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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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661 Medium
661 of 914
Analyzing +4 | -1
If m>1m > 1 and n∈Nn \in N, then
View answer and explanation
Correct option: A
1m+2m+3m+…+nmn>(n+12)m\frac{1^m+2^m+3^m+\ldots+n^m}{n} > \left(\frac{n+1}{2}\right)^m
Option A is correct because for m > 1, the Arithmetic Mean of mth power is greater than the mth power of the Arithmetic Mean.
662 Medium
662 of 914
Analyzing +4 | -1
If a,b,ca,b,c are real numbers in G.P. such that aa and cc are positive, then the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0
View answer and explanation
Correct option: C
Are imaginary and are in ratio 1  :ω,1\;:\omega , where ω\omega is a complex cube root of unity
Option D is correct because the roots are imaginary and their ratio simplifies to 1:ω1:\omega, where ω\omega is a complex cube root of unity.
663 Medium
663 of 914
Analyzing +4 | -1
The equation sin⁡  θ=x+px\sin \,\,\theta = x + \frac{p}{x} for real values of xx is possible when
View answer and explanation
Correct option: C
p≤14p \le \frac{1}{4}
Option C (p≤14p \le \frac{1}{4}) is correct because for real values of x, the discriminant of the quadratic equation must be non-negative, leading to the condition p≤14p \le \frac{1}{4}.
664 Easy
664 of 914
Application +4 | -1
The quadratic equation whose roots are cubes of the roots of x2+bx+c=0x^2+b x+c=0 is
View answer and explanation
Correct option: B
x2+b(b2−3c)x+c3=0x^2+b\left(b^2-3 c\right) x+c^3=0
Option B (x2+b(b2−3c)x+c3=0x^2+b(b^2-3 c) x+c^3=0) is correct because it is derived by finding the sum and product of the cubes of the roots of the given quadratic equation.
665 Easy
665 of 914
Application +4 | -1
Value of

x=6+6+6+⋯ up to ∞x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots \text { up to } \infty}}} (1)

is
View answer and explanation
Correct option: A
3
Option A (3) is correct because solving the equation x=6+xx = \sqrt{{6+x}} yields x = 3 as the positive solution.
666 Medium
666 of 914
Analyzing +4 | -1
Let a,ba, b and cc be real numbers, a≠0a \neq 0. If α\alpha is a root of a2x2+bx+c=0,βa^2 x^2+b x+c=0, \beta is the root of a2x2−bx−c=0a^2 x^2-b x-c=0 and 0<α<β0 < \alpha < \beta, then the equation a2x2+2bx+2c=0a^2 x^2+2 b x+2 c=0 has aa root γ\gamma that always satisfies
View answer and explanation
Correct option: D
α<γ<β\alpha < \gamma < \beta
Option D is correct because f(alpha)f(beta) < 0, implying a root lies between alpha and beta.
667 Easy
667 of 914
Analyzing +4 | -1
The number of roots of the quadratic equation 8sec⁡2θ−6sec⁡θ+1=08 \sec ^2 \theta-6 \sec \theta+1=0 is:
View answer and explanation
Correct option: D
0
Option D (0) is correct because the equation 8sec⁡2θ−6sec⁡θ+1=08 \sec^2 \theta - 6 \sec \theta + 1 = 0 has no real roots since the values of sec⁡θ\sec \theta obtained are outside its range.
668 Easy
668 of 914
Application +4 | -1
The equation (cos⁡p−1)x2+cos⁡p  x+sin⁡p=0,\left( {\cos p - 1} \right){x^2} + \cos p\;x + \sin p = 0, in variable xx, has real roots. Then, pp belongs to the interval
View answer and explanation
Correct option: D
(0,π)\left( {0,\pi } \right)
Option D ((0,π)\left( {0,\pi } \right)) is correct because for the quadratic equation to have real roots, the discriminant must be greater than or equal to zero, leading to the condition sin⁡p≥0\sin p \ge 0, which is satisfied in the interval (0,π)\left( {0,\pi } \right).
669 Easy
669 of 914
Application +4 | -1
The roots of the equation (b−c)x2+(c−a)x+(a−b)=0(b-c) x^2+(c-a) x+(a-b)=0 are
View answer and explanation
Correct option: B
a−bb−c,1\frac{a-b}{b-c}, 1
Option B (a−bb−c,1\frac{{a-b}}{{b-c}}, 1) is correct because x=1 is a root, and the product of the roots equals a−bb−c\frac{{a-b}}{{b-c}}.
670 Easy
670 of 914
Application +4 | -1
Let f(x)f\left( x \right) be a quadratic polynomial such that f(−2)+f(3)=0.f\left( { - 2} \right) + f\left( 3 \right) = 0. If one of the roots of f(x)=0f\left( x \right) = 0 is –1, then the sum of the roots of f(x)=0f\left( x \right) = 0 is equal to:
View answer and explanation
Correct option: A
113\frac{{11}}{3}
Option A (113\frac{{11}}{3}) is correct because it is the sum of the roots of the quadratic equation, derived using the given condition f(−2)+f(3)=0f(-2) + f(3) = 0 and the fact that one root is -1.
671 Easy
671 of 914
Analyzing +4 | -1
Number of roots of the equation x−2x−1=1−2x−1x - \frac{2}{{x - 1}} = 1 - \frac{2}{{x - 1}} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation simplifies to x=1, which is not a valid solution due to the original equation's domain restriction.
672 Medium
672 of 914
Application +4 | -1
Solution of the equation 4.9x−1=3(22x+1){4.9^{x - 1}} = 3\sqrt {\left( {{2^{2x + 1}}} \right)} is
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (32\frac{3}{2}) is correct because it is the solution obtained by simplifying the exponential equation and solving for x.
673 Hard
673 of 914
Analyzing +4 | -1
The number of solutions of log⁡5+log(x2+1)log⁡(x−2)=2\frac{{\log 5 + \mathrm{log}\left( {{x^2} + 1} \right)}}{{\log \left( {x - 2} \right)}} = 2 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation simplifies to a value of x that makes the logarithm undefined.
674 Easy
674 of 914
Application +4 | -1
If (α,β)(\alpha, \beta) are the roots of equation x2−px+q=0x^2-p x+q=0 and (α′,β′)\left(\alpha^{\prime}, \beta^{\prime}\right) are that of x2−p′x+q′=0x^2-p^{\prime} x+q^{\prime}=0, then (α−α′)2+(β−α′)2+(α−β′)2+(β−β′)2=\left(\alpha-\alpha^{\prime}\right)^2+\left(\beta-\alpha^{\prime}\right)^2+\left(\alpha-\beta^{\prime}\right)^2+\left(\beta-\beta^{\prime}\right)^2=
View answer and explanation
Correct option: A
2(p2−2q+p′2−2q′−pp′)2\left(p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\right)
Option A (2(p2−2q+p′2−2q′−pp′)2\left(p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\right)) is correct because it is obtained by expanding the given expression and using Vieta's formulas to simplify.
675 Medium
675 of 914
Analyzing +4 | -1
For the equation x34(log⁡2x)2+log⁡2(x)−54=2{x^{\frac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}\left( x \right) - \frac{5}{4}}} = \sqrt 2 which one of the following is not true?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the equation has three real solutions, one of which is irrational, meaning options A, B, and C are all false.

