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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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151 Medium
151 of 362
Application +4 | -1
If x2−3x+2x^2-3 x+2 is one of the factors of the expression x4−px2+qx^4-p x^2+q, then
View answer and explanation
Correct option: B
p=5p=5 and q=4q=4
Option B (p=5 and q=4) is correct because when x4−px2+qx^4-px^2+q is divided by x2−3x+2x^2-3x+2, the remainder must be zero, leading to p=5 and q=4.
152 Medium
152 of 362
Analyzing +4 | -1
If a,b,c,d\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} and p\mathrm{p} are distinct non zero real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p +(b2+c2+d2)≤0+\left(b^2+c^2+d^2\right) \leq 0 then a,b,c,da, b, c, d are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the given condition leads to the ratios of consecutive terms being equal, indicating a geometric progression.
153 Medium
153 of 362
Analyzing +4 | -1
The set of all real numbers satisfying the inequation x(log⁡10x)2−3(log⁡10x)+1>1000,{x^{{{\left( {{{\log }_{10}}x} \right)}^2} - 3\left( {{{\log }_{10}}x} \right) + 1}} > 1000, is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these. After solving the inequality using logarithms, the set of real numbers satisfying the inequation is (0,1) U (1000, ∞).
154 Medium
154 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of equation ax2+bx+c=0a x^2+b x+c=0, then the sum of the roots of the equation a2x2+(b2−2ac)x+b2−4ac=0a^2 x^2+\left(b^2-2 a c\right) x+b^2-4 a c=0 is given by -
View answer and explanation
Correct option: D
−(α2+β2)-\left(\alpha^2+\beta^2\right)
Option D (−(α2+β2)-\left(\alpha^2+\beta^2\right)) is correct because it represents the sum of the roots of the transformed quadratic equation in terms of the original roots.
155 Medium
155 of 362
Application +4 | -1
The values of aa, for which the quadratic equation 3x2+2(a2+1)x+(a2−3a+2)=03 x^2+2\left(a^2+1\right) x+\left(a^2-3 a+2\right)=0 possesses roots of opposite sign, are
View answer and explanation
Correct option: A
1<a<21 < a < 2
Option A (1<a<21 < a < 2) is correct because the roots have opposite signs when the product of the roots is negative, leading to the inequality (a−1)(a−2)<0(a-1)(a-2) < 0.
156 Medium
156 of 362
Application +4 | -1
If the first, second and the last terms of an A.P. are a,b,ca, b, c respectively, then the sum is
(a+b)(a+c−2b)2(b−a)\frac{(a+b)(a+c-2 b)}{2(b-a)}
(b+c)(a+b−2c)2(b−a)\frac{(b+c)(a+b-2 c)}{2(b-a)}
View answer and explanation
Correct option: C
(b+c)(a+b−2c)2(b−a)\frac{{\left( {b + c} \right)\left( {a + b - 2c} \right)}}{{2\left( {b - a} \right)}}
Option C ((b+c)(a+b−2c)2(b−a)\frac{{\left( {b + c} \right)\left( {a + b - 2c} \right)}}{{2\left( {b - a} \right)}}) is correct because it represents the sum of an arithmetic progression with given first, second, and last terms.
157 Medium
157 of 362
Application +4 | -1
If α,β\alpha, \beta are the roots of quadratic equation x2+px+q=0x^2+p x+q=0 and γ,δ\gamma, \delta are the roots of x2+px−r=x^2+p x-r= 0 , then (α−γ)⋅(α−δ)(\alpha-\gamma) \cdot(\alpha-\delta) is equal to :
View answer and explanation
Correct option: C
−(q+r)-(q+r)
Option C (−(q+r)-(q+r)) is correct because by substituting α\alpha into the second equation and using the relationships between roots and coefficients, the expression simplifies to −(q+r)-(q+r).
158 Medium
158 of 362
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the roots of the equation x3+ax+bx^3+a x+b =0=0, then α3+β3+γ3α2+β2+γ2=?\frac{\alpha^3+\beta^3+\gamma^3}{\alpha^2+\beta^2+\gamma^2}=?
