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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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751 Medium
751 of 914
Application +4 | -1
If α,β\alpha ,\beta are roots of the equation ax2+bx+c=0,a{x^2} + bx + c = 0, then the equation whose roots are 2  α+3  β2\;\alpha + 3\;\beta and 3  α+2  β3\;\alpha + 2\;\beta is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derived equation doesn't match any of the provided options.
752 Medium
752 of 914
Application +4 | -1
The number of integral values of kk, for which one root of the equation 2x2−8x+k=02 x^{2}-8 x+k=0 lies in the interval (1,2)(1,2) and its other root lies in the interval (2,3)(2,3) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one integer value of k satisfies the given conditions based on the roots lying in specified intervals.
753 Medium
753 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px−12p2=0x^2+p x-\frac{1}{2 p^2}=0, where p∈Rp \in R, then the minimum value of α4+β4\alpha^4+\beta^4 is
View answer and explanation
Correct option: B
2+22+\sqrt{2}
Option B (2+22+\sqrt{2}) is correct because by completing the square, the minimum value of the expression α4+β4\alpha^4 + \beta^4 is found to be 2+22+\sqrt{2}.
754 Easy
754 of 914
Application +4 | -1
Let a,ba, b and cc be real numbers such that 4a+2b+c=04 a+2 b+c=0 and ab>0a b>0. Then the equation ax2+bx+c=0a x^2+b x+c=0 has
View answer and explanation
Correct option: A
real roots
Option A (real roots) is correct because the discriminant D is greater than 0, indicating real roots.
755 Easy
755 of 914
Analyzing +4 | -1
The set of values of xx which satisfy the inequations 5x+2<3x+85 x+2<3 x+8 and x+2x−1<4\frac{x+2}{x-1} < 4 is
View answer and explanation
Correct option: D
(−∞,1)∪(2,3)(-\infty, 1) \cup(2,3)
Option D ((−∞,1)∪(2,3))((-\infty, 1) \cup(2,3)) is correct because it represents the intersection of the solution sets for both inequalities.
756 Medium
756 of 914
Application +4 | -1
If the sum of the roots of the equation ax2+2x+3a=0a{x^2} + 2x + 3a = 0 is equal to their products, then the value of aa is
View answer and explanation
Correct option: A
−23- \frac{2}{3}
Option A (−23\frac{-2}{3}) is correct because by setting the sum of the roots equal to the product of the roots, we can solve for the value of 'a'.
757 Easy
757 of 914
Application +4 | -1
The value of kk for which the number 3 lies between the roots of the equation x2+(1−2k)x+(k2−k−2)x^2+(1-2 k) x+\left(k^2-k-2\right) =0=0 is given by
View answer and explanation
Correct option: A
2<k<52 < k < 5
Option A (2 < k < 5) is correct because it satisfies the condition f(3) < 0, which ensures that 3 lies between the roots of the equation.
758 Easy
758 of 914
Application +4 | -1
If a,b,ca, b, c are non-zero rational numbers such that a+b+c=0a+b+c=0, then the roots of the equation (b+c−a)x2+(c+a−b)x+(a+b−c)=0(b+c-a) x^2+(c+a-b) x+(a+b-c)=0 are
View answer and explanation
Correct option: A
both rational
Option A (both rational) is correct because substituting a+b+c=0 into the equation simplifies it, revealing that x=1 is a root, and the other root is also rational since a, b, and c are rational.
759 Easy
759 of 914
Analyzing +4 | -1
If 3x2−7x−30+2x2−7x−5=x+5\sqrt {3{x^2} - 7x - 30} + \sqrt {2{x^2} - 7x - 5} = x + 5, then xx is equal to
View answer and explanation
Correct option: C
6
Option C (6) is the correct answer because it satisfies the given equation after squaring and simplifying the terms.
760 Medium
760 of 914
Application +4 | -1
Two non-integer roots of

(3x−12x+3)4−5(3x−12x+3)2+4=0\left(\frac{3 x-1}{2 x+3}\right)^4-5\left(\frac{3 x-1}{2 x+3}\right)^2+4=0 (1)

are
View answer and explanation
Correct option: A
−5/7,−2/5-5 / 7,-2 / 5
Option A (-5/7, -2/5) is correct because these are the two non-integer roots obtained by solving the equation after reducing it to a quadratic form.
761 Easy
761 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 2x2−35x+2=02 x^2-35 x+2= 0 then the value of (2a−35)3(2β−35)3(2 a-35)^3(2 \beta-35)^3 is equal to
View answer and explanation
Correct option: C
64
Option C (64) is correct because by using the relationship between roots and coefficients, the expression simplifies to a constant value.
