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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 241-270 of 362 questions

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241 Medium
241 of 362
Application +4 | -1
Given that tan⁡A\tan A and tan⁡B\tan B are the roots of x2−px+q=0{x^2} - px + q = 0, then the value of sin⁡2(A+B){\sin ^2}\left( {A + B} \right) is
View answer and explanation
Correct option: A
p2p2(1−q)2\frac{{{p^2}}}{{{p^2}{{\left( {1 - q} \right)}^2}}}
Option D (p2p2+(1−q)2)\frac{{{p^2}}}{{{p^2} + {{\left( {1 - q} \right)}^2}}}) is correct because using the relationship between the roots and coefficients of the quadratic equation, we can find tan⁡(A+B)\tan(A+B) and then use trigonometric identities to find sin⁡2(A+B)\sin^2(A+B).
242 Medium
242 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 then, (1+α+α2)  (1+β+β2)\left( {1 + \alpha + {\alpha ^2}} \right)\;\left( {1 + \beta + {\beta ^2}} \right) is equal to
View answer and explanation
Correct option: B
Positive
The correct answer is A (x=4,  9x = 4,\;9) because these are the only values that satisfy the given absolute value equation.
243 Medium
243 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a x^2+b x+c=0, then the equation whose roots are 1α+β,1α+1β\frac{1}{\alpha+\beta}, \frac{1}{\alpha}+\frac{1}{\beta} is
View answer and explanation
Correct option: B
bcx2+(b2+ac)x+ab=0b c x^2+\left(b^2+a c\right) x+a b=0
Option B (bcx2+(b2+ac)x+ab=0b c x^2+\left(b^2+a c\right) x+a b=0) is correct because it is derived from the sum and product of the new roots formed from the original roots α and β.
244 Medium
244 of 362
Application +4 | -1
The value of aa for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0 assumes the least value, is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because the minimum value of the sum of the squares of the roots occurs when a = 1.
245 Medium
245 of 362
Application +4 | -1
If ai>0a_i > 0 for i=1,2,…,ni=1,2, \ldots, n and a1a2…an=1a_1 a_2 \ldots a_n=1, then minimum value of (1+a1)(1+a2)…(1+an)\left(1+a_1\right)\left(1+a_2\right) \ldots\left(1+a_n\right) is
View answer and explanation
Correct option: B
2n2^n
The correct answer is B (2n2^n) because it is found by applying the AM-GM inequality to each (1+ai)(1+a_i) term and then multiplying the resulting inequalities.
246 Medium
246 of 362
Application +4 | -1
If aa and bb are the non-zero distinct roots of x2+ax+b=0,{x^2} + ax + b = 0, then the least value of x2+ax+b{x^2} + ax + b is
View answer and explanation
Correct option: C
−94- \frac{9}{4}
Option C (-9/4) is correct because it is the minimum value of the quadratic x2+ax+b{x^2} + ax + b given the conditions on a and b.
247 Medium
247 of 362
Application +4 | -1
Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3)(4,3). Rahul made a mistake in writing down coefficient of xx to get roots (3,2)(3,2). The correct roots of equation are
View answer and explanation
Correct option: B
6,1
Option B (6,1) is correct because Sachin's mistake affects the product of roots, and Rahul's mistake affects the sum of roots, allowing us to reconstruct the correct quadratic equation.
248 Medium
248 of 362
Application +4 | -1
If   sec⁡  α    and    cosec  α  \mathrm{\;}\sec \,\,\mathrm{\alpha \;}\,\,\mathrm{and\;}\,\,\mathrm{cosec}\,\,\mathrm{\alpha \;} are the root of the equation x2−px+q=0{x^2} - px + q = 0,then
View answer and explanation
Correct option: B
p2=q(q+2){p^2} = q\left( {q + 2} \right)
Option A (1\mathrm{1}) is correct because by substituting sin⁡x=cos⁡2x\sin x = {\cos ^2}x into the expression and simplifying, we obtain 1.
249 Medium
249 of 362
Analyzing +4 | -1
If x1,x2,x3,…,xnx_1, x_2, x_3, \ldots, x_n are the roots of xn+ax+b=0x^n+a x+b=0, then the value of (x1−x2)(x1−x3)(x1−x4)…(x1−xn)\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right) is
View answer and explanation
Correct option: C
n(x1)n−1+an\left(x_1\right)^{n-1}+a
Option C (n(x1)n−1+an\left(x_1\right)^{n-1}+a) is correct because it is the result of differentiating the polynomial and evaluating it at x1x_1.