View answer and explanation
Correct option: A
3b/2a3 b / 2 a
The correct answer is A (3b/2a3 b / 2 a) because α+β+γ=0\alpha+\beta+\gamma=0 and α3+β3+γ3=3αβγ\alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma, which simplifies the expression to 3b/2a3b/2a.
159 Medium
159 of 362
Application +4 | -1
a,b,ca, b, c are integers, not all simultaneously equal and ω\omega is cube root of unity (ω≠1)(\omega \neq 1), then minimum value of ∣a+bω+cω2∣\left|a+b \omega+\mathrm{c} \omega^2\right| is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the minimum value of the expression is achieved when two of the integers are zero and the third is 1.
160 Medium
160 of 362
Application +4 | -1
If a,b,c>0a,b,c > 0 and if abc=1abc = 1, then the value of a+b+c+ab+bc+caa + b + c + ab + bc + ca lies in the interval
View answer and explanation
Correct option: D
(6,  ∞)\left( {6,\;\infty } \right)
Option D (6,  ∞)\left( {6,\;\infty } \right) is correct because by AM-GM inequality, a+b+c>3a+b+c > 3 and ab+bc+ca>3ab+bc+ca > 3, thus a+b+c+ab+bc+ca>6a+b+c+ab+bc+ca > 6.
161 Medium
161 of 362
Application +4 | -1
The value of aa for which one root of the quadratic equation (a2−5a+3)x2+(3a−1)x+2=0\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0 is twice as large as the other, is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (23\frac{2}{3}) is correct because it satisfies the condition where one root of the quadratic equation is twice the other, derived by relating the roots to the coefficients.
162 Medium
162 of 362
Application +4 | -1
If a,ba, b and cc are three positive real numbers such that a+b≥ca+b \geq c, then
View answer and explanation
Correct option: A
a1+a+b1+b≥c1+c\frac{a}{1+a}+\frac{b}{1+b} \geq \frac{c}{1+c}
The correct answer is A (a1+a+b1+b≥c1+c\frac{a}{1+a}+\frac{b}{1+b} \geq \frac{c}{1+c}) because it uses the given condition a+b≥ca+b \geq c to prove the inequality.
163 Medium
163 of 362
Application +4 | -1
The value of xx for which 12x−6<0,12−3x<012 x-6 < 0,12-3 x < 0, is
View answer and explanation
Correct option: A
ϕ\phi
Option A (ϕ\phi) is correct because the inequalities 12x−6<012x - 6 < 0 and 12−3x<012 - 3x < 0 lead to contradictory conditions on xx, resulting in no solution.
164 Medium
164 of 362
Application +4 | -1
If the roots of the quadratic equation x2−4x−log⁡3a=0{x^2} - 4x - {\log _3}a = 0 are real, then the least value of aa is
View answer and explanation
Correct option: B
181\frac{1}{{81}}
Option B (181\frac{1}{{81}}) is correct because the discriminant of the quadratic equation must be non-negative for real roots, leading to the inequality a >= 1/81.
165 Medium
165 of 362
Application +4 | -1
If tan⁡α\tan \alpha and tan⁡β\tan \beta are roots of the equation x2+px+q=0{x^2} + px + q = 0 with p≠0,p \ne 0, then
View answer and explanation
Correct option: A
sin⁡2(α+β)+psin⁡(α+β)cos⁡(α+β)+qcos⁡2(α+β)=q{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q
Option A (sin⁡2(α+β)+psin⁡(α+β)cos⁡(α+β)+qcos⁡2(α+β)=q{\sin ^2}\left( {\alpha + \beta } \right) + p\sin \left( {\alpha + \beta } \right)\cos \left( {\alpha + \beta } \right) + q{\cos ^2}\left( {\alpha + \beta } \right) = q) is correct because it is derived by substituting the sum and product of roots into the trigonometric identity.

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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
166 Medium
166 of 362
Analyzing +4 | -1
If the ratio of the equation x2+px+q=0{x^2} + px + q = 0 be equal to the ratio of the roots of x2+lx+m=0,{x^2} + lx + m = 0, then
View answer and explanation
Correct option: D
p2m=l2q{p^2}m = {l^2}q
Option C (p2m=l2q{p^2}m = {l^2}q) is correct because it is derived from equating the ratios of the roots and manipulating the equations relating roots and coefficients.
