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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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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 301-330 of 362 questions

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301 Medium

301 of 362

Analyzing +4 | -1

If m is chosen in the quadratic equation

({m^2} + 1){x^2} - 3x + {({m^2} + 1)^2} = 0

such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is

View answer and explanation

Correct option: B

8\sqrt 5

Option B (

8\sqrt{5}

) is correct because it represents the absolute difference of the cubes of the roots when m=0, which maximizes the sum of the roots.

302 Medium

302 of 362

Application +4 | -1

\alpha ,\beta

are roots of the equation

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

then the equation whose roots are

2\;\alpha + 3\;\beta

and

3\;\alpha + 2\;\beta

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derived equation doesn't match any of the provided options.

303 Medium

303 of 362

Application +4 | -1

The number of integral values of

k

, for which one root of the equation

2 x^{2}-8 x+k=0

lies in the interval

(1,2)

and its other root lies in the interval

(2,3)

View answer and explanation

Correct option: A

Option A (1) is correct because only one integer value of k satisfies the given conditions based on the roots lying in specified intervals.

304 Medium

304 of 362

Analyzing +4 | -1

\alpha

and

\beta

are the roots of the equation

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

, where

p \in R

, then the minimum value of

\alpha^4+\beta^4

View answer and explanation

Correct option: B

2+\sqrt{2}

Option B (

2+\sqrt{2}

) is correct because by completing the square, the minimum value of the expression

\alpha^4 + \beta^4

is found to be

2+\sqrt{2}

305 Medium

305 of 362

Application +4 | -1

If the sum of the roots of the equation

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

is equal to their products, then the value of

a

View answer and explanation

Correct option: A

- \frac{2}{3}

Option A (

\frac{-2}{3}

) is correct because by setting the sum of the roots equal to the product of the roots, we can solve for the value of 'a'.

306 Medium

306 of 362

Application +4 | -1

Two non-integer roots of

\left(\frac{3 x-1}{2 x+3}\right)^4-5\left(\frac{3 x-1}{2 x+3}\right)^2+4=0

(1)

are

View answer and explanation

Correct option: A

-5 / 7,-2 / 5

Option A (-5/7, -2/5) is correct because these are the two non-integer roots obtained by solving the equation after reducing it to a quadratic form.

307 Medium

307 of 362

Analyzing +4 | -1

x-1=\sqrt{a-x^2}

has real solution for

x

then

View answer and explanation

Correct option: B

a \geq 1

Option B (

a \geq 1

) is correct because it ensures that the equation has a real solution for x, considering the non-negativity of the square root and the condition x >= 1.

308 Medium

308 of 362

Analyzing +4 | -1

If the sum of the roots of the equation

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

is equal to the sum of the squares of their reciprocals of their reciprocals, then

View answer and explanation

Correct option: A

{c^2}b,{a^2}c,{b^2}a

are in A.P.

Option A (

{c^2}b,{a^2}c,{b^2}a

are in A.P.) is correct because the given condition leads to the relationship

2a^2c = ab^2 + bc^2

, which is the condition for an arithmetic progression.

309 Medium

309 of 362

Analyzing +4 | -1

x^2+(a-b) x+(1-a-b)=0

, where

a, b \in R

, then the values of

a

for which equation has unequal roots for all values of

b

, is given by

View answer and explanation

Correct option: B

a > 1

Option B (

a > 1

) is correct because the discriminant of the quadratic in

b

must be negative to ensure unequal roots for all

b

310 Medium

310 of 362

Analyzing +4 | -1

{b^2} \ge 4ac

for the equation

a{x^4} + b{x^2} + c = 0

, then all the roots of the equation will be real positive of

View answer and explanation

Correct option: B

b < 0,a > 0,c > 0

Option B (

b < 0, a > 0, c > 0

) is correct because it ensures the quadratic in

x^2

has positive roots, which then lead to real roots for x.

311 Medium

311 of 362

Analyzing +4 | -1

The number of solution(s) of the inequation

\sqrt {3{x^2} + 6x + 7} + \sqrt {5{x^2} + 10x + 14} \le 4 - 2x - {x^2},

View answer and explanation

Correct option: A

Option A (1) is correct because the inequation holds only when both sides are equal to 5, which occurs at x = -1.

312 Medium

312 of 362

Application +4 | -1

If set

S=\left\{(\sqrt{3}+\sqrt{2})^{x^2-4}+(\sqrt{3}-\sqrt{2})^{x^2-4}=10\right\}

then

n(S)

equals:

View answer and explanation

Correct option: C

Option C (4) is correct because solving the equation leads to four distinct solutions for x.