250 Medium
250 of 362
Analyzing +4 | -1
If cos⁡4θ+α,sin⁡4θ+α\cos ^4 \theta+\alpha, \sin ^4 \theta+\alpha are the roots of the equation x2+2bx+b=0x^2+2 b x+b=0 and cos⁡2θ+β,sin⁡2θ+β\cos ^2 \theta+\beta, \sin ^2 \theta+\beta are the roots of x2+4x+2=0x^2+4 x+2=0 the value of b{b} is
View answer and explanation
Correct option: C
−1-1
Option C (-1) is correct because the difference of roots of the given equations is the same, leading to a quadratic equation in 'b' which gives b = -1 as a valid solution.
251 Medium
251 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation (5+2)x2−(5+\sqrt{2}) x^2- (4+5)x+8+25=0(4+\sqrt{5}) x+8+2 \sqrt{5}=0, then 2(1/α)+(1/β)\frac{2}{(1 / \alpha)+(1 / \beta)} is equal to
View answer and explanation
Correct option: B
4
Option B (4) is correct because it is calculated by using Vieta's formulas to find the sum and product of the roots and then simplifying the given expression.
252 Medium
252 of 362
Analyzing +4 | -1
If the roots of a1x2+b1x+c1=0a_1 x^2+b_1 x+c_1=0 are α1,β1\alpha_1, \beta_1 and that of a2x2+b2x+c2=0a_2 x^2+b_2 x+c_2=0 are α2\alpha_2, β2\beta_2 such that α1α2=β1β2=1\alpha_1 \alpha_2=\beta_1 \beta_2=1, then
View answer and explanation
Correct option: B
a1c2=b1b2=c1a2\frac{a_1}{c_2}=\frac{b_1}{b_2}=\frac{c_1}{a_2}
Option B is correct because the roots of the second equation are reciprocals of the roots of the first equation.
253 Medium
253 of 362
Analyzing +4 | -1
If x2+6x−27>0x^2+6 x-27 > 0 and x2−3x−4<0x^2-3 x-4 < 0, then
View answer and explanation
Correct option: C
3<x<43 < x < 4
Option C (3<x<43 < x < 4) is correct because it represents the intersection of the solution sets for the two given inequalities.
254 Medium
254 of 362
Analyzing +4 | -1
The number of solutions of the equation (1+ex2)1+x21+x4−x2=1+cos⁡x,\frac{{\left( {1 + {e^{{x^2}}}} \right)\sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^4}} - {x^2}}} = 1 + \cos x, is
View answer and explanation
Correct option: A
1
The correct answer is A. By analyzing the LHS and RHS of the equation, we find that the only possible solution is x=0.
255 Medium
255 of 362
Application +4 | -1
If the least value of the expression y=x2−ax+ay=x^2-a x+a is 1; then the value of aa is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the least value of the quadratic expression is found by completing the square or using the vertex formula, leading to a = 2.
256 Medium
256 of 362
Application +4 | -1
If a,ba, b and cc are the sides of a triangle, then ab+c−a+bc+a−b+ca+b−c\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}
View answer and explanation
Correct option: B
≥3\geq 3
The correct answer is B (≥3\geq 3) because it is proven by substituting x=b+c−a,y=c+a−b,z=a+b−cx=b+c-a, y=c+a-b, z=a+b-c and applying the AM-GM inequality.
257 Medium
257 of 362
Application +4 | -1
The number of solutions of the equation ∣x∣2−2∣x∣−8=0|x|^2-2|x|-8=0 which belong to the domain of the function f(x)=5−2xf(x)=\sqrt{5-2 x} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution of the quadratic equation satisfies the domain restriction of the square root function.
258 Medium
258 of 362
Application +4 | -1
If one root of the equation x2+px+q=0x^2+p x+q=0 is the square of the other, then:
View answer and explanation
Correct option: D
p3+q2+q(1−3p)=0p^3+q^2+q(1-3 p)=0
Option D (p3+q2+q(1−3p)=0p^3+q^2+q(1-3 p)=0) is correct because by expressing the relationship between the roots and coefficients and manipulating the equations, we can derive the given relation.
