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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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391 Easy
391 of 914
Application +4 | -1
The least value of 5sin⁡x−1+5−sin⁡x−1{5^{\sin x - 1}} + {5^{ - \sin x - 1}} is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C 25\frac{2}{5} is correct because it is derived using the AM-GM inequality.
392 Medium
392 of 914
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the roots of the equation x3+ax+bx^3+a x+b =0=0, then α3+β3+γ3α2+β2+γ2=?\frac{\alpha^3+\beta^3+\gamma^3}{\alpha^2+\beta^2+\gamma^2}=?
View answer and explanation
Correct option: A
3b/2a3 b / 2 a
The correct answer is A (3b/2a3 b / 2 a) because α+β+γ=0\alpha+\beta+\gamma=0 and α3+β3+γ3=3αβγ\alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma, which simplifies the expression to 3b/2a3b/2a.
393 Easy
393 of 914
Application +4 | -1
If tan⁡A\tan A and tan⁡B\tan B are the roots of the quadratic equation, 3x2−10x−25=03 x^{2}-10 x-25=0 then the value of 3sin⁡2(A+B)−10sin⁡(A+B)⋅cos⁡(A+B)−25cos⁡2(A+B)3 \sin ^{2}(A+B)-10 \sin (A+B) \cdot \cos (A+B)-25 \cos ^{2}(A+B) is
View answer and explanation
Correct option: B
-25
Option A (34\frac{{3}}{{4}}) is correct because after solving the given equation for x, substituting x into the expression 4x2−1\sqrt{{4x^2 - 1}} yields 34\frac{{3}}{{4}}.
394 Easy
394 of 914
Analyzing +4 | -1
If the sum of the roots of the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is equal to the sum of the square of their reciprocals, then ac,  ba\frac{a}{c},\;\frac{b}{a} and cb\frac{c}{b} are in
View answer and explanation
Correct option: C
Harmonic progression
Option A (Φ) is correct because there is no real solution that satisfies both inequalities simultaneously.
395 Easy
395 of 914
Application +4 | -1
If one root of the equation x2−λx+12=0x^2-\lambda x+12=0 is even prime while x2+λx+μ=0x^2+\lambda x+\mu=0 has equal roots, then μ\mu is
View answer and explanation
Correct option: B
16
Option B (16) is correct because the value of μ\mu is found by first determining λ\lambda from the first equation and then using the condition for equal roots in the second equation.
396 Medium
396 of 914
Application +4 | -1
a,b,ca, b, c are integers, not all simultaneously equal and ω\omega is cube root of unity (ω≠1)(\omega \neq 1), then minimum value of ∣a+bω+cω2∣\left|a+b \omega+\mathrm{c} \omega^2\right| is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the minimum value of the expression is achieved when two of the integers are zero and the third is 1.
397 Medium
397 of 914
Application +4 | -1
If a,b,c>0a,b,c > 0 and if abc=1abc = 1, then the value of a+b+c+ab+bc+caa + b + c + ab + bc + ca lies in the interval
View answer and explanation
Correct option: D
(6,  ∞)\left( {6,\;\infty } \right)
Option D (6,  ∞)\left( {6,\;\infty } \right) is correct because by AM-GM inequality, a+b+c>3a+b+c > 3 and ab+bc+ca>3ab+bc+ca > 3, thus a+b+c+ab+bc+ca>6a+b+c+ab+bc+ca > 6.
398 Medium
398 of 914
Application +4 | -1
The value of aa for which one root of the quadratic equation (a2−5a+3)x2+(3a−1)x+2=0\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0 is twice as large as the other, is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (23\frac{2}{3}) is correct because it satisfies the condition where one root of the quadratic equation is twice the other, derived by relating the roots to the coefficients.
399 Easy
399 of 914
Application +4 | -1
The roots of the equation 2x+233x/(x−1)=9{2^{x + 2}}{3^{3x/\left( {x - 1} \right)}} = 9 are given by
View answer and explanation
Correct option: D
−2,  1−log3log2- 2,\;1 - \frac{{\mathrm{log}3}}{{\mathrm{log}2}}
Option D (−2,  1−log3log2- 2,\;1 - \frac{{\mathrm{log}3}}{{\mathrm{log}2}}) is correct because by taking the logarithm of both sides of the equation and solving for x, we obtain the given roots.
400 Easy
400 of 914
Application +4 | -1
{x∈R:∣x−2∣=x2}=\left\{x \in R:|x-2|=x^2\right\}= ?
View answer and explanation
Correct option: D
{1,−2}\{1,-2\}
Option D ({1,−2}\{1,-2\}) is correct because these are the real solutions obtained by solving the absolute value equation.
401 Medium
401 of 914
Application +4 | -1
If a,ba, b and cc are three positive real numbers such that a+b≥ca+b \geq c, then
View answer and explanation
Correct option: A
a1+a+b1+b≥c1+c\frac{a}{1+a}+\frac{b}{1+b} \geq \frac{c}{1+c}
The correct answer is A (a1+a+b1+b≥c1+c\frac{a}{1+a}+\frac{b}{1+b} \geq \frac{c}{1+c}) because it uses the given condition a+b≥ca+b \geq c to prove the inequality.