762 Medium
762 of 914
Analyzing +4 | -1
If x−1=a−x2x-1=\sqrt{a-x^2} has real solution for xx then
View answer and explanation
Correct option: B
a≥1a \geq 1
Option B (a≥1a \geq 1) is correct because it ensures that the equation has a real solution for x, considering the non-negativity of the square root and the condition x >= 1.
763 Easy
763 of 914
Application +4 | -1
The values of 'aa' for which (a2−1)x2+2(a−1)x+2\left( {{a^2} - 1} \right){x^2} + 2\left( {a - 1} \right)x + 2 is positive for any xx, are
View answer and explanation
Correct option: D
a<−3  or  a>1a < - 3\,\,\mathrm{or}\,\,a > 1
Option D (a<−3  or  a>1a < - 3\,\,\mathrm{or}\,\,a > 1) is correct because the expression is positive for all xx if a2−1>0a^2 - 1 > 0 and the discriminant is negative, leading to the solution a<−3a < -3 or a>1a > 1.
764 Easy
764 of 914
Application +4 | -1
Let for a≠a1≠0,f(x)=ax2+bx+ca \neq a_1 \neq 0, f(x)=a x^2+b x+c, g(x)=a1x2+b1x+c1g(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=(−1)x=(-1) and p(−2)=2p(-2)=2, the value of p(2)p(2) is
View answer and explanation
Correct option: A
18
Option A (18) is correct because by using the given information to determine the quadratic p(x)p(x), we can then evaluate p(2)p(2) to find the answer.
765 Medium
765 of 914
Analyzing +4 | -1
If the sum of the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 is equal to the sum of the squares of their reciprocals of their reciprocals, then
View answer and explanation
Correct option: A
c2b,a2c,b2a{c^2}b,{a^2}c,{b^2}a are in A.P.
Option A (c2b,a2c,b2a{c^2}b,{a^2}c,{b^2}a are in A.P.) is correct because the given condition leads to the relationship 2a2c=ab2+bc22a^2c = ab^2 + bc^2, which is the condition for an arithmetic progression.

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766 Easy
766 of 914
Analyzing +4 | -1
If the roots of ax2+bx+c=0  are  α,  βa{x^2} + bx + c = 0\;\mathrm{are}\;\alpha ,\;\beta and the roots of Ax2+Bx+C=0  are  α−k,  β−k,  then  B2−4ACb2−4acA{x^2} + Bx + C = 0\;\mathrm{are}\;\alpha - k,\;\beta - k,\;\mathrm{then}\;\frac{{{B^2} - 4AC}}{{{b^2} - 4ac}} is equal to
View answer and explanation
Correct option: C
(Aa)2{\left( {\frac{A}{a}} \right)^2}
Option C ((Aa)2{\left( {\frac{A}{a}} \right)^2}) is correct because the discriminant ratio is the square of the ratio of the leading coefficients, derived from the relationship between roots and coefficients.
767 Easy
767 of 914
Analyzing +4 | -1
Let a,  b∈Ra,\,\,b \in R be such that the equation ax2−2bx+15=0a{x^2} - 2bx + 15 = 0 has a repeated root α.\alpha . If α   and   β\alpha {\text{ }}\,\,{\text{and}}\,\,{\text{ }}\beta are the roots of the equation x2−2bx+21=0,{x^2} - 2bx + 21 = 0, then α2+β2{\alpha ^2} + {\beta ^2} is equal to
View answer and explanation
Correct option: B
58
Option B (58) is correct because by solving the repeated root equation and substituting into the second equation, we find the value of α² + β².
768 Medium
768 of 914
Analyzing +4 | -1
If x2+(a−b)x+(1−a−b)=0x^2+(a-b) x+(1-a-b)=0, where a,b∈Ra, b \in R, then the values of aa for which equation has unequal roots for all values of bb, is given by
View answer and explanation
Correct option: B
a>1a > 1
Option B (a>1a > 1) is correct because the discriminant of the quadratic in bb must be negative to ensure unequal roots for all bb.
769 Medium
769 of 914
Analyzing +4 | -1
If b2≥4ac{b^2} \ge 4ac for the equation ax4+bx2+c=0a{x^4} + b{x^2} + c = 0, then all the roots of the equation will be real positive of
View answer and explanation
Correct option: B
b<0,a>0,c>0b < 0,a > 0,c > 0
Option B (b<0,a>0,c>0b < 0, a > 0, c > 0) is correct because it ensures the quadratic in x2x^2 has positive roots, which then lead to real roots for x.
