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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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271 Medium
271 of 362
Analyzing +4 | -1
If a,b,ca,b,c are real numbers in G.P. such that aa and cc are positive, then the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0
View answer and explanation
Correct option: C
Are imaginary and are in ratio 1  :ω,1\;:\omega , where ω\omega is a complex cube root of unity
Option D is correct because the roots are imaginary and their ratio simplifies to 1:ω1:\omega, where ω\omega is a complex cube root of unity.
272 Medium
272 of 362
Analyzing +4 | -1
The equation sin⁡  θ=x+px\sin \,\,\theta = x + \frac{p}{x} for real values of xx is possible when
View answer and explanation
Correct option: C
p≤14p \le \frac{1}{4}
Option C (p≤14p \le \frac{1}{4}) is correct because for real values of x, the discriminant of the quadratic equation must be non-negative, leading to the condition p≤14p \le \frac{1}{4}.
273 Medium
273 of 362
Analyzing +4 | -1
Let a,ba, b and cc be real numbers, a≠0a \neq 0. If α\alpha is a root of a2x2+bx+c=0,βa^2 x^2+b x+c=0, \beta is the root of a2x2−bx−c=0a^2 x^2-b x-c=0 and 0<α<β0 < \alpha < \beta, then the equation a2x2+2bx+2c=0a^2 x^2+2 b x+2 c=0 has aa root γ\gamma that always satisfies
View answer and explanation
Correct option: D
α<γ<β\alpha < \gamma < \beta
Option D is correct because f(alpha)f(beta) < 0, implying a root lies between alpha and beta.
274 Medium
274 of 362
Application +4 | -1
Solution of the equation 4.9x−1=3(22x+1){4.9^{x - 1}} = 3\sqrt {\left( {{2^{2x + 1}}} \right)} is
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (32\frac{3}{2}) is correct because it is the solution obtained by simplifying the exponential equation and solving for x.
275 Medium
275 of 362
Analyzing +4 | -1
For the equation x34(log⁡2x)2+log⁡2(x)−54=2{x^{\frac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}\left( x \right) - \frac{5}{4}}} = \sqrt 2 which one of the following is not true?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the equation has three real solutions, one of which is irrational, meaning options A, B, and C are all false.
276 Medium
276 of 362
Application +4 | -1
If x2−2xcos⁡θ+1=0x^2-2 x \cos \theta+1=0, then the value of x2n−2xncos⁡nθ+1,n∈Nx^{2 n}-2 x^n \cos n \theta+1, n \in N is equal to -
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting x=cos⁡θ+isin⁡θx = \cos \theta + i \sin \theta into the expression and using De Moivre's theorem results in the expression simplifying to 0.
277 Medium
277 of 362
Application +4 | -1
If ecos⁡x−e−cos⁡x=4{e^{\cos x}} - {e^{ - \cos x}} = 4 then the value of cos⁡x\cos x is
View answer and explanation
Correct option: A
log⁡(2+5)\log \left( {2 + \sqrt 5 } \right)
Option A (log⁡(2+5)\log \left( {2 + \sqrt 5 } \right)) is correct because by substituting y=ecos⁡xy = e^{\cos x} into the equation and solving the quadratic, we find that cos⁡x=log⁡(2+5)\cos x = \log(2 + \sqrt{5}).
278 Medium
278 of 362
Analyzing +4 | -1
If α\alpha and β\beta are roots of the equation x2+ax+b=0x^2+a x+b=0, then maximum value of the expression −(x2+ax+b)−(α−β)24-\left(x^2+a x+b\right)-\frac{(\alpha-\beta)^2}{4} will be
View answer and explanation
Correct option: C
0
Option C (0) is correct because the maximum value of the given expression is 0, which is obtained by substituting the maximum value of the quadratic expression.
279 Medium
279 of 362
Application +4 | -1
The number of real roots of the equation 2  x4+5  x2+3=0,2\;{x^4} + 5\;{x^2} + 3 = 0, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the equation has no real roots, as the expression is always positive for real x.
280 Medium
280 of 362
Application +4 | -1
If a+b+c=6a+b+c=6, then 4a+1+4b+1+4c+1\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}
View answer and explanation
Correct option: A
≤9\leq 9
The correct answer is A (≤9\leq 9) because it is derived using the Cauchy-Schwarz inequality with the given condition a+b+c=6a+b+c=6.
281 Medium
281 of 362
Analyzing +4 | -1
Let pp and qq be two positive numbers, such that p+q=2p + q = 2 and p4+q4=272.{p^4} + {q^4} = 272. Then pp and qq are root of the equation
View answer and explanation
Correct option: B
x2−2x+16=0{x^2} - 2x + 16 = 0
Option B (x2−2x+16=0{x^2} - 2x + 16 = 0) is correct because it represents the quadratic equation whose roots are p and q, derived from the given conditions p+q=2p + q = 2 and p4+q4=272{p^4} + {q^4} = 272.
