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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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91 Easy
91 of 914
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of ax2+c=bx,a{x^2} + c = bx, then the equation (a+cy)2=b2y{\left( {a + cy} \right)^2} = {b^2}y in yy has the roots
View answer and explanation
Correct option: B
α2,β2{\alpha ^2},{\beta ^2}
Option B (α², β²) is correct because substituting y = x² into the transformed equation yields the roots as the squares of the original roots.
92 Medium
92 of 914
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation a1x2+2b1x+c1=0a_1 x^2+2 b_1 x+c_1=0 and γ,δ\gamma, \delta be the roots of a2x2+2b2x+c2=0a_2 x^2+2 b_2 x+c_2=0. If the four points A(α,0),B(β,0),C(γ,0)A(\alpha, 0), B(\beta, 0), C(\gamma, 0) and D(δ,0)D(\delta, 0) are such that the sum of the ratios in which CC and DD divide ABA B is zero, then a1c2+a2c1a_1 c_2+a_2 c_1 is
View answer and explanation
Correct option: A
2b1b22 b_1 b_2
Option A (2b1b22 b_1 b_2) is correct because the given condition about the ratios in which C and D divide AB leads to this relationship between the coefficients of the quadratic equations.
93 Easy
93 of 914
Analyzing +4 | -1
The number of real solutions of the equation (910)x=−3+x−x2\left(\frac{9}{10}\right)^x=-3+x-x^2 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the left-hand side of the equation is always positive, while the right-hand side is always negative, resulting in no real solutions.
94 Easy
94 of 914
Application +4 | -1
If tan⁡θ\tan \theta and cot⁡θ\cot \theta are roots of x2+2ax+b=0x^2+2 a x+b=0, then least value of ∣a∣|a| is
View answer and explanation
Correct option: B
1
Option A (1) is correct. Using the AM-GM inequality, we can find the minimum value of |a| to be 1.
95 Easy
95 of 914
Understanding +4 | -1
The number of real roots of f(x)=0f\left( x \right) = 0, where f(x)=(x−1)(x−2)(x−3)(x−4)f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) lying in the interval (1,3)\left( {1,3} \right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only the root x=2 lies within the interval (1,3).
96 Medium
96 of 914
Application +4 | -1
The solution set of the equation
[4(1−13+19−127+...)]log⁡2x=[54(1+13+19+127+...)]log⁡x2{\left[ {4\left( {1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{{27}} + ...} \right)} \right]^{{{\log }_2}x}} = {\left[ {54\left( {1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + ...} \right)} \right]^{{{\log }_x}2}} is
View answer and explanation
Correct option: A
{4,  14}\left\{ {4,\;\frac{1}{4}} \right\}
Option A {4,  14}\left\{{4,\;\frac{{1}}{{4}}}\right\} is the correct solution set because it satisfies the given logarithmic equation after simplification using geometric series and logarithm properties.
97 Easy
97 of 914
Application +4 | -1
The roots of the equation ix2−4x−4i=0i x^2-4 x-4 i=0 are:
View answer and explanation
Correct option: C
−2i,−2i-2 i,-2 i
Option C (−2i,−2i-2i, -2i) is correct because the quadratic equation has a repeated root of -2i.
98 Easy
98 of 914
Analyzing +4 | -1
The minimum value of e(x4−x3+x2){e^{\left( {{x^4} - {x^3} + {x^2}} \right)}}, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the minimum value of the exponent is 0, resulting in a minimum value of 1 for the exponential function.
99 Easy
99 of 914
Application +4 | -1
The equation (cos⁡p−1)x2+cos⁡px+sin⁡p=0(\cos p-1) x^{2}+\cos p x+\sin p=0 in xx has real roots. Then the set of values of pp is
View answer and explanation
Correct option: D
[0,π][0, \pi]
Option D ([0,π][0, \pi]) is correct because the discriminant of the quadratic equation must be non-negative for real roots, which leads to the condition that sin⁡p\sin p must be non-negative, restricting p to the interval [0,π][0, \pi].
100 Easy
100 of 914
Application +4 | -1
If the sum of the roots of a quadratic equation is 3 and the sum of their cubes is 7 , then the equation is
View answer and explanation
Correct option: D
9x2−27x+20=09 x^2-27 x+20=0
Option D (9x2−27x+20=09 x^2-27 x+20=0) is correct because it is obtained by using the given sum of roots and sum of their cubes to find the product of roots and then constructing the quadratic equation.
101 Easy
101 of 914
Application +4 | -1
Let α\alpha and β\beta be the roots of x2−6x−2=0x^2-6 x-2=0, with α>β\alpha>\beta. If an=αn−βna_n=\alpha^n-\beta^n for n≥1n \geq 1, the value of a10−2a82a9\frac{a_{10}-2 a_8}{2 a_9} is
View answer and explanation
Correct option: C
3
Option C (3) is correct because by substituting the roots into the equation and simplifying the expression, we find that the value is 3.
