Skip to main content
Back to QUADRATIC EQUATIONS

Your Progress

0 / 914 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 331-360 of 914 questions

Preparing practice and saved progress…

331 Easy
331 of 914
Analyzing +4 | -1
x2−11x+ax^2-11 x+a and x2−14x+2ax^2-14 x+2 a will have a common factor, if a=a= ?
View answer and explanation
Correct option: B
0, 24
Option B (0, 24) is correct because setting the two quadratic expressions equal to each other to find a common factor leads to the solutions a=0a=0 and a=24a=24.
332 Easy
332 of 914
Application +4 | -1
The number of rational roots of 81(2x−53x+1)4−45(2x−53x+1)2+4=0,x≠1/381\left(\frac{2 x-5}{3 x+1}\right)^4-45\left(\frac{2 x-5}{3 x+1}\right)^2+4=0, x \neq 1 / 3 is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the equation has four rational roots after solving for x.
333 Easy
333 of 914
Application +4 | -1
If x2/3−7x1/3+10=0{x^{2/3}} - 7{x^{1/3}} + 10 = 0 then the set of values of xx, is
View answer and explanation
Correct option: D
{8,  125}\left\{ {8,\;125} \right\}
Option D ({8,  125}\left\{ {8,\;125} \right\}) is correct because these are the solutions obtained by solving the given equation by substitution.
334 Easy
334 of 914
Application +4 | -1
The value of kk for which the equation
3x2+2x(k2+1)+k2−3k+2=03{x^2} + 2x\left( {{k^2} + 1} \right) + {k^2} - 3k + 2 = 0
has roots of opposite signs, lies in the interval
View answer and explanation
Correct option: C
(1,  2)\left( {1,\;2} \right)
The correct answer is B (A circle) because the locus of z satisfying the given condition is a circle.
335 Easy
335 of 914
Analyzing +4 | -1
sin⁡x+cos⁡x=y2−y+a\sin x + \cos x = {y^2} - y + a has no value of xx for any yy, if 'a' belongs to
View answer and explanation
Correct option: D
(3,∞)\left( {\sqrt {3,} \infty } \right)
Option D ((3∞)\left( {\sqrt {3} \infty } \right)) is correct because it ensures that the range of the quadratic in 'y' does not overlap with the range of sin(x) + cos(x).
336 Easy
336 of 914
Application +4 | -1
Suppose α,β\alpha, \beta are roots of 8x2−10x+3=08 x^2-10 x+3=0, then ∑n=0∞(αx+βn)\sum\limits_{{n=0}}^{\infty}\left(\alpha^x+\beta^n\right) is
View answer and explanation
Correct option: C
6
Option C (6) is correct because by finding the roots and using the formula for the sum of an infinite geometric series, the sum converges to 6.
337 Easy
337 of 914
Application +4 | -1
The value of 7+7−7+7−…\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\ldots}}}} upto ∞\infty is
View answer and explanation
Correct option: C
3
Option C (3) is correct because after setting up the equation and solving for x, the valid solution is 3, considering the nested radical structure.
338 Medium
338 of 914
Analyzing +4 | -1
If x2−(a+b+c)x+(ab+bc+ca)=0x^2-(a+b+c) x+(a b+b c+c a)=0 has imaginary roots, where a,b,c∈R+a, b, c \in R^{+}, then a,b\sqrt{a}, \sqrt{b} and c\sqrt{c}
View answer and explanation
Correct option: A
can be sides of triangle
Option A (can be sides of triangle) is correct because the given condition implies that the square roots of a, b, and c satisfy the triangle inequality.
339 Medium
339 of 914
Analyzing +4 | -1
The absolute minimum value, of the function f(x)=∣x2−x+1∣+[x2−x+1]f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right], where [t][t] denotes the greatest integer function, in the interval [−1.2],[-1.2], is
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{3}{4}) is correct because the minimum value of the function occurs at x=1/2, and the greatest integer function evaluates to 0 at that point.
340 Easy
340 of 914
Application +4 | -1
If the difference of the roots of x2−px+q=0x^2-p x+q=0 is unity, then
View answer and explanation
Correct option: C
p2+4q2=(1+2q)2p^2+4 q^2=(1+2 q)^2
The correct answer is C (p2+4q2=(1+2q)2p^2+4 q^2=(1+2 q)^2) because it is derived from the relationship between the roots and coefficients of a quadratic equation and the given condition that the difference of the roots is unity.
341 Medium
341 of 914
Analyzing +4 | -1
If roots of x2−(a−3)x+a=0x^2-(a-3) x+a=0 are such that both of them is greater than 2 , then
View answer and explanation
Correct option: C
a∈[9,10)a \in[9,10)
Option C (a∈[9,10)a \in[9,10)) is correct because it satisfies the conditions for both roots to be greater than 2, considering the discriminant, f(2) > 0, and -B/2A > 2.