313 Medium

313 of 362

Analyzing +4 | -1

f\left( x \right) = \left( {ab - {b^2} - 2} \right)x + \int \nolimits_0^x \left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)d\theta

is decreasing function of

x

for all

x \in R

and

b \in R,\;b

being independent of

x

, then

View answer and explanation

Correct option: B

a \in \left( { - \sqrt 6 ,\;\sqrt 6 } \right)

Option B (

\frac{1}{4}

) is correct because it represents the point where the function attains its maximum value within the given interval.

314 Medium

314 of 362

Application +4 | -1

\mathrm{x}

is a solution of the equation,

\sqrt{2 x+1}-\sqrt{2 x-1}=1,\left(x \geq \frac{1}{2}\right)

, then

\sqrt{4 x^{2}-1}

is equal to :

View answer and explanation

Correct option: A

\frac{3}{4}

Option A (

\frac{3}{4}

) is correct because solving the given equation for x and substituting it into

\sqrt{4 x^{2}-1}

yields

\frac{3}{4}

315 Medium

315 of 362

Application +4 | -1

a,\;b,\;c > 0

, the minimum value of

\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}

View answer and explanation

Correct option: B

\frac{3}{2}

Option B

\frac{3}{2}

is correct because the minimum value is obtained using AM-GM inequality.

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316 Medium

316 of 362

Analyzing +4 | -1

\alpha

and

\beta(\alpha<\beta)

, are the roots of the equation

x^2+b x+c=0

, where

c < 0 < b

, then

View answer and explanation

Correct option: B

\alpha<0<\beta<|\alpha|

Option C (

n(x_1)^{n-1}+a

) is correct. By differentiating the polynomial and evaluating at

x_1

, we find the value of the expression.

317 Medium

317 of 362

Analyzing +4 | -1

x,a,b,c

are real and

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

then

a,b,c

are in

View answer and explanation

Correct option: C

A.P.

Option C (A.P.) is correct because the given condition implies that a, b, and c form an arithmetic progression.

318 Medium

318 of 362

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.

319 Medium

319 of 362

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.

320 Medium

320 of 362

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.

321 Medium

321 of 362

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]

322 Medium

322 of 362

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.

323 Medium

323 of 362

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

324 Medium

324 of 362

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.

325 Medium

325 of 362

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)

326 Medium

326 of 362

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.

327 Medium

327 of 362

Analyzing +4 | -1

\alpha ,\;\beta

be the roots of

{x^2} - a\left( {x - 1} \right) + b = 0

, then the value of

\frac{1}{{{\alpha ^2} - a\alpha }} + \frac{1}{{{\beta ^2} + a\beta }} + \frac{2}{{a + b}}

View answer and explanation

Correct option: C

Option C (0) is correct because by using the properties of roots and simplifying the expression, the value becomes zero.

328 Medium

328 of 362

Analyzing +4 | -1

If the equation

a{x^2} + 2bx + 3c = 0\;\,\mathrm{and}\;3{x^2} + 8x + 15 = 0

have a common root, where

a,\;b,\;c

are the lengths of the sides of a

\Delta ABC

, then

{\sin ^2}A + {\sin ^2}B + {\sin ^2}C

is equal to

View answer and explanation

Correct option: D

Option B

\left( {x - c} \right)\left( {x - d} \right) + k = 0

is correct because if c and d are roots of (x-a)(x-b) - k = 0, then (x-a)(x-b) - k = (x-c)(x-d). Thus, (x-c)(x-d) + k = (x-a)(x-b), implying a and b are roots of (x-c)(x-d) + k = 0.

329 Medium

329 of 362

Analyzing +4 | -1

If the difference between the corresponding roots of

x^2+

a x+b=0

and

x^2+b x+a=0

is same and

a \neq b

, then:

View answer and explanation

Correct option: A

a+b+4=0

Option A (

a+b+4=0

) is correct because it is the result of equating the difference between the roots of the two quadratic equations and simplifying the expression.

330 Medium

330 of 362

Analyzing +4 | -1

Let

S=\left\{\begin{array}{c}x: x \in \mathbb{R} \text { and }(\sqrt{3}+\sqrt{2})^{x^{2}-4} +(\sqrt{3}-\sqrt{2})^{x^{2}-4}=10\end{array}\right\}

. Then

n(S)

is equal to

View answer and explanation

Correct option: D

Option B (4) is correct because solving the equation for x yields four distinct real solutions.

Revision notes and key takeaways

Review QUADRATIC EQUATIONS in your textbook or class notes, then use this set to check your understanding.

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