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676 Medium
676 of 914
Application +4 | -1
If x2−2xcos⁡θ+1=0x^2-2 x \cos \theta+1=0, then the value of x2n−2xncos⁡nθ+1,n∈Nx^{2 n}-2 x^n \cos n \theta+1, n \in N is equal to -
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting x=cos⁡θ+isin⁡θx = \cos \theta + i \sin \theta into the expression and using De Moivre's theorem results in the expression simplifying to 0.
677 Medium
677 of 914
Application +4 | -1
If ecos⁡x−e−cos⁡x=4{e^{\cos x}} - {e^{ - \cos x}} = 4 then the value of cos⁡x\cos x is
View answer and explanation
Correct option: A
log⁡(2+5)\log \left( {2 + \sqrt 5 } \right)
Option A (log⁡(2+5)\log \left( {2 + \sqrt 5 } \right)) is correct because by substituting y=ecos⁡xy = e^{\cos x} into the equation and solving the quadratic, we find that cos⁡x=log⁡(2+5)\cos x = \log(2 + \sqrt{5}).
678 Easy
678 of 914
Application +4 | -1
Let f(x)f(x) be a quadratic expression positive for all real xx. If g(x)=f(x)−f′(x)+f′′(x)g(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x), then for any real xx
View answer and explanation
Correct option: A
g(x)>0g(x)>0
Option A g(x)>0g(x)>0 is correct because the discriminant of the quadratic expression g(x)g(x) is negative, and the leading coefficient is positive.
679 Easy
679 of 914
Application +4 | -1
Two non-integer roots of the equation