259 Medium
259 of 362
Application +4 | -1
If the roots of the equation x2+px+q=0{x^2} + px + q = 0 are tan⁡300\tan {30^0} and tan⁡150\tan {15^0} respectively, then the value of 2+q−p2 + q - p is
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the sum and product of roots, and the tangent addition formula, we can find the value of the expression 2 + q - p.
260 Medium
260 of 362
Analyzing +4 | -1
If 0<c<a<b<d0< c < a < b < d, then roots of the equation bx2+{1−b(c+d)}x+bcd−a=0b x^2+\{1-b(c+d)\} x+b c d-a=0 are
View answer and explanation
Correct option: B
real and distinct in which one lie between cc and dd
Option B (real and distinct in which one lie between cc and dd) is correct because f(c)f(c) is negative and f(d)f(d) is positive, indicating a root between cc and dd.
261 Medium
261 of 362
Analyzing +4 | -1
If a(p+q)2+2b  pq+c=0a{\left( {p + q} \right)^2} + 2b\;pq + c = 0 and a(p+r)2+2b  pr+c=0,a{\left( {p + r} \right)^2} + 2b\;pr + c = 0, then qr=qr =
View answer and explanation
Correct option: A
p2+ca{p^2} + \frac{c}{a}
Option A (p2+ca{p^2} + \frac{c}{a}) is correct because it represents the product of the roots qq and rr of the quadratic equation derived from the given conditions.
262 Medium
262 of 362
Application +4 | -1
The value of 'a' for which one root of quadratic equation (a2−5a+3)x2+(3a−1)x+2=0\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0 is twice as large as other is
View answer and explanation
Correct option: A
2/32 / 3
Option A (2/3) is correct because it satisfies the condition derived from the relationship between the roots of the quadratic equation.
263 Medium
263 of 362
Understanding +4 | -1
If 0<a<b<c0 < a < b < c and the roots α\alpha and β\beta of the equation ax2+bx+c=0a x^2+b x+c=0 are imaginary, then
View answer and explanation
Correct option: A
∣α∣=∣β∣|\alpha|=|\beta|
Option A (∣α∣=∣β∣\left|\alpha\right|=\left|\beta\right|) is correct because when the roots of a quadratic equation are imaginary, they are complex conjugates, and complex conjugates have the same magnitude.
264 Medium
264 of 362
Application +4 | -1
If α,β\alpha, \beta are roots of x2−2x−1=0x^2-2 x-1=0, then value of 5α4+12β35 \alpha^4+12 \beta^3 is
View answer and explanation
Correct option: B
169
Option B (169) is correct because substituting the roots into the given equation and simplifying the expression 5α4+12β35 \alpha^4 + 12 \beta^3 yields 169.
265 Medium
265 of 362
Application +4 | -1
The real 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 signs lie in
View answer and explanation
Correct option: C
(1,2)(1,2)
Option C ((1,2)) is correct because the product of roots must be negative, leading to the interval (1,2).
266 Medium
266 of 362
Application +4 | -1
If one root is square of the other root of the equation x2+px+q=0x^2+p x+q=0, then the relation between pp and qq is
View answer and explanation
Correct option: A
p3−(3p−1)q+q2=0p^3-(3 p-1) q+q^2=0
Option A (p3−(3p−1)q+q2=0p^3-(3 p-1) 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 of the given quadratic equation.
267 Medium
267 of 362
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 equation whose roots are 2α+3β2\alpha + 3\beta and 3α+2β3\alpha + 2\beta, is
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the derived equation does not match any of the provided options.
268 Medium
268 of 362
Application +4 | -1
If both the roots of the equation x2+ax+b+1=0x^2+a x+b+1=0 are positive integers, then a possible value of a2+b2a^2+b^2 is
View answer and explanation
Correct option: A
50
Option A (50) is correct because it is a possible value of a2+b2a^2 + b^2 when both roots of the equation are positive integers.
269 Medium
269 of 362
Application +4 | -1
If the sum of the greatest integer less than or equal to xx and the least integer greater than or equal to xx is 5, then the solution set for xx is
View answer and explanation
Correct option: A
(2,3)\left( {2,3} \right)
Option A ((2,3)\left( {2,3} \right)) is correct because the sum of the greatest integer less than or equal to xx and the least integer greater than or equal to xx is 5 only when 2<x<32 < x < 3.
270 Medium
270 of 362
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.
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