Formula involved: $(a+b)^2 = a^2 + 2ab + b^2$.
Domain of definition requirement: All radical expressions must be non-negative.

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

A

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

B

$x = 0$

C

$x = 4$

D

$x = 2$

Question 95: Palette 95/111

DPT MCQ

Algebraic substitution and identity method for cubic radicals (Irrational Equations).
Formulas involved: $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ and $(1 - x)^3 = 1 - 3x + 3x^2 - x^3$.

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

A

$x = 0$

B

$x = 2$

C

$x = 1$

D

$x = -1$

Question 96: Palette 96/111

DPT MCQ

The Method of Introducing New Variables (Substitution Method for Irrational Equations).
Formula involved: Let $y = \sqrt{f(x)} \ge 0$, then $f(x) = y^2$, transforming the irrational equation into a quadratic equation $y^2 + ay + b = 0$.

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

$x = 3, x = -4$

Question 97: Palette 97/111

DPT MCQ

System of Symmetric Equations and Substitution for Radicals (Irrational Equations).
Formulas involved: Let $u = \sqrt[4]{a(x)}$ and $v = \sqrt[4]{b(x)}$ such that $u \ge 0$ and $v \ge 0$.
$(u + v)^4 = u^4 + v^4 + 4uv(u^2 + v^2) + 6u^2v^2$ and $u^2 + v^2 = (u + v)^2 - 2uv$.

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

$x = 2, x = -14$

Question 98: Palette 98/111

DPT MCQ

System of Equations by Substitution for Radicals (Irrational Equations).
Formulas involved: Let $u = \sqrt[3]{x+1}$ and $v = \sqrt{x-3}$ where $v \ge 0$.
$u^3 - v^2 = (x + 1) - (x - 3) = 4$.

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

A

$x = 3$

B

$x = 5$

C

$x = 7$

D

$x = 11$

Question 99: Palette 99/111

DPT MCQ

Concept

+

Modulus equation of the type $|f(x)| = g(x)$ is split into two cases $f(x) = g(x)$ and $f(x) = -g(x)$, subject to the condition $g(x) \ge 0$ (since $|y| = a \Rightarrow a \ge 0$).
Quadratic roots are found using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

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

$11 - \sqrt{3}$

Question 100: Palette 100/111

DPT MCQ

Concept

+

Piecewise removal of modulus using critical points: $|x + a| = x + a$ for $x \ge -a$ and $|x + a| = -(x + a)$ for $x < -a$.
The equation is solved separately in each interval defined by the critical points, and a root is valid only if it lies in the assumed interval. Quadratic roots via $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$; no real roots if $b^2 - 4ac < 0$.

[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

Question 101: Palette 101/111

DPT MCQ

Concept

+

Sign scheme of $x^2 - 1 = (x - 1)(x + 1)$: positive on $(-\infty, -1] \cup [1, \infty)$ and negative on $(-1, 1)$, so $|x^2 - 1| = x^2 - 1$ outside $(-1, 1)$ and $|x^2 - 1| = -(x^2 - 1)$ inside $(-1, 1)$.
Solve the quadratic in each region and accept only roots lying in the assumed region; roots via $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

[JEE Adv.-2021] For $x \in R$, the number of real roots of the equation $3x^2 - 4|x^2 - 1| + x - 1 = 0$ is

A

2

B

6

C

4

D

3

Question 102: Palette 102/111

DPT MCQ

Concept

+

Absolute value equation of the form $|A| = B$ splits into two linear equations $A = B$ or $A = -B$, and since $|A| \ge 0$ we must have $B \ge 0$; every root must be checked in the original equation.

Solve $|3x - 2| = x$

A

$x = 1/2$ or $x = 1$

B

$x = 1/2$ or $x = -1$

C

$x = 1$ or $x = -1$

D

$x = 2$ or $x = -2$

Question 103: Palette 103/111

DPT MCQ

Concept

+

Sum of distances property: $|x - a| + |x - b|$ equals the fixed distance $|a - b|$ exactly when $x$ lies between $a$ and $b$ (inclusive); solved by case analysis at critical points $x = a$ and $x = b$. Formula: for $a < b$, $|x - a| + |x - b| = (b - a)$ when $a \le x \le b$, $= 2x - a - b$ when $x > b$, and $= a + b - 2x$ when $x < a$.

Solve $|x+1| + |x-1| = 2$

A

$x \in (-1, 1)$

B

$x \in [-1, 1]$

C

$x \in [1, 2]$

D

$x \in (1, 2)$

Question 104: Palette 104/111

DPT MCQ

Concept

+

Nested absolute value equation: unwrap from the outside using $|A| = c \Rightarrow A = c$ or $A = -c$ (requires $c \ge 0$), then solve each inner equation $|x - a| = b \Rightarrow x = a \pm b$ (requires $b \ge 0$); any branch giving a negative absolute value is rejected.

