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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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91 Medium
91 of 362
Understanding +4 | -1
If p,q,rp,q,r are any real numbers, then
View answer and explanation
Correct option: B
min⁡(p,q)=12(P+q−∣p−q∣)\min \left( {p,q} \right) = \frac{1}{2}\left( {P + q - \left| {p - q} \right|} \right)
Option B is correct because it provides the correct formula for calculating the minimum of two real numbers p and q using the absolute value function.
92 Medium
92 of 362
Analyzing +4 | -1
Number of non-negative integral solution of equation y4+6xy2−8x=0y^4+6 x y^2-8 x=0 is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one non-negative integer value of y (y=0) results in a non-negative integer value for x.
93 Medium
93 of 362
Analyzing +4 | -1
If 0<a<1,0 < a < 1, then the solution set of the inequation 1+(log⁡ax)21+(log⁡ax)>1,\frac{{1 + {{\left( {{{\log }_a}x} \right)}^2}}}{{1 + \left( {{{\log }_a}x} \right)}} > 1, is
View answer and explanation
Correct option: C
(1,  1/a)∪(0,a)\left( {1,\;1/a} \right) \cup \left( {0,a} \right)
Option C ((1,  1/a)∪(0,a)\left( {1,\;1/a} \right) \cup \left( {0,a} \right)) is correct because solving the logarithmic inequality results in two intervals for x: (1, 1/a) and (0, a).
94 Medium
94 of 362
Application +4 | -1
If pp and qq are non-zero real numbers and α3+β3=−p,αβ=q\alpha^3+\beta^3=-p, \alpha \beta=q, then a quadratic equation whose roots are α2β,β2α\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha} is :
View answer and explanation
Correct option: B
qx2+px+q2=0q x^{2}+p x+q^{2}=0
Option B (qx2+px+q2=0q x^{2}+p x+q^{2}=0) is correct because it's derived from finding the sum and product of the new roots and constructing the quadratic equation.
95 Medium
95 of 362
Application +4 | -1
Suppose a∈Ra \in \mathbf{R}. If 3x2+2(a2+1)x+(a23 x^2+2\left(a^2+1\right) x+\left(a^2\right. −3a+2)=0-3 a+2)=0 possesses roots of opposite signs, then aa lies in the interval:
View answer and explanation
Correct option: C
(1,2)(1,2)
Option C ((1,2)) is correct because the product of the roots must be negative for them to have opposite signs, leading to the inequality (a-1)(a-2) < 0, which is satisfied when a is between 1 and 2.
96 Medium
96 of 362
Analyzing +4 | -1
If f(x)=x2+2bx+2c2  and  g(x)=−x2−2cx+b2  such  thatmin⁡f(x)>max⁡g(x),f\left( x \right) = {x^2} + 2bx + 2{c^2}\;\mathrm{and\;g}\left( x \right) = - {x^2} - 2cx + {b^2}\;\mathrm{such\;that}\min f\left( x \right) > \max \mathrm{g}\left( x \right), then the relation between bb and cc is
View answer and explanation
Correct option: D
∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2
Option D (∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2) is correct because the condition min⁡f(x)>max⁡g(x)\min f(x) > \max g(x) leads to the inequality ∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2.
97 Medium
97 of 362
Analyzing +4 | -1
If the equation x2+5x+8=0x^2+5 x+8=0 and ax2+bx+c=0;a,b,c∈Ra x^2+b x+c=0 ; a, b, c \in R have a common root, then a:b:ca: b: c is
View answer and explanation
Correct option: B
1:5:81: 5: 8
Option B (1:5:81: 5: 8) is correct because the given quadratic equation has imaginary roots, implying the second equation shares the same roots, leading to proportional coefficients.
98 Medium
98 of 362
Analyzing +4 | -1
The equation x2⋅2∣x−3∣+4+2x−1x^2 \cdot 2^{|x-3|+4}+2^{x-1} =x2⋅2x+1+2∣x−3∣+2=x^2 \cdot 2^{x+1}+2^{|x-3|+2} has
View answer and explanation
Correct option: C
x=±12x= \pm \frac{1}{2} and x≥3x \geq 3
Option D (x=±12x= \pm \frac{1}{2} and x≥3x \geq 3) is correct because it represents the complete solution set obtained by considering both cases when x-3 is positive and negative.
99 Medium
99 of 362
Application +4 | -1
The inequality ∣2x−3∣<1|2 x-3| < 1 is valid when xx lies in the interval
View answer and explanation
Correct option: B
(1,2)(1,2)
The correct answer is B ((1,2)) because solving the absolute value inequality ∣2x−3∣<1|2x-3| < 1 results in the interval (1, 2).
100 Medium
100 of 362
Application +4 | -1
If x2+2ax+10−3a>0{x^2} + 2ax + 10 - 3a > 0 for all x∈R,x \in R, then
View answer and explanation
Correct option: A
−5<a<2- 5 < a < 2
Option A (−5<a<2- 5 < a < 2) is correct because the given quadratic expression is always positive if its discriminant is negative, leading to this range for 'a'.
