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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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Showing 781-810 of 914 questions

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781 Medium

781 of 914

Application +4 | -1

a \leq 0

then roots of

x^2-2 a|x-a|-3 a^2=0

View answer and explanation

Correct option: A

(-1+\sqrt{6}) \mathrm{a}

Option A ((-1+

\sqrt{6}

)a) is correct because it is one of the roots obtained by considering the case where x <= a and solving the quadratic equation.

782 Medium

782 of 914

Analyzing +4 | -1

Let

f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}

be real valued function defined as

f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}

. Then range of

f

View answer and explanation

Correct option: A

\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)

The correct answer is A (

\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)

) because it is the range obtained by analyzing the quadratic equation and its discriminant.

783 Easy

783 of 914

Analyzing +4 | -1

The equation

e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0

x \in \mathrm{R}

has:

View answer and explanation

Correct option: B

Two solutions and both are negative

Option B (Two solutions and both are negative) is correct because by substituting

e^{x}-e^{-x}=t

and solving the resulting quadratic equation, we find two negative solutions for x.

784 Easy

784 of 914

Application +4 | -1

x = \frac{1}{2}\left( {\sqrt 7 + \frac{1}{{\sqrt 7 }}} \right)

Then,

\frac{{\sqrt {{x^2} - 1} }}{{x - \sqrt {{x^2} - 1} }}

is equal to

View answer and explanation

Correct option: C

Option C (3) is correct because after substituting the value of x and simplifying, the expression evaluates to 3.

785 Medium

785 of 914

Analyzing +4 | -1

a, b, c > 0

and

a(1-b) > \frac{1}{4}, b(1-c) > \frac{1}{4}

and

c(1-a) > \frac{1}{4}

, then

View answer and explanation

Correct option: A

never possible

Option A (never possible) is correct because the given inequalities lead to a contradiction, making the situation impossible.

786 Easy

786 of 914

Application +4 | -1

\alpha ,\beta

are the roots of

{x^2} + bx - c = 0,

then the equation whose roots are

b

and

c

View answer and explanation

Correct option: D

{x^2} + x\left( {\alpha + \beta + \alpha \;\beta } \right) + \alpha \;\beta \left( {\alpha + \beta } \right) = 0

Option D (

{x^2} + x\left( {\alpha + \beta + \alpha \\;\beta } \right) + \alpha \\;\beta \left( {\alpha + \beta } \right) = 0

) is correct because it represents the quadratic equation with roots b and c, derived from the relationships between roots and coefficients of the original equation.

787 Easy

787 of 914

Application +4 | -1

If the roots of the equation

3{x^2} - 6x + 5 = 0

are

\alpha

and

\beta

, then the equation whose roots are

\alpha + \beta

and

\frac{2}{{\alpha + \beta }}

will be

View answer and explanation

Correct option: D

{x^2} - 3x + 2 = 0

Option D (

{x^2} - 3x + 2 = 0

) is correct because it is the quadratic equation formed with roots

\alpha + \beta

and

\frac{2}{{\alpha + \beta}}

788 Hard

788 of 914

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.

789 Easy

789 of 914

Application +4 | -1

2 + i\sqrt 3

is a root of the equation

{x^2} + px + q = 0,

where

p

and

q

are real, then

\left( {p,q} \right)

is equal to

View answer and explanation

Correct option: A

\left( { - 4,\;7} \right)

Option A (

\left( { - 4,\;7} \right)

) is correct because if

2 + i\sqrt 3

is a root, then its conjugate is also a root, and we can find p and q using the sum and product of roots.

790 Easy

790 of 914

Application +4 | -1

The number of solutions of

\sqrt{4-x}+\sqrt{x+9}=5

(1)

is:

View answer and explanation

Correct option: C

Option C (2) is correct because after solving the equation, we find two solutions, x=0 and x=-5, both of which satisfy the original equation.

791 Easy

791 of 914

Application +4 | -1

y=2[x]+3

and

y=3[x-2]

, then

[x+y]=?

View answer and explanation

Correct option: C

Option C (30) is correct because solving the given equations for [x] and y, we find [x] = 9 and y = 21, so [x+y] = [9+21] = 30.

792 Easy

792 of 914

Analyzing +4 | -1

The number of solutions of

{3^{\left| x \right|}} = \left| {2 - \left| x \right|} \right|

View answer and explanation

Correct option: B

Option B (2) is correct because the graphs of the exponential and absolute value functions intersect at two points.

793 Easy

793 of 914

Application +4 | -1

If the roots of the quadratic equation

x^2+p x+q=0

are

\tan 30^{\circ}

and

\tan 15^{\circ}

, respectively, the value of

2+q-p

View answer and explanation

Correct option: B

Option B (3) is correct because using the tangent addition formula, we can relate the sum and product of the roots to find the value of

2+q-p

794 Medium

794 of 914

Application +4 | -1

Consider the equation

x^2+2 x-n=0

, where

n \in N

and

n \in[5,100]

Total number of different values of

n

so that the given equation has integral roots, is

View answer and explanation

Correct option: B

The correct answer is B (8) because we need to find the number of values of

n

such that

n+1

is a perfect square and

n \in [5, 100]

795 Easy

795 of 914

Analyzing +4 | -1

\frac{{{x^2} - bx}}{{ax - c}} = \frac{{\lambda - 1}}{{\lambda + 1}}

has roots equal in magnitude and opposite in sign then the value of

\lambda

View answer and explanation

Correct option: A

\frac{{a - b}}{{a + b}}

Option A is correct because the sum of the roots being zero leads to the given expression for lambda.

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796 Easy

796 of 914

Application +4 | -1

Product of roots of the equation

\sqrt{13-x^2}=x+5

View answer and explanation

Correct option: C

Option C (6) is correct because the product of the roots of the equation is 6.