282 Medium
282 of 362
Analyzing +4 | -1
If 16−x2>∣x−a∣16-x^2 > |x-a| is to be satisfied by atleast one negative value of xx, then complete set of values of aa is
View answer and explanation
Correct option: A
(−16,654)\left(-16, \frac{65}{4}\right)
Option A (−16,654)\left(-16, \frac{{65}}{{4}}\right) is correct because the inequality 16−x2>∣x−a∣16-x^2 > |x-a| is satisfied by at least one negative value of xx when aa lies in this interval.
283 Medium
283 of 362
Analyzing +4 | -1
Let α\alpha be a root of the equation (a−c)x2+(b−a)x(a-c) x^{2}+(b-a) x +(c−b)=0+(c-b)=0 where a,b,ca, b, c are distinct real numbers such that the matrix [α2α1111abc]\left[\begin{array}{ccc}\alpha^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{array}\right] is singular. Then, the value of (a−c)2(b−a)(c−b)+(b−a)2(a−c)(c−b)+(c−b)2(a−c)(b−a) is \frac{(a-c)^{2}}{(b-a)(c-b)}+\frac{(b-a)^{2}}{(a-c)(c-b)}+\frac{(c-b)^{2}}{(a-c)(b-a)} \text { is }
View answer and explanation
Correct option: D
3
Option D (3) is correct because the expression simplifies to 3 using the identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) when a+b+c=0a+b+c=0, then a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc.
284 Medium
284 of 362
Application +4 | -1
The sum of an infinite geometric series is 2 and the sum of geometric series made from the cube of this infinite series 24 . Then the series is -
View answer and explanation
Correct option: C
3−32+34−38+…3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+\ldots
Option C (3−32+34−38+…3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+\ldots) is correct because it satisfies the given conditions for the sum of the infinite geometric series and the sum of the cubes of the series.
285 Medium
285 of 362
Application +4 | -1
For the equation ∣x2∣+∣x∣−6=0\left|x^2\right|+|x|-6=0, the roots are
View answer and explanation
Correct option: C
Real with sum zero
Option C (Real with sum zero) is correct because the roots are 2 and -2, which are real and sum to zero.

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286 Medium
286 of 362
Analyzing +4 | -1
If a,ba, b and cc are distinct positive integers, then axb−c+bxc−a+cxa−ba x^{b-c}+b x^{c-a}+c x^{a-b} is
View answer and explanation
Correct option: C
>(a+b+c)> (a+b+c)
Option C (>(a+b+c)> (a+b+c)) is correct because it follows from applying AM-GM inequality to the given expression.
287 Medium
287 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 8x2−3x+27=08 x^2-3 x+27=0, then (α2β)1/3+(β2α)1/3\left(\frac{\alpha^2}{\beta}\right)^{1 / 3}+\left(\frac{\beta^2}{\alpha}\right)^{1 / 3} is equal to
View answer and explanation
Correct option: A
1/41/4
Option A (1/4) is correct because it results from substituting the sum and product of roots into the simplified expression.
288 Medium
288 of 362
Analyzing +4 | -1
The sum of all the roots of the equation ∣x2−8x+15∣−2x+7=0\left|x^{2}-8 x+15\right|-2 x+7=0 is:
View answer and explanation
Correct option: B
9+39+\sqrt{3}
Option B (9+39+\sqrt{3}) is correct because it is the sum of the roots of the given equation after considering both cases of the absolute value.
289 Medium
289 of 362
Application +4 | -1
If p,q∈{1,2,3,4}p, q \in\{1,2,3,4\}, then the number of equations of the form px2+qx+1=0p x^2+q x+1=0 having real roots is
View answer and explanation
Correct option: C
7
The correct answer is 7 because we need to find the number of pairs (p, q) such that q2≥4pq^2 \geq 4p where p,q∈{1,2,3,4}p, q \in \{1, 2, 3, 4\}.
290 Medium
290 of 362
Analyzing +4 | -1
If roots of x2−(a−3)x+a=0x^2-(a-3) x+a=0 are such that at least one root is greater than 2, then aa belongs to
View answer and explanation
Correct option: B
[9,∞)[9, \infty)
The correct answer is B ([9,∞)[9, \infty)) because it represents the range of 'a' for which at least one root of the quadratic equation is greater than 2.
