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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Showing 481-510 of 914 questions

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481 Medium
481 of 914
Analyzing +4 | -1
If x2+ax+1x^2+a x+1 is a factor of ax3+bx+ca x^3+b x+c, then
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the correct relationships between a, b, and c are b - a + a3 = 0 and c + a2 = 0, which are not present in the other options.
482 Easy
482 of 914
Application +4 | -1
The product of the real roots of the equation ∣2x+3∣2−3∣2x+3∣+2=0,{\left| {2x + 3} \right|^2} - 3\left| {2x + 3} \right| + 2 = 0, is
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B (52\frac{5}{2}) is correct because it is the product of the real roots obtained by solving the absolute value equation.
483 Medium
483 of 914
Analyzing +4 | -1
The number of triangles formed by the lines represented by x3−x2−x−2=0x^3-x^2-x-2=0 and xy2+2xy+4x−2y2−4y−8=0x y^2+2 x y+4 x-2 y^2-4 y-8=0 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given equations represent the same line, so no triangle can be formed.
484 Easy
484 of 914
Application +4 | -1
The real root of the equation x3−6x+9=0{x^3} - 6x + 9 = 0 is
View answer and explanation
Correct option: D
−3- 3
Option D (−3- 3) is correct because it is the only real root of the given cubic equation, found by factoring.
485 Easy
485 of 914
Application +4 | -1
The least integer satisfying 49.4−(27−x10)<47.4−(27−9x10)49.4-\left(\frac{27-x}{10}\right) < 47.4-\left(\frac{27-9 x}{10}\right) is
View answer and explanation
Correct option: B
3
Option B (3) is correct because it is the smallest integer greater than the solution to the inequality.
486 Easy
486 of 914
Analyzing +4 | -1
If a,ba,b are odd integers, then the roots of the equation 2  ax2+(2  a+b)x+b=0,a≠02\;a{x^2} + \left( {2\;a + b} \right)x + b = 0,a \ne 0 are
View answer and explanation
Correct option: A
Rational
Option A is correct because the roots of the given quadratic equation are rational, as they can be expressed as -1 and -b/2a.
487 Easy
487 of 914
Analyzing +4 | -1
The number of real solutions of 1+∣ex−1∣=ex(ex−2),1 + \left| {{e^x} - 1} \right| = {e^x}\left( {{e^x} - 2} \right), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation simplifies to a single valid solution for x.
488 Easy
488 of 914
Application +4 | -1
If α,β\alpha ,\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0, then the value of 1aα+b+1aβ+b\frac{1}{{a\alpha + b}} + \frac{1}{{a\beta + b}} is equal to
View answer and explanation
Correct option: D
bac\frac{b}{{ac}}
The correct answer is D (bac\frac{{b}}{{ac}}) because by using Vieta's formulas and simplifying the expression, we arrive at this result.
489 Easy
489 of 914
Application +4 | -1
The minimum value of ∣x−3∣+∣x−2∣+∣x−5∣|x-3|+|x-2|+|x-5| is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the minimum value of the sum of absolute differences occurs at the median, which is 3, and the minimum value is |3-3|+|3-2|+|3-5| = 0+1+2 = 3.
490 Medium
490 of 914
Application +4 | -1
The number of real solutions of the equation (5+26)x2−3+(5−26)x2−3=10,{\left( {5 + 2\sqrt 6 } \right)^{{x^2} - 3}} + {\left( {5 - 2\sqrt 6 } \right)^{{x^2} - 3}} = 10, is
View answer and explanation
Correct option: B
4
Option B (4) is correct because solving the given equation by recognizing the reciprocal relationship leads to four real solutions for x.
491 Medium
491 of 914
Analyzing +4 | -1
If sec⁡α\sec \alpha and tan⁡α\tan \alpha are the roots of ax2+bx+c=0,a{x^2} + bx + c = 0, then
View answer and explanation
Correct option: C
a4+4ab2c=b4{a^4} + 4a{b^2}c = {b^4}
Option D (a4+4ab2c=b4{a^4} + 4a{b^2}c = {b^4}) is correct because it is derived from the relationship between the roots (sec⁡α\sec \alpha and tan⁡α\tan \alpha) and coefficients of the quadratic equation, along with the trigonometric identity sec⁡2α−tan⁡2α=1{\sec ^2}\alpha - {\tan ^2}\alpha = 1.
492 Easy
492 of 914
Analyzing +4 | -1
If xx is real, then the maximum and minimum values of the expression x2−3x+4x2+3x+4\frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}} will be
View answer and explanation
Correct option: C
7,177,\frac{1}{7}
Option C (7, 1/7) is correct because the maximum and minimum values are found by setting the discriminant greater than or equal to zero and solving for y.