402 Easy
402 of 914
Application +4 | -1
The sum of the real roots of the equation ∣x−2∣2+∣x−2∣−2=0,{\left| {x - 2} \right|^2} + \left| {x - 2} \right| - 2 = 0, is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the sum of the real roots obtained from solving the absolute value equation is 4.
403 Medium
403 of 914
Application +4 | -1
The value of xx for which 12x−6<0,12−3x<012 x-6 < 0,12-3 x < 0, is
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the inequalities 12x−6<012x - 6 < 0 and 12−3x<012 - 3x < 0 lead to contradictory conditions on xx, resulting in no solution.
404 Hard
404 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots of the equation ax2+bx+c=0\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0 and Sn=αn+βn\mathrm{S}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}, then a Sn+1+cSn−1=\mathrm{S}_{\mathrm{n}+1}+\mathrm{c} \mathrm{S}_{\mathrm{n}-1}=
View answer and explanation
Correct option: D
−bSn-\mathrm{bS}_{\mathrm{n}}
Option D (-bS_n) is correct because by substituting the roots into the original equation and manipulating the expression, it simplifies to -bS_n.
405 Medium
405 of 914
Application +4 | -1
If the roots of the quadratic equation x2−4x−log⁡3a=0{x^2} - 4x - {\log _3}a = 0 are real, then the least value of aa is
View answer and explanation
Correct option: B
181\frac{1}{{81}}
Option B (181\frac{1}{{81}}) is correct because the discriminant of the quadratic equation must be non-negative for real roots, leading to the inequality a >= 1/81.

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406 Medium
406 of 914
Application +4 | -1
If tan⁡α\tan \alpha and tan⁡β\tan \beta are roots of the equation x2+px+q=0{x^2} + px + q = 0 with p≠0,p \ne 0, then
View answer and explanation
Correct option: A
sin⁡2(α+β)+psin⁡(α+β)cos⁡(α+β)+qcos⁡2(α+β)=q{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q
Option A (sin⁡2(α+β)+psin⁡(α+β)cos⁡(α+β)+qcos⁡2(α+β)=q{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q) is correct because it is derived by substituting the sum and product of roots into the trigonometric identity.
407 Easy
407 of 914
Analyzing +4 | -1
Number of integer solutions of x+2x2+1>12\frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2} is
View answer and explanation
Correct option: D
3
Option D (3) is correct because solving the inequality yields -1 < x < 3, resulting in three integer solutions: 0, 1, and 2.
408 Easy
408 of 914
Application +4 | -1
The rational values of aa and bb in ax2+bx+1=0a x^2+b x+1=0 if 14+3\frac{1}{4+\sqrt{3}} is a root, are
View answer and explanation
Correct option: A
a=13,b=−8a=13, b=-8
Option A (a=13,b=−8a=13, b=-8) is correct because it is obtained by rationalizing the given root and comparing coefficients of the quadratic equation.
409 Medium
409 of 914
Analyzing +4 | -1
If the ratio of the equation x2+px+q=0{x^2} + px + q = 0 be equal to the ratio of the roots of x2+lx+m=0,{x^2} + lx + m = 0, then
View answer and explanation
Correct option: D
p2m=l2q{p^2}m = {l^2}q
Option C (p2m=l2q{p^2}m = {l^2}q) is correct because it is derived from equating the ratios of the roots and manipulating the equations relating roots and coefficients.
410 Medium
410 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 375x2−25x−2=0375{x^2} - 25x - 2 = 0, then lim⁡n→∞ ∑r=1nαr+lim⁡n→∞  ∑r=1nβr\lim\limits_{n \to \infty } \,\sum\limits_{r = 1}^n {{\alpha ^r} + } \lim\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^n {{\beta ^r}} is equal to
View answer and explanation
Correct option: C
112\frac{{1}}{{12}}
Option C (112\frac{{1}}{{12}}) is correct because it is the result of summing the infinite geometric series formed by the roots of the given quadratic equation.
411 Medium
411 of 914
Analyzing +4 | -1
Consider the two sets
A={m ∈ RA = \{ m\, \in \,R: both the roots of x2−(m+1)x+m+4=0{x^2} - (m + 1)x + m + 4 = 0 are real }\} and B=[−3, 5)B = \left[ { - 3,\,5} \right).
Which of the following is not true?
View answer and explanation
Correct option: A
A−B=(−∞,−3)∪(5,∞)A - B = ( - \infty , - 3) \cup (5,\infty )
Option A A−B=(−∞,−3)∪(5,∞)A - B = ( - \infty , - 3) \cup (5,\infty ) is correct because it incorrectly represents the set difference between A and B. The correct set difference is A−B=(−∞,−3)∪[5,∞)A - B = ( - \infty , - 3) \cup [5,\infty ).