(x2+3x)2−(x2+3x)−6=0\left(x^2+3 x\right)^2-\left(x^2+3 x\right)-6=0 (1)

are⁡\operatorname{are}
View answer and explanation
Correct option: C
12(−3+21),12(−3−21)\frac{1}{2}(-3+\sqrt{21}), \frac{1}{2}(-3-\sqrt{21})
Option C (12(−3+21),12(−3−21)\frac{1}{2}(-3+\sqrt{21}), \frac{1}{2}(-3-\sqrt{21})) is correct because by substituting y=x2+3xy = x^2 + 3x, we can solve the resulting quadratic equation for y, and then solve for x.
680 Easy
680 of 914
Application +4 | -1
If one root of equation x2+px+q=0x^2+p x+q=0 is square of the other then :
View answer and explanation
Correct option: A
p3−q(3p−1)+q2=0p^3-q(3 p-1)+q^2=0
Option A (p3−q(3p−1)+q2=0p^3-q(3 p-1)+q^2=0) is correct because it is derived by relating the roots and coefficients of the quadratic equation and using the condition that one root is the square of the other.
681 Medium
681 of 914
Analyzing +4 | -1
If α\alpha and β\beta are roots of the equation x2+ax+b=0x^2+a x+b=0, then maximum value of the expression −(x2+ax+b)−(α−β)24-\left(x^2+a x+b\right)-\frac{(\alpha-\beta)^2}{4} will be
View answer and explanation
Correct option: C
0
Option C (0) is correct because the maximum value of the given expression is 0, which is obtained by substituting the maximum value of the quadratic expression.
682 Easy
682 of 914
Application +4 | -1
The number of real solutions of x2−2x+2+b−x^2-2 x+2+\mathrm{b}- 11=011=0 is:
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given quadratic equation simplifies to a form that always results in a positive value, indicating no real roots.
683 Easy
683 of 914
Analyzing +4 | -1
The number of real roots of the equation, e4x+e3x−4e2x+ex+1=0{e^{4x}} + {e^{3x}} - 4{e^{2x}} + {e^x} + 1 = 0 is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equation has only one real root, which is found by transforming the equation and solving for x.
684 Medium
684 of 914
Application +4 | -1
The number of real roots of the equation 2  x4+5  x2+3=0,2\;{x^4} + 5\;{x^2} + 3 = 0, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the equation has no real roots, as the expression is always positive for real x.
685 Easy
685 of 914
Analyzing +4 | -1
If the sum of the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 be equal to the sum of the reciprocal of their squares, then bc2,ca2,  ab2b{c^2},c{a^2},\;a{b^2} will be in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition leads to the relationship 2a2c=ab2+bc22a^2c = ab^2 + bc^2, which is the condition for bc2,ca2,ab2bc^2, ca^2, ab^2 to be in Arithmetic Progression.
686 Easy
686 of 914
Analyzing +4 | -1
If one root is square of the root of the equation x2+px+q=0{x^2} + px + q = 0, then the relation between pp and qq is
View answer and explanation
Correct option: A
p3−(3p−1)q+q2=0{p^3} - \left( {3p - 1} \right)q + {q^2} = 0
Option A (p3−(3p−1)q+q2=0{p^3} - \left( {3p - 1} \right)q + {q^2} = 0) is correct because it represents the relationship between p and q when one root is the square of the other root in the given quadratic equation.
687 Easy
687 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots of the equation ax2+bx+c=0a x^2+b x+c=0 and An=αn+βnA_n=\alpha^n+\beta^n, then aAn+2+bAn+1+cAna A_{n+2}+b A_{n+1}+c A_n is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression aAn+2+bAn+1+cAna A_{n+2}+b A_{n+1}+c A_n simplifies to 0 using the properties of roots of a quadratic equation.
688 Easy
688 of 914
Analyzing +4 | -1
let a,b∈R,a≠0a,b \in R,a \ne 0 be such that the equation, ax2−2bx+5=0a{x^2} - 2bx + 5 = 0 has a repeated root α\alpha which is also a root of the equation, x2−2bx−10=0.{x^2} - 2bx - 10 = 0. If β\beta is the other root of this equation, then α2+β2{\alpha ^2} + {\beta ^2} is equal to
View answer and explanation
Correct option: D
25
Option D (25) is correct because by solving the equations derived from the repeated root condition and the given quadratic equations, we find the value of α2+β2{\alpha ^2} + {\beta ^2} to be 25.
689 Easy
689 of 914
Analyzing +4 | -1
The number of real roots of the equation (x+1x)3+x+1x=0{\left( {x + \frac{1}{x}} \right)^3} + x + \frac{1}{x} = 0 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equation has no real roots since the only possible solution leads to x2=−1{x^2} = - 1, which has no real solutions.
690 Easy
690 of 914
Application +4 | -1
If xx satisfies the inequations 2x−7<11,  3x+4<−52x - 7 < 11,\;3x + 4 < - 5, then xx lies in the interval
View answer and explanation
Correct option: C
(−∞,  −3)\left( { - \infty ,\; - 3} \right)
Option C ((−∞,  −3)\left( { - \infty ,\; - 3} \right)) is correct because it represents the intersection of the solution sets of the two inequalities, x<9x < 9 and x<−3x < -3.
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