167 Medium
167 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 375x2−25x−2=0375{x^2} - 25x - 2 = 0, then lim⁡n→∞ ∑r=1nαr+lim⁡n→∞  ∑r=1nβr\lim\limits_{n \to \infty } \,\sum\limits_{r = 1}^n {{\alpha ^r} + } \lim\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^n {{\beta ^r}} is equal to
View answer and explanation
Correct option: C
112\frac{{1}}{{12}}
Option C (112\frac{{1}}{{12}}) is correct because it is the result of summing the infinite geometric series formed by the roots of the given quadratic equation.
168 Medium
168 of 362
Analyzing +4 | -1
Consider the two sets
A={m ∈ RA = \{ m\, \in \,R: both the roots of x2−(m+1)x+m+4=0{x^2} - (m + 1)x + m + 4 = 0 are real }\} and B=[−3, 5)B = \left[ { - 3,\,5} \right).
Which of the following is not true?
View answer and explanation
Correct option: A
A−B=(−∞,−3)∪(5,∞)A - B = ( - \infty , - 3) \cup (5,\infty )
Option A A−B=(−∞,−3)∪(5,∞)A - B = ( - \infty , - 3) \cup (5,\infty ) is correct because it incorrectly represents the set difference between A and B. The correct set difference is A−B=(−∞,−3)∪[5,∞)A - B = ( - \infty , - 3) \cup [5,\infty ).
169 Medium
169 of 362
Analyzing +4 | -1
If a,b,c∈Ia, b, c \in I and ax2+bx+c=0a x^2+b x+c=0 has irrational root and λ∈Q=pq\lambda \in Q=\frac{p}{q}, where f(x)=ax2+bx+cf(x)=a x^2+b x+c, then ∣f(λ)∣|f(\lambda)|
View answer and explanation
Correct option: A
≥1q2\geq \frac{1}{q^2}
Option A (≥1q2\geq \frac{1}{q^2}) is correct because the absolute value of the quadratic function evaluated at a rational number is always greater than or equal to 1q2\frac{1}{q^2}.
170 Medium
170 of 362
Application +4 | -1
If α\alpha is one root of the equation 4x2+2x−1=04 x^2+2 x-1=0, then its other root is given by
View answer and explanation
Correct option: A
4α3−3α4 \alpha^3-3 \alpha
Option A (4α3−3α4 \alpha^3-3 \alpha) is the correct answer because it represents the other root of the quadratic equation, derived using the relationship between the roots and the given equation.
171 Medium
171 of 362
Analyzing +4 | -1
For what values of K∈RK \in R the expression 2x2+Kxy+3y2−5y−22 x^2+K x y+3 y^2-5 y-2 can be expressed as (a1x+b1y+c1)⋅(a2x+b2y+c2)\left(a_1 x+b_1 y+c_1\right) \cdot\left(a_2 x+b_2 y+c_2\right) ?
View answer and explanation
Correct option: D
7,−77,-7
Option D (7,−77, -7) is correct because these values of K allow the given quadratic expression to be factored into two linear expressions.
172 Medium
172 of 362
Analyzing +4 | -1
If the equations k(6x2+3)+rx+2x2−1=0k\left( {6{x^2} + 3} \right) + rx + 2{x^2} - 1 = 0 and 6k(2x2+1)+px+4x2−2=06k\left( {2{x^2} + 1} \right) + px + 4{x^2} - 2 = 0 have both roots common, then the value of (2r−p)\left( {2r - p} \right) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because by comparing the coefficients of the two equations after making them proportional, we find that 2r - p = 0.
173 Medium
173 of 362
Application +4 | -1
Let f(x)=ax2+bx+cf(x)=a x^2+b x+c, where a,b,c∈Ra, b, c \in R and a≠0a \neq 0, is such that 2a+b+c=02 \mathrm{a}+\mathrm{b}+\mathrm{c}=0, then the quadratic equation f(x)=0f(x)=0 has :
View answer and explanation
Correct option: D
two distinct real roots
Option D (two distinct real roots) is correct because the given condition implies that either 0 and 2 are roots or one root lies in (0,2), indicating two distinct real roots.