101 Medium
101 of 362
Application +4 | -1
If a2+b2+c2=1,{a^2} + {b^2} + {c^2} = 1, then ab+bc+ca  ab + bc + ca\; lies in the interval
View answer and explanation
Correct option: A
[−12,1]\left[ { - \frac{1}{2},1} \right]
Option A [−12,1]\left[ { - \frac{1}{2},1} \right] is correct because it represents the interval in which ab+bc+caab + bc + ca lies, derived from the given condition and algebraic manipulations.
102 Medium
102 of 362
Analyzing +4 | -1
If a>0,a≠1a > 0, a \neq 1, then the equation 2log⁡xa+log⁡axa+3log⁡a2xa=02 \log _x a+\log _{a x} a+3 \log _{a^2 x} a=0 has
View answer and explanation
Correct option: B
two real roots
Option B (two real roots) is correct because solving the logarithmic equation leads to two distinct real solutions for x, given the constraints on a.
103 Medium
103 of 362
Analyzing +4 | -1
If the equation x3−3x+a=0{x^3} - 3x + a = 0 has distinct roots between 0 and 1, then the value of aa is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derivative of the given function does not have any roots between 0 and 1, implying that the function cannot have distinct roots between 0 and 1.
104 Medium
104 of 362
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of the equation λ(x2−x)+x+5=0\lambda \left( {{x^2} - x} \right) + x + 5 = 0 and if λ1  and  λ2{\lambda _1}\;\mathrm{and}\;{\lambda _2} are two values of λ\lambda obtained from αβ+βα=45\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{4}{5} then λ1λ22+λ2λ12\frac{{{\lambda _1}}}{{\lambda _2^2}} + \frac{{{\lambda _2}}}{{\lambda _1^2}} is equal to
View answer and explanation
Correct option: D
4048
Option D (4048) is correct because it is the result of substituting the values of λ1+λ2\lambda_1 + \lambda_2 and λ1λ2\lambda_1 \lambda_2 into the expression after simplifying the given equation.
105 Medium
105 of 362
Application +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−2x+2=0x^{2}-\sqrt{2} x+2=0. Then α14+β14\alpha^{14}+\beta^{14} is equal to
View answer and explanation
Correct option: C
−128-128
Option C (-128) is correct because after finding the roots in polar form and raising them to the power of 14, the sum simplifies to -128.

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106 Medium
106 of 362
Application +4 | -1
The solution of the equation  (3+22)x2−8+(3+22)8−x2=6\,{\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} + {\left( {3 + 2\sqrt 2 } \right)^{8 - {x^2}}} = 6 are
View answer and explanation
Correct option: D
±3,  ±7\pm 3,\; \pm \sqrt 7
Option D (±3,;±7\pm 3,\\; \pm \sqrt 7) is correct because by substituting y=(3+22)x2−8y = {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} and solving the quadratic equation, we find the possible values of x.
107 Medium
107 of 362
Analyzing +4 | -1
Let f(x)=x2+ax+b;a,  b∈R.  If  f(1)+f(2)+f(3)=0,f\left( x \right) = {x^2} + ax + b;a,\;b \in R.\,\,\mathrm{If}\;f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) = 0, then the roots of the equation f(x)=0f\left( x \right) = 0
View answer and explanation
Correct option: D
Real and distinct
Option D (Real and distinct) is correct. If f(1) + f(2) + f(3) = 0, the values of f(1), f(2), and f(3) cannot all have the same sign, implying the roots are real and distinct.
108 Medium
108 of 362
Application +4 | -1
If α,β\alpha, \beta be the roots of x2−px+q=0x^2-p x+q=0 and α′,β′\alpha^{\prime}, \beta^{\prime} be the roots of x2−p′x+q′=0x^2-p^{\prime} x+q^{\prime}=0, then the value of (α−α′)2+(β−α′)2+(a−β′)2+(β−β′)2\left(\alpha-\alpha^{\prime}\right)^2+\left(\beta-\alpha^{\prime}\right)^2+\left(a-\beta^{\prime}\right)^2+\left(\beta-\beta^{\prime}\right)^2 is:
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\{p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\}) is correct because it is derived by expanding the given expression and substituting the sum and product of roots for both quadratic equations.
109 Medium
109 of 362
Analyzing +4 | -1
The number of real solutions of the equation 3(x2+1x2)−2(x+1x)+5=03\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0, is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the resulting quadratic equation in 't' has no real solutions within the allowed range for t = x + 1/x.
110 Medium
110 of 362
Application +4 | -1
If one root of the equation x2+px+12=0{x^2} + px + 12 = 0 is 4, while the equation x2+px+q=0{x^2} + px + q = 0 has equal roots, then the value of qq is
View answer and explanation
Correct option: D
494\frac{{49}}{4}
Option D (494\frac{49}{4}) is correct because by using the given root and the properties of quadratic equations, we can find the value of p and then use that to find q, which equals 494\frac{49}{4}.