493 Easy
493 of 914
Application +4 | -1
The equation x2−22kx+2e2log⁡k−1=0{x^2} - 2\sqrt 2 kx + 2{e^{2\log k}} - 1 = 0 has the product of roots equal to 31, then for what value of kk it has real roots?
View answer and explanation
Correct option: D
4
Option D (4) is correct because it is the only value of kk that satisfies both the product of roots condition and the real roots condition.
494 Easy
494 of 914
Application +4 | -1
Let for a,a1≠0,a≠a1a, a_1 \neq 0, a \neq a_1 f(x)=ax2+bx+c,g(x)=a1x2+b1x+c1f(x)=a x^2+b x+\mathrm{c}, g(x)=a_1 x^2+b_1 x+c_1 and p(x)=f(x)−g(x)p(x)=f(x)-g(x). If p(x)=0p(x)=0 only for x=−1x=-1 and p(−2)=2p(-2)=2, then the value of p(2)p(2) is
View answer and explanation
Correct option: D
18
Option D (18) is correct because the given conditions allow us to express p(x) as k(x+1)2, solve for k using p(-2)=2, and then evaluate p(2).
495 Medium
495 of 914
Application +4 | -1
If 3x/2+2x>25,{3^{x/2}} + {2^x} > 25, then
View answer and explanation
Correct option: B
(4,∞)\left( {4,\infty } \right)
The correct answer is B ((4,∞)\left( {4,\infty } \right)) because by substituting x=4 into the inequality, it becomes an equality, and for x > 4, the inequality holds.

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496 Easy
496 of 914
Application +4 | -1
If x=7−43x=\sqrt{7-4 \sqrt{3}}, then x+1xx+\frac{1}{x} is equal to:
View answer and explanation
Correct option: C
4
Option C (4) is correct because simplifying the square root and then adding x and 1/x results in the cancellation of the square root term, leaving 4.
497 Easy
497 of 914
Application +4 | -1
The set of values of xx satisfying 2≤∣x−3∣<42 \le \left| {x - 3} \right| < 4 is
View answer and explanation
Correct option: A
(−1,  1]∪[5,  7)\left( { - 1,\;1} \right] \cup \left[ {5,\;7} \right)
Option A (−1,  1]∪[5,  7)\left( { - 1,\;1} \right] \cup \left[ {5,\;7} \right) is correct because it represents the solution set of the compound inequality involving absolute value.
498 Easy
498 of 914
Application +4 | -1
If the equation ax2+bx+6=0a x^{2}+b x+6=0 does not have two distinct real roots, then the least value of 3a+b3 a+b is
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because the equation not having two distinct real roots implies that the discriminant is non-negative, leading to the minimum value of 3a+b3a+b.
499 Easy
499 of 914
Application +4 | -1
If two equations
a1  x2+b1  x+c1=0{a_1}\;{x^2} + {b_1}\;x + {c_1} = 0 and a2x2+b2x+c2=0{a_2}{x^2} + {b_2}x + {c_2} = 0 have a common root, then the value of (a1b2−a2b1)(b1c2−b2c1),\left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{b_1}{c_2} - {b_2}{c_1}} \right), is
View answer and explanation
Correct option: D
(a1c2−c1a2)2{\left( {{a_1}{c_2} - {c_1}{a_2}} \right)^2}
Option D ((a1c2−c1a2)2{\left( {{a_1}{c_2} - {c_1}{a_2}} \right)^2}) is correct because the condition for a common root leads to this relationship between the coefficients.
500 Easy
500 of 914
Analyzing +4 | -1
The solution set of the in equation ∣x+1x∣<4,\left| {x + \frac{1}{x}} \right| < 4, is
View answer and explanation
Correct option: A
(2−3,  2+3)∪(−2−3,  −2+3)\left( {2 - \sqrt 3 ,\;2 + \sqrt 3 } \right) \cup \left( { - 2 - \sqrt 3 ,\; - 2 + \sqrt 3 } \right)
Option A ((2−3,  2+3)∪(−2−3,  −2+3)\left( {2 - \sqrt 3 ,\;2 + \sqrt 3 } \right) \cup \left( { - 2 - \sqrt 3 ,\; - 2 + \sqrt 3 } \right)) is correct because it represents the solution set obtained by solving the inequality ∣x+1x∣<4\left| {x + \frac{1}{x}} \right| < 4 by considering both positive and negative cases and finding the intersection of the solution sets.