102 Easy
102 of 914
Application +4 | -1
Let xx be an integer and x2+x+1x^2+x+1 is divisible by 3. When xx is divided by 3 , it leaves remainder
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Correct option: B
1
Option B (1) is correct because when x is divided by 3, the remainder must be 1 for x2 + x + 1 to be divisible by 3.
103 Medium
103 of 914
Application +4 | -1
Let α\alpha and β\beta be the roots of the equation x2+x+1=0x^2+x+1=0, the equation whose roots are α19,β7\alpha^{19}, \beta^7 is:
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Correct option: D
x2+x+1=0x^2+x+1=0
Option D (x2+x+1=0x^2+x+1=0) is correct because the roots of the given equation are complex cube roots of unity, and raising them to certain powers results in the same roots.
104 Medium
104 of 914
Application +4 | -1
In a triangle PQR,  ∠R=π2PQR,\;\angle R = \frac{\pi }{2}. If tan⁡(P2)\tan \left( {\frac{P}{2}} \right) and tan⁡(Q2)\tan \left( {\frac{Q}{2}} \right) are the roots of ax2+bx+c=0,  a≠0a{x^2} + bx + c = 0,\;a \ne 0 then
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Correct option: C
c=a+bc = a + b
Option C (c=a+bc = a + b) is correct because it is derived from the trigonometric identity of tan⁡(P2+Q2)\tan(\frac{P}{2} + \frac{Q}{2}) and the given condition that ∠R=π2\angle R = \frac{\pi}{2}.
105 Medium
105 of 914
Analyzing +4 | -1
The number of values of kk for which the equation (x2+(2k−6)x+7−3k)(x2+(2k−2)x+3k−5)=0\left( {{x^2} + \left( {2k - 6} \right)x + 7 - 3k} \right)\left( {{x^2} + \left( {2k - 2} \right)x + 3k - 5} \right) = 0 has two different pairs of equal roots, is equal to:
View answer and explanation
Correct option: B
1
Option B, 1, is correct because the condition for two different pairs of equal roots leads to k = 2 as the only valid solution.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
106 Easy
106 of 914
Application +4 | -1
If x=1+1+1+…∞,  x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \ldots \infty } } } ,\; then xx is equal to
View answer and explanation
Correct option: A
1+52\frac{{1 + \sqrt 5 }}{2}
The correct answer is A (1+52\frac{{1 + \sqrt 5 }}{2}) because it is the positive solution to the equation x=1+xx = \sqrt {1 + x} which arises from the given infinite nested radical.
107 Medium
107 of 914
Application +4 | -1
The number of real solutions of x2+5∣x∣+x^2+5|x|+ 4=04=0 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equation x2+5∣x∣+4=0x^2+5|x|+4=0 has no real solutions, as the left-hand side is always greater than or equal to 4 for any real x.
108 Medium
108 of 914
Application +4 | -1
If the roots of the equation x2−px+q=0{x^2} - px + q = 0 differ by unity, then
View answer and explanation
Correct option: B
p2=4q+1{p^2} = 4q + 1
Option B (p2=4q+1{p^2} = 4q + 1) is correct because it relates the coefficients pp and qq based on the condition that the roots of the quadratic equation differ by unity.
109 Easy
109 of 914
Application +4 | -1
If a,b,ca, b, c are integers and b2=4(ac+5d2),d∈Nb^2=4\left(a c+5 d^2\right), d \in N, then roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0 are
View answer and explanation
Correct option: A
Irrational
Option A (Irrational) is correct because the discriminant D=b2−4ac=20d2D = b^2 - 4ac = 20d^2, and since d∈Nd \in N, D=2d5\sqrt{D} = 2d\sqrt{5} which is irrational.
110 Medium
110 of 914
Analyzing +4 | -1
Let α\alpha and β\beta be the roots of the equation x2−x−1=0x^{2}-x-1=0. If pk=(α)k+(β)k,k≥1p_{k}=(\alpha)^{k}+(\beta)^{k}, k \geq 1, then which one of the following statements is not true?
View answer and explanation
Correct option: D
p5=p2⋅p3p_{5}=p_{2} \cdot p_{3}
Option D (p5=p2⋅p3p_{5}=p_{2} \cdot p_{3}) is not true because calculating p5p_5, p2p_2, and p3p_3 shows that p5=11p_5 = 11 while p2⋅p3=12p_2 \cdot p_3 = 12, thus they are not equal.
111 Medium
111 of 914
Analyzing +4 | -1
If cos⁡α,sin⁡β,sin⁡α\cos \alpha, \sin \beta, \sin \alpha are in increasing G.P., then roots of x2−2cot⁡βx+1=0x^2-2 \cot \beta x+1=0 are: (where α,β∈R\alpha, \beta \in R )
View answer and explanation
Correct option: A
Real and distinct
Option A (Real and distinct) is correct because the discriminant of the quadratic equation is greater than 0, indicating real and distinct roots.