342 Medium
342 of 914
Application +4 | -1
If one root of the equation is ix2−2(i+1)x+(2−i)=0i x^2-2(i+1) x+(2-i)=0 is 2−i2-i, then the other root is
View answer and explanation
Correct option: A
−i-i
The correct answer is A (-i) because the coefficients are not real, so we cannot assume conjugate roots, and we must use the product of roots formula.
343 Easy
343 of 914
Analyzing +4 | -1
Let f(x)=x2+ax+bf\left( x \right) = {x^2} + ax + b, where a,  b∈Ra,\;b \in R. If f(x)=0f\left( x \right) = 0 has all its roots imaginary, then the roots of f(x)+f′(x)+f′′(x)=0f\left( x \right) + f'\left( x \right) + f''\left( x \right) = 0 are
View answer and explanation
Correct option: B
Imaginary
Option B (Imaginary) is correct because if f(x)=0f(x) = 0 has imaginary roots, then f(x)+f′(x)+f′′(x)=0f(x) + f'(x) + f''(x) = 0 also has imaginary roots.
344 Medium
344 of 914
Analyzing +4 | -1
If the roots of the equation ax2+x+b=0a x^2+x+b=0 are real, then the roots of the equation x2−4abx+1=0x^2-4 \sqrt{a b} x+1=0 will be
View answer and explanation
Correct option: D
imaginary
Option D is correct because the discriminant of the second equation is negative, implying imaginary roots.
345 Medium
345 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of ax2+bx+c=0a x^2+b x+c=0 and pn=αn+βnp_n=\alpha^n+\beta^n, then apn+1+bpn+cpn−1a p_{n+1}+b p_n+c p_{n-1} is equal to
View answer and explanation
Correct option: B
0
The correct answer is B (0) because substituting pnp_n and simplifying using the properties of roots results in zero.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
346 Easy
346 of 914
Application +4 | -1
If f(x)=(x−a1)2+(x−a2)2+⋯+(x−an)2f(x)=\left(x-a_{1}\right)^{2}+\left(x-a_{2}\right)^{2}+\cdots+\left(x-a_{n}\right)^{2}, find xx where f(x)f(x) is minimum.
View answer and explanation
Correct option: B
a1+a2+⋯+ann\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
Option B (a1+a2+⋯+ann\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}) is correct because the minimum value of f(x) occurs at the average of the aia_i values.
347 Medium
347 of 914
Application +4 | -1
If x2+y2+z2=ax^2+y^2+z^2=a, then the value of xy+zy+zxx y+z y+z x lies in the interval
View answer and explanation
Correct option: C
[−a2,a]\left[\frac{-a}{2}, a\right]
The correct answer is C ([−a2,a\frac{-a}{2}, a]) because using (x+y+z)2≥0(x+y+z)^2 \geq 0 and (x−y)2+(y−z)2+(z−x)2≥0(x-y)^2+(y-z)^2+(z-x)^2 \geq 0 allows us to find the interval for xy+yz+zxxy+yz+zx.
348 Easy
348 of 914
Application +4 | -1
If 2−3x−2x2≥0,2 - 3x - 2{x^2} \ge 0, then
View answer and explanation
Correct option: B
−2≤x≤12- 2 \le x \le \frac{1}{2}
Option B (−2≤x≤12- 2 \le x \le \frac{1}{2}) is correct because it represents the interval where the quadratic expression 2−3x−2x22 - 3x - 2{x^2} is non-negative.
349 Hard
349 of 914
Application +4 | -1
Let a,b,ca, b, c be the sides of a triangle. No two of them are equal and λ∈R\lambda \in R. If the roots of the equation x2+2(a+b+c)x+3λ(ab+bc+ca)=0x^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
λ<43\lambda<\frac{4}{3}
Option A (λ<43\lambda<\frac{4}{3}) is correct because the discriminant of the quadratic equation must be non-negative, leading to the inequality λ<43\lambda < \frac{4}{3}.
350 Easy
350 of 914
Analyzing +4 | -1
If the difference between the roots of x2+ax+b=0 and x2+bx+a=0{x^2} + ax + b = 0\,\mathrm{and}\,{x^2} + bx + a = 0 is same and a≠b,a \ne b, then
View answer and explanation
Correct option: A
a+b+4=0a + b + 4 = 0
Option A (a+b+4=0a + b + 4 = 0) is correct because the condition on the difference of roots leads to this relationship between a and b.
351 Medium
351 of 914
Application +4 | -1
If α\alpha and β\beta are roots of ax2−2bx+c=0a x^2-2 b x+c=0, then α3β3+α2β3+α3β2\alpha^3 \beta^3+\alpha^2 \beta^3+\alpha^3 \beta^2 is
View answer and explanation
Correct option: A
c2(c+2b)a3\frac{c^2(c+2 b)}{a^3}
Option A (c2(c+2b)/a3c^2(c+2 b)/a^3) is correct because it results from substituting the sum and product of the roots into the simplified expression.
352 Easy
352 of 914
Analyzing +4 | -1
If ax2+bx−ca{x^2} + bx - c is divisible by x2+bx+c,{x^2} + bx + c, then 'a' is a root of the equation
View answer and explanation
