If $\alpha ,\;\beta $ are the roots of ${x^2} - ax + b = 0\;\,{\rm{and\;if}}\,\;{\alpha ^n} + {\beta ^n} = {V_n},$ then

${V_{n + 1}} = a{V_n} + a{V_{n - 1}}$

${V_{n + 1}} = a{V_{n - 1}} - b{V_n}$

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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Hard

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Practice format and chapter quiz

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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1 Hard

1 of 24

Analyzing +4 | -1

Root(s) of the equation

9{x^2} - 18\left| x \right| + 5 = 0

belonging to the domain of definition of the function

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)

is (are)

View answer and explanation

Correct option: C

\frac{{ - 5}}{3}

Option C (

\frac{{ - 5}}{3}

) is correct because it is the only root of the given equation that also satisfies the domain of the logarithmic function.

2 Hard

2 of 24

Analyzing +4 | -1

f\left( x \right) = \sum \limits_{k = 2}^n \left( {x - \frac{1}{{k - 1}}} \right)\left( {x - \frac{1}{k}} \right),

then the product of roots of

f\left( x \right) = 0

n \to \infty ,

View answer and explanation

Correct option: B

Option B (0) is correct because as n approaches infinity, the product of the roots of the given function approaches 0.

3 Hard

3 of 24

Analyzing +4 | -1

Let

n

be an odd positive integer. Then, the number of real roots of the polynomial

{P_n}\left( x \right) = 1 + 2x + 3{x^2} + \; \ldots \ldots + \left( {n + 1} \right){x^n},

View answer and explanation

Correct option: C

Option C (1) is correct because the polynomial has exactly one real root, which lies between -1 and 0, due to the change of sign and the odd degree of the polynomial.

4 Hard

4 of 24

Analyzing +4 | -1

The solution of the quadratic equation

{\left( {3\left| x \right| - 3} \right)^2} = \left| x \right| + 7

which belongs to the domain of definition of the function

y = \sqrt {x\left( {x - 3} \right)}

are given by

View answer and explanation

Correct option: D

- \frac{1}{9},\; - 2

Option D (

- \frac{1}{9},\; - 2

) is correct because these are the only solutions to the quadratic equation that also satisfy the domain of the function

y = \sqrt {x(x - 3)}

5 Hard

5 of 24

Analyzing +4 | -1

For real values of

x

, the expression

\frac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {x - a} \right)}}

will assume all real values provided

View answer and explanation

Correct option: B

b \ge a \ge c

Option B (

b \ge a \ge c

) is correct because it ensures the discriminant of the quadratic equation is non-negative, guaranteeing real values for x.

6 Hard

6 of 24

Analyzing +4 | -1

The number of negative integral solutions of

{x^2}{.2^{x + 1}} + {2^{\left| {x - 3} \right| + 2}} = {x^2}{.2^{\left| {x - 3} \right| + 4}} + {2^{x - 1}},

View answer and explanation

Correct option: A

None

Option A (None) is correct because after analyzing the equation for both

x \ge 3

and

x < 3

, there are no negative integral solutions.

7 Hard

7 of 24

Application +4 | -1

The minimum value of

P = bcx + cay + abz,\;

when

\;xyz = abc,\;

View answer and explanation

Correct option: A

3abc

Option A (

3abc

) is correct because the minimum value of P is found using AM ≥ GM inequality, and given that

xyz = abc

8 Hard

8 of 24

Application +4 | -1

Let

a, b, c

be the sides of a triangle. No two of them are equal and

\lambda \in R

. If the roots of the equation

x^2+2(a+b+c) x+3 \lambda(a b+b c+c a)=0

are real, then

View answer and explanation

Correct option: A

\lambda<\frac{4}{3}

Option A (

\lambda<\frac{4}{3}

) is correct because the discriminant of the quadratic equation must be non-negative, leading to the inequality

\lambda < \frac{4}{3}

9 Hard

9 of 24

Understanding +4 | -1

If roots of

a{x^2} + bx + c = 0,a,b,c \in R,a \ne 0

are imaginary then

View answer and explanation

Correct option: A

ac > 0

Option C (A straight line) is correct because the equation |z - 3| = |z - 5| represents the perpendicular bisector of the line segment joining the points 3 and 5 on the complex plane, which is a straight line.

10 Hard

10 of 24

Application +4 | -1

\alpha, \beta

are the roots of the equation

\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0

and

\mathrm{S}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}

, then a

\mathrm{S}_{\mathrm{n}+1}+\mathrm{c} \mathrm{S}_{\mathrm{n}-1}=

View answer and explanation

Correct option: D

-\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.

11 Hard

11 of 24

Analyzing +4 | -1

p,q,r,s,t

are numbers such that

p + q < r + s,\;q + r < s + t,\;r + s < t + p,\;s + t < p + q,

then the largest and smallest numbers are

View answer and explanation

Correct option: A

p

and

q

respectively

Option A (

p

and

q

respectively) is correct because the inequalities lead to the conclusion that p is the largest and q is the smallest.

12 Hard

12 of 24

Analyzing +4 | -1

\alpha ,\beta

are the roots of the equation

{x^2} - \left( {5 + {3^{\sqrt {\log _3^5} }} - {5^{\sqrt {\log _5^3} }}} \right)x + 3\left( {{3^{{{\left( {{{\log }_3}5} \right)}^{\frac{1}{3}}}}} - {5^{{{\left( {\log _5^3} \right)}^{\frac{2}{3}}}}} - 1} \right) = 0

then the equation, whose roots are

\alpha+\frac{1}{\beta}

and

\beta+\frac{1}{\alpha}

View answer and explanation

Correct option: B

3 x^2-10 x-4=0

Option B (

3 x^2-10 x-4=0

) is correct because it is derived from substituting the roots into the given equation and simplifying to find the new equation.

13 Hard

13 of 24

Application +4 | -1

\alpha ,\;\beta

are the roots of

{x^2} - ax + b = 0\;\,\mathrm{and\;if}\,\;{\alpha ^n} + {\beta ^n} = {V_n},

then

View answer and explanation

Correct option: C

{V_{n + 1}} = a{V_n} - b{V_{n - 1}}

Option C (

{V_{n + 1}} = a{V_n} - b{V_{n - 1}}

) is correct because it is derived by using the given recurrence relation and properties of roots of the quadratic equation.

14 Hard

14 of 24

Analyzing +4 | -1

Let

\left[ x \right]

denote the greatest integer less than or equal to

x

. Then, in

\left[ {0,3} \right]

the number of solutions of the equation

{x^2} - 3x + \left[ x \right] = 0,

View answer and explanation

Correct option: C

Option C (2) is correct because by analyzing the equation in different intervals of [x], we find two solutions.

15 Hard

15 of 24

Analyzing +4 | -1

The number of solutions of

\frac{{\log 5 + \mathrm{log}\left( {{x^2} + 1} \right)}}{{\log \left( {x - 2} \right)}} = 2

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the equation simplifies to a value of x that makes the logarithm undefined.

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16 Hard

16 of 24

Application +4 | -1

e_1

and

e_2

are the roots of the equation

x^2-a x+2=0

where

e_1

and

e_2

eccentricities of an ellipse and hyperbola respectively then the complete set of values of

a

View answer and explanation

Correct option: A

(3, \infty)

Option A (

(3, \infty)

) is correct because the eccentricity of an ellipse is less than 1 and the eccentricity of a hyperbola is greater than 1, leading to the condition a > 3.

17 Hard

17 of 24

Analyzing +4 | -1

a,b,c

are different positive real number such that

\left( {b + c - a} \right),\left( {c + a - b} \right)

and

\;\left( {a + b - c} \right)

are positive, then

\left( {a + b - c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right) - abc

View answer and explanation

Correct option: B

Negative

Option B (Negative) is correct because, by AM > GM, the product (a+b-c)(b+c-a)(c+a-b) is less than abc, making the expression negative.