797 Easy

797 of 914

Analyzing +4 | -1

The number of real solutions of the equation

{\left( {\frac{9}{{10}}} \right)^x} = - 3 + x - {x^2}

View answer and explanation

Correct option: A

None

Option A (None) is correct because the left-hand side (LHS) is always positive, while the right-hand side (RHS) is always negative, so there are no real solutions.

798 Easy

798 of 914

Application +4 | -1

Let

\alpha

and

\beta

be two real roots of the equation

(k + 1){\tan ^2}x - \sqrt 2 \cdot \mathrm{ }\lambda \tan x = (1 - k),

where

k( \ne - 1)

and

\mathrm{ }\lambda

are real numbers. If

{\tan ^2}(\alpha \mathrm{ } + \mathrm{ }\beta ) = 50

, then a value of

\lambda

View answer and explanation

Correct option: B

Option B (10) is correct because it is one of the values of

\lambda

obtained by solving the trigonometric equation using the given condition

{\tan ^2}(\alpha + \mathrm{ }\beta ) = 50

799 Medium

799 of 914

Application +4 | -1

The equation

{x^2} - 3\left| x \right| + 2 = 0

has

View answer and explanation

Correct option: D

Four real roots

Option D (Four real roots) is correct because by substituting |x| with a variable and solving the resulting quadratic equation, we find four real solutions: x = ±1 and x = ±2.

800 Easy

800 of 914

Application +4 | -1

The values of

a

which make the expression

x^2-a x+

1-2 a^2

always positive for real values of

x

, are

View answer and explanation

Correct option: A

-\frac{2}{3} < a < \frac{2}{3}

Option A (-2/3 < a < 2/3) is correct because it ensures the discriminant of the quadratic is negative, making the expression always positive.

801 Easy

801 of 914

Analyzing +4 | -1

a,b,c

are positive real numbers such that

a + b + c = 2

then, which one of the following is true?

View answer and explanation

Correct option: A

\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc

Option A (

\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc

) is correct because it is derived from applying the AM-GM inequality to the given conditions.

802 Medium

802 of 914

Analyzing +4 | -1

Let

f(x)=a x^2+b x+c, a \neq 0

Suppose,

f(-1) < 1

f(1) > -1

and

f(3) < -4

. Then,

View answer and explanation

Correct option: B

b+1 > 0

Option C (

b+1 < 0

) is incorrect. Option B (

b+1 > 0

) is correct because from the given inequalities, we can derive that

b+1 > 0

803 Easy

803 of 914

Analyzing +4 | -1

\left( {1 + k} \right){\tan ^2}x - 4\tan x - 1 + k = 0

has real roots

\tan {x_1}

and

\tan {x_2},

then

View answer and explanation

Correct option: A

{k^2} \le 5

Option A (

{k^2} \le 5

) is correct because it represents the condition for the quadratic equation to have real roots, based on the discriminant.

804 Easy

804 of 914

Application +4 | -1

Suppose

p \in \mathbf{R}

and

\alpha, \beta

are roots of

x^2-p x+\frac{1}{2 p^2}=0

, then the minimum value of

\alpha^4+\beta^4

View answer and explanation

Correct option: C

\sqrt{2}-2

Option C (

\sqrt{2}-2

) is correct because by using AM-GM inequality, we can find the minimum value of the expression, which occurs when

p^4 = \frac{1}{2p^4}

805 Medium

805 of 914

Application +4 | -1

Let

a

and

b

be two integers such that

10 a+b=5

and

p(x)=x^2+a x+b

. Let integer

n

be such that

p(10) p(11)=p(n)

then value of

\frac{n}{23}

is.

View answer and explanation

Correct option: A

Option A (5) is correct because substituting the given conditions and factoring the polynomial leads to n/23 = 5.

806 Medium

806 of 914

Analyzing +4 | -1

f\left( x \right) = {x^2} + 2bx + 2{c^2}

and

g\left( x \right) = - {x^2} - 2cx + {b^2}

such that

\min f\left( x \right) > \max g\left( x \right)

, then the relation between

b

and

c

, is

View answer and explanation

Correct option: D

\left| c \right| > \left| b \right|\sqrt 2

Option D

\left| c \right| > \left| b \right|\sqrt 2

is correct because it represents the relationship between

b

and

c

derived from the condition

\min f(x) > \max g(x)

807 Easy

807 of 914

Analyzing +4 | -1

The minimum value of the sum of the squares of the roots of

x^2+(3-a) x+1=2 a

View answer and explanation

Correct option: C

Option C (6) is correct because the minimum value of the sum of squares of the roots is found by completing the square and finding the vertex of the resulting quadratic.

808 Easy

808 of 914

Analyzing +4 | -1

x \in R,

the least value of the expression

\frac{{{x^2} - 6x + 5}}{{{x^2} + 2x + 1}},

View answer and explanation

Correct option: C

\frac{{ - 1}}{3}

Option C (-1/3) is correct because the minimum value of the expression is found by setting the discriminant greater than or equal to zero and solving for y.

809 Medium

809 of 914

Analyzing +4 | -1

Roots of the equation

a x^3+b x^2+c x+d=0

remain unchanged by increasing each coefficient by one unit, then

View answer and explanation

Correct option: A

a=b=c=d \neq 0

Option D (

a=b=c=d \neq 0

) is correct because for the roots to remain unchanged, all coefficients must be equal.

810 Easy

810 of 914

Analyzing +4 | -1

The roots of the quadratic equation

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

are

View answer and explanation

Correct option: D

None of these

Option B (At least two real roots) is correct because the product of the discriminants of the two quadratic factors is always non-negative, implying at least one factor has real roots.

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