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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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Showing 181-210 of 362 questions

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181 Medium
181 of 362
Analyzing +4 | -1
If both the roots of the quadratic equation x2−2kx+k2+k−5=0{x^2} - 2kx + {k^2} + k - 5 = 0 are less than 5, then kk lies in the interval
View answer and explanation
Correct option: B
(−∞,  4)\left( { - \infty ,\;4} \right)
Option B ((−∞,;4)\left( { - \infty ,\\;4} \right)) is correct because the roots of the quadratic equation are less than 5 only when k is less than 4.
182 Medium
182 of 362
Analyzing +4 | -1
Solution set of log⁡3(x2−2)<log⁡3(32∣x∣−1)\log _3\left(x^2-2\right)<\log _3\left(\frac{3}{2}|x|-1\right) is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the solution set obtained by solving the logarithmic inequality does not match any of the given options.
183 Medium
183 of 362
Application +4 | -1
In a triangle ABC,∠C=π2\mathrm{ABC}, \angle \mathrm{C}=\frac{\pi}{2}. If tan⁡(A2)\tan \left(\frac{\mathrm{A}}{2}\right) and tan⁡(B2)\tan \left(\frac{\mathrm{B}}{2}\right) are the roots of the equation ax2+bx+c=0(a≠0)a x^2+b x+c=0(a \neq 0) then&nbsp;
View answer and explanation
Correct option: A
a+b=c\mathrm{a}+\mathrm{b}=\mathrm{c}
Option A (a+b=c) is correct because by using the given condition ∠ C = π2\frac{π}{2} and the properties of tangent function, we can derive the relationship a+b=c.
184 Medium
184 of 362
Application +4 | -1
The solution set of ∣3−4x∣≥9|3-4 x| \geq 9 is
View answer and explanation
Correct option: B
(−∞,−32]∪[3,∞)\left(-\infty,-\frac{3}{2}\right] \cup[3, \infty)
Option B ((−∞,−32]∪[3,∞)\left(-\infty,-\frac{{3}}{{2}}\right] \cup[3, \infty)) is correct because it represents the solution set where the absolute value expression is greater than or equal to 9.
185 Medium
185 of 362
Analyzing +4 | -1
The system y(x2+7x+12)=1   and  x+y=6,y>0{y^{\left( {{x^2} + 7x + 12} \right)}} = 1\;\,\mathrm{and}\;x + y = 6,y > 0 has
View answer and explanation
Correct option: D
More than 2 solutions
Option D (More than 2 solutions) is correct because by solving the equations y(x2+7x+12)=1{y^{\left( {{x^2} + 7x + 12} \right)}} = 1 and x+y=6x + y = 6, we find three distinct solutions.
186 Medium
186 of 362
Application +4 | -1
If the sum of the roots of ax2+bx+c=0a x^2+b x+c=0 is equal to the sum of the squares, then
View answer and explanation
Correct option: A
2ac=ab+b22 a c=a b+b^2
Option A (2ac = ab + b²) is correct because it is derived from equating the sum of roots to the sum of squares of roots.
187 Medium
187 of 362
Analyzing +4 | -1
The number of real solutions of the equation 6−xx2−4=2+x(x+2)\frac{6-x}{x^2-4}=2+\frac{x}{(x+2)} is
View answer and explanation
Correct option: B
one
Option B (one) is correct because after simplifying the equation and excluding the extraneous solution, only one real solution remains.
188 Medium
188 of 362
Application +4 | -1
The set of values of xx satisfying log⁡∣sin⁡x∣∣cos⁡x∣\log _{|\sin x|}|\cos x| +log⁡∣cos⁡x∣∣sin⁡x∣=2+\log _{|\cos x|}|\sin x|=2, when x∈(0,π2)x \in\left(0, \frac{\pi}{2}\right), is
View answer and explanation
Correct option: A
{π4}\left\{\frac{\pi}{4}\right\}
Option A ({π4}\{\frac{{\pi}}{{4}}\}) is correct because it is the only solution to the given logarithmic equation within the specified interval.
189 Medium
189 of 362
Application +4 | -1
Let α,  β\alpha ,\;\beta be the roots of the equation x2−ax+b=0{x^2} - ax + b = 0 and An=αn+βn{A_n} = {\alpha ^n} + {\beta ^n}. Then, An+1−aAn+bAn−1{A_{n + 1}} - a{A_n} + b{A_{n - 1}} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because substituting the roots α and β into the given expression results in zero, due to the properties of the roots satisfying the original quadratic equation.
190 Medium
190 of 362
Analyzing +4 | -1
If α,β\alpha ,\beta are roots of the equation 375x2−25x−2=0375{x^2} - 25x - 2 = 0 and Sn=an+βn,{S_n} = {a^n} + {\beta ^n}, then lim⁡n→∞∑r=1nSr\lim\limits_{n \to \infty } \sum \limits_{r = 1}^n {S_r} is equal to
View answer and explanation
Correct option: C
29348\frac{{29}}{{348}}
Option C (29348\frac{{29}}{{348}}) is correct because by finding the sum of the infinite geometric series formed by the roots and using the relationships between the roots and coefficients, the limit simplifies to the given fraction.
191 Medium
191 of 362
Application +4 | -1
The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2 e^{-\sin x}=2, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equation has no solution as the value of sin x obtained is greater than 1.