501 Medium
501 of 914
Analyzing +4 | -1
For any real value of θ≠π\theta \neq \pi, the value of the expression y=cos⁡2θ−1cos⁡2θ+cos⁡θy=\frac{\cos ^2 \theta-1}{\cos ^2 \theta+\cos \theta} is
View answer and explanation
Correct option: B
y<0y < 0 and y>2y > 2
Option B (y<0y < 0 and y>2y > 2) is correct because it represents the range of y values derived from the given expression after considering the possible values of cos⁡θ\cos \theta.
502 Easy
502 of 914
Application +4 | -1
If α\alpha is a root of x4+x2−1=0x^4+x^2-1=0, the value of (α6+2α4)2012\left(\alpha^6+2 \alpha^4\right)^{2012} is
View answer and explanation
Correct option: C
1
Option C (1) is correct. By using the given equation and substituting, we can simplify the expression to 1.
503 Medium
503 of 914
Analyzing +4 | -1
Let α\alpha and β\beta are roots of the equation x2−6x+12=0x^2-6 x+12=0. The value of the expression (α−2)12+(β−6)12α12−1(\alpha-2)^{12}+\frac{(\beta-6)^{12}}{\alpha^{12}}-1 is:
View answer and explanation
Correct option: C
2122^{12}
Option C (2122^{12}) is correct because the expression simplifies to 848^4 after substituting the roots of the quadratic equation and using algebraic manipulations.
504 Easy
504 of 914
Application +4 | -1
The set of values of pp for which the roots of the equation 3  x2+2  x+p(p−1)=03\;{x^2} + 2\;x + p\left( {p - 1} \right) = 0 are of opposite signs is
View answer and explanation
Correct option: B
(0,  1)\left( {0,\;1} \right)
Option B ((0, 1)) is correct because for roots to have opposite signs, the product of roots must be negative, leading to the inequality p(p−1)3<0\frac{{p(p - 1)}}{3} < 0 which is satisfied when p∈(0,1)p \in (0, 1).
505 Easy
505 of 914
Analyzing +4 | -1
The number of solution of log⁡4(x−1)=log⁡2(x−3){\log _4}\left( {x - 1} \right) = {\log _2}\left( {x - 3} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because after solving the logarithmic equation and checking for extraneous solutions, only one valid solution remains.
506 Easy
506 of 914
Application +4 | -1
The number of real roots of 32x2−7x+7=9{3^{2{x^2} - 7x + 7}} = 9 is
View answer and explanation
Correct option: B
2
Option B (2) is correct because simplifying the exponential equation leads to a quadratic equation with a positive discriminant, indicating two real roots.
507 Medium
507 of 914
Application +4 | -1
If α,β\alpha, \beta are roots of the equation x2+px−q=0x^2+p x-q=0 and γ,δ\gamma, \delta are roots of x2+px+r=0,x^2+p x+r=0, then the value of (α−γ)(α−δ)(\alpha-\gamma)(\alpha-\delta) is:
View answer and explanation
Correct option: D
q+rq+r
Option D (q+rq+r) is correct because substituting the root α\alpha into the quadratic equation and simplifying leads to this result.
508 Easy
508 of 914
Application +4 | -1
If x=2+32−3x = \sqrt {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }}} then x2(x−4)2{x^2}{\left( {x - 4} \right)^2} is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because simplifying the expression for x and then substituting into the given expression results in a value of 1.
509 Medium
509 of 914
Application +4 | -1
If tan⁡θi;i=1,2,3,4\tan \theta_i ; i=1,2,3,4 are the roots of equation x4−x3sin⁡2β+x2cos⁡2β−xcos⁡β−sin⁡β=0x^4-x^3 \sin 2 \beta+x^2 \cos 2 \beta-x \cos \beta-\sin \beta=0, then tan⁡(θ1+θ2+θ3+θ4)=\tan \left(\theta_1+\theta_2+\theta_3+\theta_4\right)=
View answer and explanation
Correct option: D
cot⁡β\cot \beta
Option D (cot⁡β\cot \beta) is correct because the tangent of the sum of angles can be found using Vieta's formulas and trigonometric identities.
510 Easy
510 of 914
Application +4 | -1
If the equations x2+ax+b=0{x^2} + ax + b = 0 and x2+bx+a=0{x^2} + bx + a = 0 have a common root, then the numerical value of a+ba + b is
View answer and explanation
Correct option: C
−1- 1
The correct answer is C (-1) because substituting the common root into both equations and solving leads to a + b = -1.
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