412 Medium
412 of 914
Analyzing +4 | -1
If a,b,c∈Ia, b, c \in I and ax2+bx+c=0a x^2+b x+c=0 has irrational root and λ∈Q=pq\lambda \in Q=\frac{p}{q}, where f(x)=ax2+bx+cf(x)=a x^2+b x+c, then ∣f(λ)∣|f(\lambda)|
View answer and explanation
Correct option: A
≥1q2\geq \frac{1}{q^2}
Option A (≥1q2\geq \frac{1}{q^2}) is correct because the absolute value of the quadratic function evaluated at a rational number is always greater than or equal to 1q2\frac{1}{q^2}.
413 Easy
413 of 914
Application +4 | -1
The least value of n∈Nn \in \mathbf{N} for which (n−4)x2+8x(n-4) x^2+8 x +n+2>0∀x∈R+n+2 > 0 \forall x \in \mathbf{R}, is
View answer and explanation
Correct option: D
7
Option D (7) is correct because it is the smallest natural number that satisfies the conditions for the quadratic expression to be always positive.
414 Easy
414 of 914
Application +4 | -1
If α≠β\alpha \ne \beta and α2=5α−3,  β2=5β−3{\alpha ^2} = 5\alpha - 3,\;{\beta ^2} = 5\beta - 3, then the equation having αβ\frac{\alpha }{\beta } and βα\frac{\beta }{\alpha } as its roots is
View answer and explanation
Correct option: B
3x2−19x+3=03{x^2} - 19x + 3 = 0
Option B (3x2−19x+3=03{x^2} - 19x + 3 = 0) is correct because it represents the quadratic equation with roots αβ\frac{\alpha}{\beta} and βα\frac{\beta}{\alpha} derived from the given conditions.
415 Easy
415 of 914
Application +4 | -1
The roots of the quadratic equation x2−23x−22=0{x^2} - 2\sqrt 3 x - 22 = 0 are
View answer and explanation
Correct option: D
Real, rational and unequal
The roots are Real, rational and unequal because the discriminant is a perfect square (100).
416 Medium
416 of 914
Application +4 | -1
If α\alpha is one root of the equation 4x2+2x−1=04 x^2+2 x-1=0, then its other root is given by
View answer and explanation
Correct option: A
4α3−3α4 \alpha^3-3 \alpha
Option A (4α3−3α4 \alpha^3-3 \alpha) is the correct answer because it represents the other root of the quadratic equation, derived using the relationship between the roots and the given equation.
417 Medium
417 of 914
Analyzing +4 | -1
For what values of K∈RK \in R the expression 2x2+Kxy+3y2−5y−22 x^2+K x y+3 y^2-5 y-2 can be expressed as (a1x+b1y+c1)⋅(a2x+b2y+c2)\left(a_1 x+b_1 y+c_1\right) \cdot\left(a_2 x+b_2 y+c_2\right) ?
View answer and explanation
Correct option: D
7,−77,-7
Option D (7,−77, -7) is correct because these values of K allow the given quadratic expression to be factored into two linear expressions.
418 Medium
418 of 914
Analyzing +4 | -1
If the equations k(6x2+3)+rx+2x2−1=0k\left( {6{x^2} + 3} \right) + rx + 2{x^2} - 1 = 0 and 6k(2x2+1)+px+4x2−2=06k\left( {2{x^2} + 1} \right) + px + 4{x^2} - 2 = 0 have both roots common, then the value of (2r−p)\left( {2r - p} \right) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because by comparing the coefficients of the two equations after making them proportional, we find that 2r - p = 0.
419 Easy
419 of 914
Application +4 | -1
If α\alpha and β\beta the roots of x2−x−1=0   and  An=αn+βn,   then  AM  of  An−1  and  An{x^2} - x - 1 = 0\;\,\mathrm{and}\;{A_n} = {\alpha ^n} + {\beta ^n},\;\,\mathrm{then\;AM}\;\mathrm{of}\;{A_{n - 1}}\,\,\mathrm{and}\;{A_n} is
View answer and explanation
Correct option: B
An+1  2\frac{{{A_{n + 1\;}}}}{2}
Option B (An+1  2\frac{{{{A_{n + 1\;}}}}}{{{{2}}}}) is correct because it is derived by using the properties of roots of the given quadratic equation and simplifying the expression for the arithmetic mean.
420 Medium
420 of 914
Application +4 | -1
Let f(x)=ax2+bx+cf(x)=a x^2+b x+c, where a,b,c∈Ra, b, c \in R and a≠0a \neq 0, is such that 2a+b+c=02 \mathrm{a}+\mathrm{b}+\mathrm{c}=0, then the quadratic equation f(x)=0f(x)=0 has :
View answer and explanation
Correct option: D
two distinct real roots
Option D (two distinct real roots) is correct because the given condition implies that either 0 and 2 are roots or one root lies in (0,2), indicating two distinct real roots.
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