192 Medium
192 of 362
Analyzing +4 | -1
The number of positive real roots of x4−4x−1=0,{x^4} - 4x - 1 = 0, is
View answer and explanation
Correct option: C
1
The correct answer is C. By analyzing the sign changes in the polynomial and evaluating the function at specific points, we can conclude that there is only one positive real root.
193 Medium
193 of 362
Application +4 | -1
If x2/3−7x1/3+10=0{x^{2/3}} - 7{x^{1/3}} + 10 = 0, then x=?
View answer and explanation
Correct option: D
{125,8}\left\{ {125,8} \right\}
Option D ({125,8}\{125, 8\}) is correct because both 125 and 8 satisfy the given equation.
194 Medium
194 of 362
Analyzing +4 | -1
If a,b,ca,b,c are in G.P., then the equations ax2+2bx+c=0a{x^2} + 2bx + c = 0 and dx2+2ex+f=0d{x^2} + 2ex + f = 0 have a common root if da,eb,fc\frac{d}{a},\frac{e}{b},\frac{f}{c} are in
View answer and explanation
Correct option: A
A.P.
Option A (A.P.) is correct because if the given equations have a common root, then da\frac{d}{a}, eb\frac{e}{b}, fc\frac{f}{c} must be in Arithmetic Progression.
195 Medium
195 of 362
Analyzing +4 | -1
If 2x2x2+5x+2>1x+1\frac{{2x}}{{2{x^2} + 5x + 2}} > \frac{1}{{x + 1}}, then
View answer and explanation
Correct option: C
−2<x<−1- 2 < x < - 1
Option C (−2<x<−1-2 < x < -1) is correct because solving the inequality leads to the solution set where x is between -2 and -1.

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196 Medium
196 of 362
Application +4 | -1
The product of real roots of the equation ∣2x+3∣2−3∣2x+3∣|2 x+3|^2-3|2 x+3| +2=0+2=0 is -
View answer and explanation
Correct option: B
5/25 / 2
Option B (5/2) is correct because after solving for the roots of the equation, the product of the real roots is calculated to be 5/2.
197 Medium
197 of 362
Application +4 | -1
The least value of ∣a∣\left| a \right| for which tan⁡θ\tan \mathrm{\theta } and cot⁡θ\cot \mathrm{\theta } are roots of the equation x2+ax+1=0{x^2} + ax + 1 = 0, is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the least value of |a| for which tan θ and cot θ are roots of the equation is 2, derived from the condition for real roots.
198 Medium
198 of 362
Analyzing +4 | -1
Let α,β\alpha ,\beta be the roots of the equation x2−x+p=0   and  γ,δ{x^2} - x + p = 0\,\;\mathrm{and}\;\gamma ,\delta be the roots of x2−4x+q=0{x^2} - 4x + q = 0. If α,  β,  γ,  δ\alpha ,\;\beta ,\;\gamma ,\;\delta are in GP, then integral values of p,qp,q are respectively
View answer and explanation
Correct option: A
−2,  −32- 2,\; - 32
Option A (−2,−32- 2, - 32) is correct because these values satisfy the given conditions for the roots to be in GP.
199 Medium
199 of 362
Application +4 | -1
For what value of λ\lambda the sum of the squares of the roots of x2+(2+λ)x−12(1+λ)=0{x^2} + \left( {2 + \lambda } \right)x - \frac{1}{2}\left( {1 + \lambda } \right) = 0 is minimum?
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because this value of λ\lambda minimizes the sum of the squares of the roots of the given quadratic equation.
200 Medium
200 of 362
Application +4 | -1
The sum of all the solutions of the equation (8)2x−16⋅(8)x+48=0(8)^{2 x}-16 \cdot(8)^x+48=0 is :
View answer and explanation
Correct option: C
1+log⁡8(6)1+\log _8(6)
Option C (1+log⁡8(6)1+\log _8(6)) is correct because by substituting 8x=t8^x = t, solving the quadratic equation, and summing the logarithmic solutions, we arrive at the given answer.
201 Medium
201 of 362
Application +4 | -1
If a,b,c∈Ra, b, c \in \mathbf{R} and a+b+c=0a+b+c=0, then the quadratic equation 3ax2+2bx+c=03 a x^2+2 b x+c=0 has
View answer and explanation
Correct option: A
at least one root in [0,1][0,1]
Option A (at least one root in [0,1]) is correct because Rolle's theorem guarantees the existence of a root within the interval (0,1) for the derivative of a function that is zero at both endpoints.
202 Medium
202 of 362
Analyzing +4 | -1
If x,y,zx,y,z are three real numbers such that x+y+z=4x + y + z = 4 and x2+y2+z2=6,{x^2} + {y^2} + {z^2} = 6, then the exhaustive set of values of xx, is
View answer and explanation
Correct option: A
[2/3,  2]\left[ {2/3,\;2} \right]
Option A [2/3,  2]\left[ {2/3,\;2} \right] is correct because it represents the range of possible values for xx given the constraints on x,y,x, y, and zz.
203 Medium
203 of 362
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.
204 Medium
204 of 362
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.
205 Medium
205 of 362
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.
206 Medium
206 of 362
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.
207 Medium
207 of 362
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.
208 Medium
208 of 362
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.
209 Medium
209 of 362
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.
210 Medium
210 of 362
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.
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