111 Medium
111 of 362
Analyzing +4 | -1
Given, n4<10nn^4<10^n for a fixed positive integer n≥2n \geq 2, then
View answer and explanation
Correct option: A
(n+1)4<10n+1(n+1)^4 < 10^{n+1}
Option A ((n+1)4<10n+1(n+1)^4 < 10^{n+1}) is correct because it demonstrates that if n4<10nn^4 < 10^n for n≥2n \geq 2, then (n+1)4(n+1)^4 will also be less than 10n+110^{n+1}.
112 Medium
112 of 362
Analyzing +4 | -1
The solution set of x2+x+∣x∣+1<0,{x^2} + x + \left| x \right| + 1 < 0, is
View answer and explanation
Correct option: D
ϕ\phi
Option D (ϕ\phi) is correct because the inequality x2+x+∣x∣+1<0{x^2} + x + |x| + 1 < 0 has no real solutions for any value of x.
113 Medium
113 of 362
Analyzing +4 | -1
x4−4x−1=0x^4-4 x-1=0
View answer and explanation
Correct option: C
exactly two real roots
Option C (exactly two real roots) is correct because the function f(x)=x4−4x−1f(x) = x^4 - 4x - 1 has only two real roots, one positive and one negative, as determined by its derivative and the function's values.
114 Medium
114 of 362
Application +4 | -1
The value of 'cc' for which ∣α2−β2∣=7/4,\left| {{\alpha ^2} - {\beta ^2}} \right| = 7/4, where α\alpha and β\beta are the roots of 2  x2+7  x+c=0,2\;{x^2} + 7\;x + c = 0, is
View answer and explanation
Correct option: C
6
Option C (6) is the closest to the calculated value of 6, although the provided solution has an error. The correct value of c should be approximately 6.
115 Medium
115 of 362
Analyzing +4 | -1
If x2+2x+2xy+my−3=0{x^2} + 2x + 2xy + my - 3 = 0 has two rational factors, then the values of mm will be
View answer and explanation
Correct option: C
6,  −26,\; - 2
Option C (6, -2) is correct because the given equation has two rational factors only when m = 6 or m = -2.
116 Medium
116 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of x2+bx+c=0x^2+b x+c=0 and are positive, then α+β\sqrt{\alpha}+\sqrt{\beta} is
View answer and explanation
Correct option: B
−b+2c\sqrt{-b+2 \sqrt{c}}
Option B (−b+2c\sqrt{{-b+2 \sqrt{{c}}}}) is correct because it is derived by using Vieta's formulas and algebraic manipulation.
117 Medium
117 of 362
Application +4 | -1
If 3p2=5p+23{p^2} = 5p + 2 and 3q2=5q+23{q^2} = 5q + 2 where p≠qp \ne q, then the equation whose roots are 3p−2q3p - 2q and 3q−2p3q - 2p is
View answer and explanation
Correct option: A
3x2−5x−100=03{x^2} - 5x - 100 = 0
Option A (3x2−5x−100=03{x^2} - 5x - 100 = 0) is correct because it is the quadratic equation formed with roots 3p−2q3p-2q and 3q−2p3q-2p derived from the given equations.
118 Medium
118 of 362
Analyzing +4 | -1
If xx is real, the function (x−a)(x−b)(x−c)\frac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {x - c} \right)}} will assume all real values, provided
View answer and explanation
Correct option: D
a≤c≤ba \le c \le b
Option D (a≤c≤ba \le c \le b) is correct because it ensures the discriminant of the quadratic equation is non-negative for all real values of y, guaranteeing real roots for x.
119 Medium
119 of 362
Application +4 | -1
Let a,b,ca,b,c be positive real numbers and ax2+b/x2≥2a{x^2} + b/{x^2} \ge 2 for all x∈R+x \in {R^ + }. Then,
View answer and explanation
Correct option: A
4ab≥c24ab \ge {c^2}
Option A (4ab≥c24ab \ge {c^2}) is correct because it ensures that the given inequality holds for all positive real numbers x, derived from the minimum value of the expression.
120 Medium
120 of 362
Application +4 | -1
If tan⁡α,tan⁡β,tan⁡γ\tan \alpha, \tan \beta, \tan \gamma are the roots of the equation x3−px2−r=0x^3-p x^2-r=0, then the value of (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)\left(1+\tan ^2 \alpha\right)\left(1+\tan ^2 \beta\right)\left(1+\tan ^2 \gamma\right) is equal to
View answer and explanation
Correct option: B
1+(p−r)21+(p-r)^2
Option B (1+(p−r)21+(p-r)^2) is correct because by using Vieta's formulas to relate the roots to the coefficients of the polynomial, and then expanding the expression (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma), we arrive at the result 1+(p−r)21+(p-r)^2.
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