112 Medium
112 of 914
Analyzing +4 | -1
If the function f(x)=x3+3(a−7)x2+3(a2−9)x−1f\left( x \right) = {x^3} + 3\left( {a - 7} \right){x^2} + 3\left( {{a^2} - 9} \right)x - 1 has positive points of extremum then
View answer and explanation
Correct option: B
a∈(−∞,  −3)∪(3,  29/7)a \in \left( { - \infty ,\; - 3} \right) \cup \left( {3,\;29/7} \right)
Option B (a∈(−∞,  −3)∪(3,  29/7)a \in \left( { - \infty ,\; - 3} \right) \cup \left( {3,\;29/7} \right)) is correct because it satisfies the conditions for the function to have positive points of extremum.
113 Medium
113 of 914
Application +4 | -1
If (α,β)(\alpha, \beta) are roots of ax2+2bx−a=0\mathrm{ax}^2+2 \mathrm{bx}-\mathrm{a}=0 and quadratic equation whose roots are (2α−1β)\left(2 \alpha-\frac{1}{\beta}\right) and (2β−1α)\left(2 \beta-\frac{1}{\alpha}\right) is px2+qx+r=0\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0, then p+q+r=\mathrm{p}+\mathrm{q}+\mathrm{r}=
View answer and explanation
Correct option: C
6 b−8a6 \mathrm{~b}-8 \mathrm{a}
Option C (6 b−8a6 \mathrm{~b}-8 \mathrm{a}) is correct because it is the value of p+q+r obtained after finding the quadratic equation with the given roots and comparing coefficients.
114 Easy
114 of 914
Application +4 | -1
Let p,q∈{1,  2,  3,  4}.p,q \in \left\{ {1,\;2,\;3,\;4} \right\}. The number of equations of the form px2+qx+1=0p{x^2} + qx + 1 = 0 having real roots, is
View answer and explanation
Correct option: C
7
Option C (7) is correct because by checking each possible value of p from the set {1, 2, 3, 4}, we can determine the number of valid values of q that satisfy the condition for real roots.
115 Easy
115 of 914
Analyzing +4 | -1
Solution set of the equation 32x2−2⋅3x2+x+6+32(x+6)=03^{2 x^2}-2 \cdot 3^{x^2+x+6}+3^{2(x+6)}=0 is
View answer and explanation
Correct option: C
{−2,3}\{-2,3\}
The correct answer is C ({-2,3}) because it is the solution set obtained by simplifying the given exponential equation and solving the resulting quadratic equation.
116 Easy
116 of 914
Application +4 | -1
Let α,β\alpha, \beta be roots of ax2+bx+c=0a x^2+b x+c=0, where ac≠0a c \neq 0. Roots of cx2−bx+a=0c x^2-b x+a=0 are
View answer and explanation
Correct option: B
−1/α,−1/β-1 / \alpha,-1 / \beta
Option A (1/α,1/β1 / \alpha, 1 / \beta) is incorrect. The correct answer is that the roots of cx2−bx+a=0cx^2 - bx + a = 0 are 1/α1/\alpha and 1/β1/\beta.
117 Hard
117 of 914
Analyzing +4 | -1
Let nn be an odd positive integer. Then, the number of real roots of the polynomial Pn(x)=1+2x+3x2+  ……+(n+1)xn,{P_n}\left( x \right) = 1 + 2x + 3{x^2} + \; \ldots \ldots + \left( {n + 1} \right){x^n}, is
View answer and explanation
Correct option: C
1
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.
118 Medium
118 of 914
Analyzing +4 | -1
Let S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\} and β=∑x∈Stan⁡2(x3)\beta=\sum\limits_{{x \in S}} \tan ^2\left(\frac{x}{3}\right), then 16(β−14)2\frac{1}{6}(\beta-14)^2 is equal to
View answer and explanation
Correct option: D
32
Option D (32) is correct because solving the equation for tan2(x) leads to x=0, pi/4, -pi/4, and then substituting these values into tan2(x/3) and summing gives the result.
119 Medium
119 of 914
Application +4 | -1
Let p(x)=a2+bx,q(x)=lx2+mx+np(x)=a^2+b x, q(x)=l x^2+m x+n. If p(1)−q(1)=0,p(2)−q(2)=1p(1)-q(1)=0, p(2)-q(2)=1 and p(3)−q(3)=4p(3)-q(3)=4, then p(4)−q(4)p(4)-q(4) equals
View answer and explanation
Correct option: D
9
Option D (9) is correct because it is the value obtained by analyzing the given polynomial relationships and solving for the required difference.
120 Easy
120 of 914
Analyzing +4 | -1
The value of xx which satisfy the equation (5x2−8x+3)−(5x2−9x+4)=(2x2−2x)−(2x2−3x+1)\sqrt {\left( {5{x^2} - 8x + 3} \right)} - \sqrt {\left( {5{x^2} - 9x + 4} \right)} = \sqrt {\left( {2{x^2} - 2x} \right)} - \sqrt {\left( {2{x^2} - 3x + 1} \right)} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because by factoring and simplifying the equation, we find that x=1 is the only solution.
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