291 Medium
291 of 362
Analyzing +4 | -1
If the roots of the equation x2+ax+b=0x^2+a x+b=0 are cc and dd, then one of the roots of the equation x2+(2c+a)x+(c2+ac+b2)=0x^2+(2 c+a) x+\left(c^2+a c+b^2\right)=0 is
View answer and explanation
Correct option: B
d−cd-c
Option B (d−cd-c) is correct because substituting x = c into the transformed equation yields a root of 0, implying that d-c is also a root.
292 Medium
292 of 362
Application +4 | -1
The number which exceeds its positive square roots by 12, is
View answer and explanation
Correct option: B
16
Option B (16) is correct because 16 exceeds its positive square root (4) by 12 (16 = 4 + 12).
293 Medium
293 of 362
Application +4 | -1
If the product of the roots of the equation x2−3kx+x^2-3 k x+ 2e2log⁡k−1=02 e^{2 \log k}-1=0 is 7 , then the roots are real for k=k=
View answer and explanation
Correct option: C
2
The correct answer is C (2) because when k=2, the product of roots is 7 and the roots are real.
294 Medium
294 of 362
Application +4 | -1
If 2 lies between the roots of the equation t2−mt+2=0,(m∈R)t^2-m t+2=0,(m \in R) then the value of [(3∣x∣9+x2)m]\left[\left(\frac{3|x|}{9+x^2}\right)^m\right] is :
(where [] denotes greatest integer function)
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given condition implies m > 3, and the expression inside the greatest integer function is less than 1, resulting in 0.
295 Medium
295 of 362
Analyzing +4 | -1
If 2  a+3  b+6c=0,2\;a + 3\;b + 6c = 0, then at least one root of the equation ax2+bx+c=0a{x^2} + bx + c = 0 lies in the interval
View answer and explanation
Correct option: A
(0,1)
Option A ((0,1)) is correct because by constructing a function f(x) and applying Rolle's theorem, we can show that at least one root of the given equation lies in the interval (0,1).
296 Medium
296 of 362
Analyzing +4 | -1
For ∣x−1∣x+2<1,  x\frac{{\left| {x - 1} \right|}}{{x + 2}} < 1,\;x lies in the interval
View answer and explanation
Correct option: A
(−∞,  −2)∪(−12,  ∞)\left( { - \infty ,\; - 2} \right) \cup \left( { - \frac{1}{2},\;\infty } \right)
Option A (−∞,  −2)∪(−12,  ∞)\left( { - \infty ,\; - 2} \right) \cup \left( { - \frac{1}{2},\;\infty } \right) is correct because it represents the solution set of the inequality considering both cases when x<1x < 1 and x≥1x \ge 1.
297 Medium
297 of 362
Analyzing +4 | -1
If x1,x2,x3x_1, x_2, x_3 and x4x_4 are four positive real numbers such that x1+1x2=4,x2+1x3=1x_1+\frac{1}{x_2}=4, x_2+\frac{1}{x_3}=1, x3+1x4=4x_3+\frac{1}{x_4}=4 and x4+1x1=1x_4+\frac{1}{x_1}=1, then
View answer and explanation
Correct option: A
x1=x3x_1=x_3 and x2=x4x_2=x_4
Option A (x1=x3x_1=x_3 and x2=x4x_2=x_4) is correct because by applying AM-GM inequality and the given equations, it can be shown that x1=x3=2x_1 = x_3 = 2 and x2=x4=1/2x_2 = x_4 = 1/2.
298 Medium
298 of 362
Analyzing +4 | -1
If the equation x4−4x3+ax2+bx+1=0x^4-4 x^3+a x^2+b x+1=0 has four positive roots, then a,ba, b are
View answer and explanation
Correct option: D
a=6,b=−4a=6, b=-4
Option D (a=6,b=−4a=6, b=-4) is correct because by assuming all roots are equal and positive, we can find the values of a and b that satisfy the given conditions.
299 Medium
299 of 362
Analyzing +4 | -1
The set of all xx satisfying the equation xlog⁡3x2+(log⁡3x)2−10=1/x2x^{\log _3 x^2+\left(\log _3 x\right)^2-10}=1 / x^2 is
View answer and explanation
Correct option: B
{1,9,1/81}\{1,9,1 / 81\}
Option B ({1,9,1/81}) is correct because these values satisfy the given equation after taking logarithms and solving for x.
300 Medium
300 of 362
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the roots of the equation x3+3x−7=0x^3+3 x-7=0, then the equation whose roots are 2α+1,2β+12 \alpha+1,2 \beta+1 and 2γ+12 \gamma+1 is
View answer and explanation
Correct option: A
x3−3x2+15x−69=0x^3-3 x^2+15 x-69=0
The correct answer is A (x3−3x2+15x−69=0x^3-3 x^2+15 x-69=0) because substituting α=z−12\alpha = \frac{z-1}{2} into the original equation and simplifying yields the new equation.
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