174 Medium
174 of 362
Application +4 | -1
If p,  q,  rp,\;q,\;r are in AP and are positive, the roots of the quadratic equation px2+qx+r=0p{x^2} + qx + r = 0 are all real for
View answer and explanation
Correct option: A
∣rp−7∣≥43\left| {\frac{r}{p} - 7} \right| \ge 4\sqrt 3
Option A (∣rp−7∣≥43\left| {\frac{r}{p} - 7} \right| \ge 4\sqrt 3) is correct because it represents the condition for the roots of the quadratic equation to be real, derived from the given AP condition.
175 Medium
175 of 362
Analyzing +4 | -1
If α\alpha and β\beta are the roots of ax2+bx+c=0a{x^2} + bx + c = 0, then the equation ax2−bx(x−1)+c(x−1)2=0a{x^2} - bx\left( {x - 1} \right) + c{\left( {x - 1} \right)^2} = 0 has roots
View answer and explanation
Correct option: C
αα+1,ββ+1\frac{\alpha }{{\alpha + 1}},\frac{\beta }{{\beta + 1}}
Option C (αα+1,ββ+1\frac{\alpha }{{\alpha + 1}},\frac{\beta }{{\beta + 1}}) is correct because substituting x=y1+yx = \frac{y}{{1+y}} into the transformed equation gives the original equation with roots α and β.
176 Medium
176 of 362
Application +4 | -1
Let a,ba, b and cc be real numbers with a≠0a \neq 0 and let α,β\alpha, \beta be the roots of the equation ax2+bx+c=0a x^2+b x+c=0. Then, a3x2+abcx+c3=0a^3 x^2+a b c x+c^3=0 has roots
View answer and explanation
Correct option: A
α2β,β2α\alpha^2 \beta, \beta^2 \alpha
Option A (α2β,β2α\alpha^2 \beta, \beta^2 \alpha) is correct because these roots are derived by substituting xx with ca\frac{{c}}{{a}} times the original roots in the transformed equation.
177 Medium
177 of 362
Analyzing +4 | -1
If a,ba,b are distinct positive real numbers, then which one of the following is true?
View answer and explanation
Correct option: A
a4+b4>a3b+ab3{a^4} + {b^4} > {a^3}b + a{b^3}
Option A (a4+b4>a3b+ab3{a^4} + {b^4} > {a^3}b + a{b^3}) is correct because the expression a4+b4−a3b−ab3{a^4} + {b^4} - {a^3}b - a{b^3} can be factored into a positive quantity.
178 Medium
178 of 362
Analyzing +4 | -1
If the sum of the roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0 is equal to the sum of the squares of their reciprocals, then a/c,b/a,c/ba / c, b / a, c / b are in:
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the given condition leads to a relationship where the reciprocals are in A.P., thus the original terms are in H.P.
179 Medium
179 of 362
Analyzing +4 | -1
If cc and dd are roots of the equation (x−a)(x−b)−k=0,\left( {x - a} \right)\left( {x - b} \right) - k = 0, then a,ba,b are roots of the equation
View answer and explanation
Correct option: B
(x−c)(x−d)+k=0\left( {x - c} \right)\left( {x - d} \right) + k = 0
Option A (x−c)(x−d)−k=0\left( {x - c} \right)\left( {x - d} \right) - k = 0 is incorrect. The correct answer is B. Option B (x−c)(x−d)+k=0\left( {x - c} \right)\left( {x - d} \right) + k = 0 is correct because it is derived by substituting x=a and x=b into the equation and rearranging.
180 Medium
180 of 362
Application +4 | -1
The sum of all the real roots of the equation ∣x−2∣2+∣x−2∣−2=0{\left| {x - 2} \right|^2} + \left| {x - 2} \right| - 2 = 0 is
View answer and explanation
Correct option: D
4
The correct answer is D (4) because the sum of the real roots obtained by solving the given equation for both cases (x >= 2 and x < 2) is 4.
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