Solution of $||x-1| - 2| = 1$ is

A

$\{-1, 0, 1, 4\}$

B

$\{-2, 0, 2, 3\}$

C

$\{-2, 0, 3, 4\}$

D

$\{-2, 0, 2, 4\}$

Question 105: Palette 105/111

DPT MCQ

Concept

+

Sum of absolute values as a piecewise linear function: critical points at $x = 1, 2, 3$ split the real line into regions where each modulus opens with a fixed sign, giving a linear equation per region; equivalently $\sum |x - a_i|$ is minimised at the median of the $a_i$, so solutions lie in the outer regions when the right side exceeds that minimum.

Solve $|x - 1| + |x - 2| + |x - 3| = 4$

A

$x \in [2/3, 10/3]$

B

$x \in \{1, 3\}$

C

$x \in \{2/3, 10/3\}$

D

No solution

Question 106: Palette 106/111

DPT MCQ

Concept

+

Absolute value inequality of the form $|u| < c$ with $c > 0$ is equivalent to the compound inequality $-c < u < c$.
Formula: $|x - a| < b \iff -b < x - a < b \iff a - b < x < a + b$, written in interval notation as $(a - b, a + b)$.

Solve $|x - 5| < 2$.

A

$(-2, 2)$

B

$(3, 7)$

C

$[3, 7]$

D

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

Question 107: Palette 107/111

DPT MCQ

Concept

+

Absolute value inequality of the form $|u| \geq c$ with $c > 0$ splits into two separate inequalities: $u \leq -c$ or $u \geq c$.
Formula: $|x - a| \geq b \iff x \leq a - b$ or $x \geq a + b$, written in interval notation as $(-\infty, a - b] \cup [a + b, \infty)$.

Solve $|x + 3| \geq 7$.

A

$[-10, 4]$

B

$(-\infty, -10] \cup [4, \infty)$

C

$(-\infty, -10) \cup (4, \infty)$

D

$[4, \infty)$

Question 108: Palette 108/111

DPT MCQ

Concept

+

Absolute value inequality of the form $|u| < c$ with $c > 0$ is equivalent to the compound inequality $-c < u < c$; here $u$ is a linear fraction, so after splitting we clear the denominator and isolate $x$.
Formula: $|u| < c \iff -c < u < c$, then multiply by the denominator and solve $-c \cdot d < mx + a < c \cdot d$ for $x$.

Solve $\left| \frac{3x + 3}{5} \right| < 3$.

A

$(-4, 6)$

B

$(-6, 4)$

C

$[-6, 4]$

D

$(-18, 12)$

Question 109: Palette 109/111

DPT MCQ

Concept

+

Method of intervals for inequalities containing several modulus expressions: mark the points where each expression inside the modulus sign vanishes, split the number axis into intervals, replace each modulus by its signed linear form on each interval, solve the resulting linear systems, and unite the solutions.
Formula: $|u| = u$ if $u \geq 0$ and $|u| = -u$ if $u < 0$; here $|3 - 2x| = 2\left|x - \frac{3}{2}\right|$.

Solve the inequality $x + |3 - 2x| > |x + 1| - 1$.

A

All real numbers

B

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

C

$\left(\frac{3}{2}, \infty\right)$

D

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

Question 110: Palette 110/111

DPT MCQ

Concept

+

Inequality containing the modulus of a quadratic expression: split into cases according to the sign of the expression inside the modulus sign, solve the resulting inequality in each case, intersect with the case condition, and unite the case solutions.
Formula: $|u| = u$ if $u \geq 0$ and $|u| = -u$ if $u < 0$; here Case A gives $2x^2 + 3x - 2 \geq 0$ and Case B gives the linear inequality $-3x - 2 \geq 0$.

Solve the inequality $|x^2 + 3x| + x^2 - 2 \geq 0$.

A

$\left[-\frac{2}{3}, \frac{1}{2}\right]$

B

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

C

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

D

$\left(-3, -\frac{2}{3}\right]$

Question 111: Palette 111/111

DPT MCQ

Concept

+

Method of intervals for an inequality with three modulus expressions: mark on the number line the points where each expression under the absolute value sign vanishes, split the line into intervals, replace each modulus by its signed linear form on each interval, solve the resulting linear inequality within that interval, and unite the case solutions.
Formula: $|u| = u$ if $u \geq 0$ and $|u| = -u$ if $u < 0$.

Solve the inequality $|x - 1| - |x| + |2x + 3| > 3x + 4$.

A

$(0, \infty)$

B

$(-\infty, 0)$

C

$\left(-\frac{3}{2}, 0\right)$

D

$\left(-\infty, -\frac{3}{2}\right)$