770 Medium
770 of 914
Analyzing +4 | -1
The number of solution(s) of the inequation 3x2+6x+7+5x2+10x+14≤4−2x−x2,\sqrt {3{x^2} + 6x + 7} + \sqrt {5{x^2} + 10x + 14} \le 4 - 2x - {x^2}, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the inequation holds only when both sides are equal to 5, which occurs at x = -1.
771 Medium
771 of 914
Application +4 | -1
If set S={(3+2)x2−4+(3−2)x2−4=10}S=\left\{(\sqrt{3}+\sqrt{2})^{x^2-4}+(\sqrt{3}-\sqrt{2})^{x^2-4}=10\right\} then n(S)n(S) equals:
View answer and explanation
Correct option: C
4
Option C (4) is correct because solving the equation leads to four distinct solutions for x.
772 Medium
772 of 914
Analyzing +4 | -1
If f(x)=(ab−b2−2)x+∫0x(cos⁡4θ+sin⁡4θ)dθf\left( x \right) = \left( {ab - {b^2} - 2} \right)x + \int \nolimits_0^x \left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)d\theta is decreasing function of xx for all x∈Rx \in R and b∈R,  bb \in R,\;b being independent of xx, then
View answer and explanation
Correct option: B
a∈(−6,  6)a \in \left( { - \sqrt 6 ,\;\sqrt 6 } \right)
Option B (14\frac{1}{4}) is correct because it represents the point where the function attains its maximum value within the given interval.
773 Easy
773 of 914
Application +4 | -1
If (7+43)x2−2+(7−43)x2−2=14(7+4 \sqrt{3})^{x^2-2}+(7-4 \sqrt{3})^{x^2-2}=14, then the positive integral value of xx is
View answer and explanation
Correct option: A
1
Option A (1) is correct because solving the equation with exponents leads to x = 1 as the positive integral solution.
774 Medium
774 of 914
Application +4 | -1
If x\mathrm{x} is a solution of the equation,
2x+1−2x−1=1,(x≥12)\sqrt{2 x+1}-\sqrt{2 x-1}=1,\left(x \geq \frac{1}{2}\right), then 4x2−1\sqrt{4 x^{2}-1} is equal to :
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{3}{4}) is correct because solving the given equation for x and substituting it into 4x2−1\sqrt{4 x^{2}-1} yields 34\frac{3}{4}.
775 Easy
775 of 914
Application +4 | -1
If x2+2ax+10−3a>0x^2+2 a x+10-3 a > 0 for each x∈Rx \in \mathbf{R}, then
View answer and explanation
Correct option: B
−5<a<2-5 < a < 2
Option B (-5 < a < 2) is correct because for the quadratic to be always positive, the discriminant must be negative, leading to the given range for 'a'.
776 Medium
776 of 914
Application +4 | -1
If a,  b,  c>0a,\;b,\;c > 0, the minimum value of ab+c+bc+a+ca+b\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} is
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B 32\frac{3}{2} is correct because the minimum value is obtained using AM-GM inequality.
777 Medium
777 of 914
Analyzing +4 | -1
If α\alpha and β(α<β)\beta(\alpha<\beta), are the roots of the equation x2+bx+c=0x^2+b x+c=0, where c<0<bc < 0 < b, then
View answer and explanation
Correct option: B
α<0<β<∣α∣\alpha<0<\beta<|\alpha|
Option C (n(x1)n−1+an(x_1)^{n-1}+a) is correct. By differentiating the polynomial and evaluating at x1x_1, we find the value of the expression.
778 Easy
778 of 914
Analyzing +4 | -1
The number of solutions of the equation cos⁡x+∣x∣=0\cos x + \left| x \right| = 0 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the graphs of y=cos⁡xy = \cos x and y=−∣x∣y = -|x| do not intersect, indicating no real roots.
779 Easy
779 of 914
Application +4 | -1
The equation whose roots are reciprocal of the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 is
View answer and explanation
Correct option: D
cx2+bx+a=0c{x^2} + bx + a = 0
Option D (cx2+bx+a=0c{x^2} + bx + a = 0) is correct because the coefficients of the quadratic equation are reversed to find the equation with reciprocal roots.
780 Medium
780 of 914
Analyzing +4 | -1
If x,a,b,cx,a,b,c are real and (x−a+b)2+(x−b+c)2=0,{\left( {x - a + b} \right)^2} + {\left( {x - b + c} \right)^2} = 0, then a,b,ca,b,c are in
View answer and explanation
Correct option: C
A.P.
Option C (A.P.) is correct because the given condition implies that a, b, and c form an arithmetic progression.
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