Correct option: A
cx2−bx−1=0c{x^2} - bx - 1 = 0
Option C (cx2−bx−1=0c{x^2} - bx - 1 = 0) is correct because 'a' satisfies this equation when ax3+bx+ca{x^3} + bx + c is divisible by x2+bx+c{x^2} + bx + c.
353 Easy
353 of 914
Application +4 | -1
If 10x−1+10−x−1=13,{10^{x - 1}} + {10^{ - x - 1}} = \frac{1}{3}, then xx equals to
View answer and explanation
Correct option: A
±log⁡103\pm {\log _{10}}3
Option A (±log⁡103\pm {\log _{10}}3) is correct because solving the given equation leads to 10x=310^x = 3 or 10x=1310^x = \frac{1}{3}, which gives x=±log⁡103x = \pm {\log_{10}}3.
354 Easy
354 of 914
Application +4 | -1
If the roots of the equation (b−c)x2+(c−a)x+(a−b)=0(b-c) x^2+(c-a) x+(a-b)=0 be equal and a+c=kba+c=k b. Then the value of k=\boldsymbol{k}=
View answer and explanation
Correct option: A
2
Option A (2) is correct because, with equal roots and a+c=kb, solving the quadratic equation leads to k=2.
355 Medium
355 of 914
Application +4 | -1
If one root of a quadratic equation is 12+5\frac{1}{{2 + \sqrt 5 }}, then the equation is
View answer and explanation
Correct option: B
x2+4x−1=0x^2+4 x-1=0
Option B (x2+4x−1=0x^2+4 x-1=0) is correct because if one root is 12+5\frac{{1}}{{2 + \sqrt 5 }}, the other root is its conjugate, and we can form the quadratic equation using the sum and product of the roots.
356 Easy
356 of 914
Application +4 | -1
The roots of the equation (a+b)x2−15+(a−b)x2−15=2a,{(a + \sqrt b )^{{x^2} - 15}} + {(a - \sqrt b )^{{x^2} - 15}} = 2a, where a2−b=1a^2-b=1, are
View answer and explanation
Correct option: B
±4,±14\pm 4, \pm \sqrt{14}
Option B (±4,±14\pm 4, \pm \sqrt{14}) is correct because these values satisfy the given equation after simplification and solving for x.
357 Medium
357 of 914
Application +4 | -1
If p,q,rp, q, r are three positive real numbers are in APA P, then the roots of the quadratic equation px2+qx+r=0p x^2+q x+r=0 are all real for
View answer and explanation
Correct option: A
∣rp−7∣≥43\left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}
Option A (∣rp−7∣≥43\left|\frac{{r}}{{p}}-7\right| \geq 4 \sqrt{{3}}) is correct because the condition for real roots of the quadratic equation, along with the arithmetic progression relationship, leads to this inequality.
358 Easy
358 of 914
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are the roots of the equation, x3+ax2+bx+c=0x^3+a x^2+b x+c=0, then (1−α2)(1−β2)(1−γ2)\left(1-\alpha^2\right)\left(1-\beta^2\right)\left(1-\gamma^2\right) is equal to:
View answer and explanation
Correct option: A
(1+b)2−(a+c)2(1+b)^2-(a+c)^2
Option A ((1+b)2−(a+c)2(1+b)^2-(a+c)^2) is correct because by substituting y=1−x2y = 1-x^2 and simplifying, we can find the product of the roots of the transformed equation.
359 Easy
359 of 914
Analyzing +4 | -1
Let f(x)=ax2+bx+c,  a≠0f\left( x \right) = a{x^2} + bx + c,\;a \ne 0 and Δ=b2−4ac\Delta = {b^2} - 4ac. If α+β,  α2+β2\alpha + \beta ,\;{\alpha ^2} + {\beta ^2} and α3+β3{\alpha ^3} + {\beta ^3} are in GP, then
View answer and explanation
Correct option: C
cΔ=0c\Delta = 0
Option C (cΔ=0c\Delta = 0) is correct because it represents the condition when α+β,;α2+β2\alpha + \beta ,\\;{\alpha ^2} + {\beta ^2} and α3+β3{\alpha ^3} + {\beta ^3} are in GP, which implies either c=0 or the discriminant is zero.
360 Medium
360 of 914
Application +4 | -1
If tan⁡A\tan \mathrm{A} and tan⁡B\tan \mathrm{B} are the roots of x2+ax+b=0\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0, then the value of expression sin⁡2(A+B)+asin⁡(A+B)cos⁡(A+B)+bcos⁡2(A+B)\sin ^2(A+B)+a \sin (A+B) \cos (A+B)+b \cos ^2(A+B) is equal to -
View answer and explanation
Correct option: D
b\mathrm{b}
Option D (b) is correct because the expression simplifies to b after substituting the values of tan A + tan B and tan A * tan B and using trigonometric identities.
Revision notes and key takeaways

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

Showing 331-360 of 914 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (Easy) 528 MCQs → ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (Hard) 24 MCQs → ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (Medium) 362 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 914 answered · 914 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →