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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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Showing 31-60 of 362 questions
31 Medium
31 of 362
Application +4 | -1
If the equation (log⁡16x)2−(log⁡16x)+log⁡16k=0\left(\log _{16} x\right)^2-\left(\log _{16} x\right)+\log _{16} k=0 has a unique solution (where kk is constant), then the value of kk is
View answer and explanation
Correct option: B
2
The correct answer is B (2) because setting the discriminant of the quadratic equation to zero and solving for k gives k=2.
32 Medium
32 of 362
Analyzing +4 | -1
If the quadratic equation ax2−bx+12=0a x^2-b x+12=0 where a,ba, b are positive integers not exceeding 10 has roots both greater than 2 , then number of possible ordered pair (a,b)(a, b) is
View answer and explanation
Correct option: A
1
Option B (1) is correct because there is only one possible ordered pair (a, b) that satisfies the given conditions.
33 Medium
33 of 362
Analyzing +4 | -1
If a,ba, b and cc are three distinct positive real numbers, then the number of real roots of ax2+2b∣x∣−c=0a x^2+2 b|x|-c=0 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the equation can be transformed into a quadratic in ∣x∣|x|, which yields one positive solution for ∣x∣|x|, resulting in two real roots for xx.
34 Medium
34 of 362
Analyzing +4 | -1
The values of λ\lambda such that sum of the squares of the roots of the quadratic equation, x2+(3−λ)x+2=λ\mathrm{x}^2+(3-\lambda) \mathrm{x}+2=\lambda has the least value is :
View answer and explanation
Correct option: A
2
Option A (2) is correct because completing the square for the sum of squares of roots results in a minimum value when lambda equals 2.
35 Medium
35 of 362
Analyzing +4 | -1
The values of ' aa ' for which x2+ax+sin⁡−1(x2−4x+5)+cos⁡−1(x2−4x+5)=0x^2+a x+\sin ^{-1}\left(x^2-4 x+5\right)+\cos ^{-1}\left(x^2-4 x+5\right)=0 has atleast one solution, is
View answer and explanation
Correct option: D
−2−π4-2-\frac{\pi}{4}
Option D (−2−π4-2-\frac{\pi}{4}) is correct because it's the value of 'a' that satisfies the given equation when x=2, which is the only possible solution.
36 Medium
36 of 362
Application +4 | -1
Let α,β\alpha ,\beta are the roots of equation 2x2−(p+1)x+(p−1)=02{x^2} - \left( {p + 1} \right)x + \left( {p - 1} \right) = 0. If α−β=αβ\alpha - \beta = \alpha \beta, then what is the value of pp?
View answer and explanation
Correct option: B
2
Option B (2) is correct because it is the value of p that satisfies the given condition α−β=αβ\alpha - \beta = \alpha \beta.
37 Medium
37 of 362
Analyzing +4 | -1
The equation x2−4x+[x]+3=x[x]x^{2}-4 x+[x]+3=x[x], where [x][x] denotes the greatest integer function, has
View answer and explanation
Correct option: D
A unique solution in (−∞,∞)(-\infty, \infty)
Option D (A unique solution in (−∞,∞)(-\infty, \infty)) is correct because the equation simplifies to having only one solution, which is x=1.
38 Medium
38 of 362
Analyzing +4 | -1
The value of x,log⁡1/2x≥log⁡1/3xx, \log _{1 / 2} x \geq \log _{1 / 3} x is
View answer and explanation
Correct option: A
(0,1](0,1]
Option A ((0,1])((0,1]) is correct because the inequality holds for x in this interval, including x=1.
39 Medium
39 of 362
Application +4 | -1
If a1,a2,…,ana_1, a_2, \ldots, a_n are positive real numbers whose product is a fixed real number cc, then the minimum value of a1+a2+…+an−1+2ana_1+a_2+\ldots+a_{n-1}+2 a_n is
View answer and explanation
Correct option: A
n(2c)1/nn(2 c)^{1 / n}
The correct answer is A (n(2c)1/nn(2 c)^{1 / n}) because it is derived using the AM-GM inequality to find the minimum value of the given expression.
40 Medium
40 of 362
Analyzing +4 | -1
If a,b,c are real numbers such that a + b + c = 0, then the
quadratic equation 3ax2+2bx+c=03a{x^2} + 2bx + c = 0 has:
View answer and explanation
Correct option: A
At least one root in [0, 1]
Option A (At least one root in [0, 1]) is correct because applying Rolle's theorem to the function f(x)=ax3+bx2+cxf(x) = ax^3 + bx^2 + cx on the interval [0, 1] shows that there must be at least one root of 3ax2+2bx+c=03ax^2 + 2bx + c = 0 in the interval [0, 1].
41 Medium
41 of 362
Application +4 | -1
The roots of the equation (b−c)x2+(c−a)x+(a−b)(b-c) x^2+(c-a) x+(a-b) =0=0 are
View answer and explanation
Correct option: C
a−bb−c,1\frac{a-b}{b-c}, 1
The correct answer is C (a−bb−c,1\frac{a-b}{b-c}, 1) because 1 is a root and the product of the roots equals a−bb−c\frac{a-b}{b-c}.
42 Medium
42 of 362
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.
43 Medium
43 of 362
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.
44 Medium
44 of 362
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:
View answer and explanation
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.
45 Medium
45 of 362
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
View answer and explanation
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}.

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46 Medium
46 of 362
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.
47 Medium
47 of 362
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.
48 Medium
48 of 362
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.
49 Medium
49 of 362
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.
50 Medium
50 of 362
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.
51 Medium
51 of 362
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.
52 Medium
52 of 362
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.
53 Medium
53 of 362
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.
54 Medium
54 of 362
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.
55 Medium
55 of 362
Analyzing +4 | -1
The values of 'a' for which the roots of the equation x2+x+a=0{x^2} + x + a = 0 are real and exceed 'a' are
View answer and explanation
Correct option: C
a<−2a < - 2
Option C (a<−2a < -2) is correct because it satisfies the conditions for the roots to be real and greater than 'a'.
56 Medium
56 of 362
Analyzing +4 | -1
If a∈R−a \in R^{-}and a≠−2a \neq-2, then the equation x2+a∣x∣+1=0x^2+a|x|+1=0
View answer and explanation
Correct option: D
must have either four real roots or no real root
Option D (must have either four real roots or no real root) is correct because the equation can have either four real roots when a < -2 or no real roots depending on the value of the discriminant.
57 Medium
57 of 362
Application +4 | -1
If α,β\alpha, \beta are the roots of x2−6x−2=0.x^2-6 x-2=0. If an=αn−βna_n=\alpha^n-\beta^n for n≥1.n \geq 1. Then the value of a12−2a103a11\frac{a_{12}-2 a_{10}}{3 a_{11}} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because by using the given recurrence relation and the properties of roots, we can simplify the expression to obtain the value 2.
58 Medium
58 of 362
Analyzing +4 | -1
If α,β\alpha, \beta be the positive real roots of the equation x2−px+q=0x^2-p x+q=0, then the value of α1/4+β1/4\alpha^{1 / 4}+\beta^{1 / 4} is (p+6q+4q1/4p+2q)k\left(p+6 \sqrt{q}+4 q^{1 / 4} \sqrt{p+2 \sqrt{q}}\right)^k, where k\mathrm{k} is equal to
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (1/4) is correct because after simplifying the expression (α1/4+β1/4)4(\alpha^{1/4} + \beta^{1/4})^4, we find that k = 1/4.
59 Medium
59 of 362
Analyzing +4 | -1
The largest possible value of the expression
S=sin⁡x1⋅cos⁡x2+sin⁡x2⋅cos⁡x3+…..+sin⁡x99⋅cos⁡x100+sin⁡x100⋅cos⁡x1, where x1,x2,x3,….,x100S=\sin x_1 \cdot \cos x_2+\sin x_2 \cdot \cos x_3+\ldots . .+\sin x_{99} \cdot \cos x_{100}+\sin x_{100} \cdot \cos x_1, \text { where } x_1, x_2, x_3, \ldots ., x_{100} are arbitrary real numbers is
View answer and explanation
Correct option: A
50
Option A (50) is correct because using the inequality 2ab≤a2+b22ab \leq a^2 + b^2, the expression S can be bounded by 50.
60 Medium
60 of 362
Analyzing +4 | -1
The number of solutions of 3x2+6x+7+5x2+10x+14=4−2x−x2,\sqrt {3{x^2} + 6x + 7} + \sqrt {5{x^2} + 10x + 14} = 4 - 2x - {x^2}, is
View answer and explanation
Correct option: A
1
The correct answer is A. By completing the square and analyzing the minimum values of the square roots and the maximum value of the quadratic expression, we find that the only solution is x=-1.
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