18 Hard

18 of 24

Analyzing +4 | -1

Let

a,b,c

be real number

a \ne 0.\,\mathrm{If}\;\alpha

is a root of

{a^2}{x^2} + bx + c = 0,\;\beta

is a root of

{a^2}{x^2} - bx - c = 0

and

0 < \alpha < \beta

, then the equation

{a^2}{x^2} + 2bx + 2c = 0

has a root of

\gamma

that always satisfies

View answer and explanation

Correct option: D

\alpha < \gamma < \beta

Option D (

\alpha < \gamma < \beta

) is correct because by Intermediate Value Theorem, since f(α) and f(β) have opposite signs, there must be a root between α and β.

19 Hard

19 of 24

Application +4 | -1

{a_i} > 0\;

for

\;i = 1,2, \ldots ,n\;

and

\;{a_1}{a_2} \ldots {a_n} = 1,\;

then minimum value of

\left( {1 + {a_1}} \right)\left( {1 + {a_2}} \right) \ldots \left( {1 + {a_n}} \right)\;

View answer and explanation

Correct option: B

{2^n}

Option B (

{2^n}

) is correct because the minimum value is obtained by repeatedly applying AM ≥ GM inequality to each factor and using the given condition

{a_1}{a_2} \ldots {a_n} = 1

20 Hard

20 of 24

Analyzing +4 | -1

p\left( x \right) = a{x^2} + bx + c\,\;\mathrm{and}\;Q\left( x \right) = - a{x^2} + dx + c

, where

ac \ne 0

, then

P\left( x \right)Q\left( x \right) = 0

has at least

View answer and explanation

Correct option: B

Two real roots

Option B (Two real roots) is correct because if both quadratics had only imaginary roots, it would lead to a contradiction, thus at least one must have real roots.

21 Hard

21 of 24

Analyzing +4 | -1

a, b

and

c \in R

, then the square root of

a^2+b^2+c^2-b c-c a-a b

is greater than or equal to

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{2} \max \{|b-c|,|c-a|,|a-b|\}

Option A (

\frac{\sqrt{3}}{2} \max \\{|b-c|,|c-a|,|a-b|\\}

) is correct because it is derived by manipulating the given expression and comparing it with the maximum difference between the variables.

22 Hard

22 of 24

Application +4 | -1

\alpha ,\;\beta

are the roots of the equation

a{x^2} + bx + c = 0\;\,\mathrm{such\;that}\;\beta < \alpha < 0,

then the quadratic equation whose roots are

\left| \alpha \right|,\;\left| \beta \right|,\;

is given by

View answer and explanation

Correct option: B

a{x^2} - \left| b \right|x + c = 0

Option B (

a{x^2} - \left| b \right|x + c = 0

) is correct because it represents the quadratic equation with roots

\left| \alpha \right|

and

\left| \beta \right|

given that both

\alpha

and

\beta

are negative.

23 Hard

23 of 24

Analyzing +4 | -1

b > a,

then the equation

\left( {x - a} \right)\left( {x - b} \right) - 1 = 0

has

View answer and explanation

Correct option: C

Roots in

\left( { - \infty ,a} \right)

and other in

\left( {b,\infty } \right)

Option C (Roots in

\left( { - \infty ,a} \right)

and other in

\left( {b,\infty } \right)

) is correct because the function changes sign between a and b, indicating a root exists in each of the specified intervals.

24 Hard

24 of 24

Analyzing +4 | -1

The number of values of a for which the system of equation

{2^{\left| x \right|}} + \left| x \right| = y + {x^2} + a

and

{x^2} + {y^2} = 1

has only one solution where

a,x,y

are real, is

View answer and explanation

Correct option: A

Option A (1) is correct because by analyzing the symmetry of the equations, we find that there is only one value of 'a' for which the system has a unique solution.

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Key Takeaways

✓ This set contains 24 single-correct-answer MCQs on QUADRATIC EQUATIONS.
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