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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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121 Medium
121 of 362
Application +4 | -1
If tan⁡A\tan A and tan⁡B\tan B are the roots of the equation x2−px+q=0x^2-p x+q=0, then the value of sin⁡2(A+B)\sin ^2(A+B) is
View answer and explanation
Correct option: A
p2p2+(1−q)2\frac{p^2}{p^2+(1-q)^2}
The correct answer is A (p2p2+(1−q)2\frac{p^2}{p^2+(1-q)^2}) because it is derived by using the tangent addition formula and trigonometric identities.
122 Medium
122 of 362
Analyzing +4 | -1
The function f(x)=x3+ax2+bx+c,a2≤3bf\left( x \right) = {x^3} + a{x^2} + bx + c,{a^2} \le 3b has
View answer and explanation
Correct option: C
No extreme value
Option C (No extreme value) is correct because the condition a2≤3ba^2 \le 3b implies that the derivative has no real roots, hence no extreme values.
123 Medium
123 of 362
Analyzing +4 | -1
What can be said about the quadratic equation ax2+bx+c=0a x^2+b x+c=0 if 2a+3b+6c=0,(a,b,c∈R)?2 a+3 b+6 c=0,(a, b, c \in R) ?
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 by applying Rolle's theorem to a constructed function, we can prove the existence of a root in the given interval.
124 Medium
124 of 362
Application +4 | -1
Let a,ba, b be positive real numbers. If the equations x2+ax+2b=0x^2+a x+2 b=0 and x2+2bx+a=0x^2+2 b x+a=0 have real roots, then minimum value of a+ba+b is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the minimum value of a+b is 6, derived from the conditions for real roots of the given quadratic equations.
125 Medium
125 of 362
Analyzing +4 | -1
The number of values of k for which the equation x3−3  x+k=0{x^3} - 3\;x + k = 0 has two distinct roots lying in the interval (0,1) are
View answer and explanation
Correct option: D
New value of k satisfies the requirement
The correct answer is "New value of k satisfies the requirement" because the given cubic equation cannot have two distinct roots between 0 and 1 for any value of k.
126 Medium
126 of 362
Application +4 | -1
If α,  β\alpha ,\;\beta be the roots of the quadratic equation ax2+bx+c=0   and  ka{x^2} + bx + c = 0\;\,\mathrm{and}\;k be a real number, then the condition so that α<k<β\alpha < k < \beta is given by
View answer and explanation
Correct option: D
a2k2+abk+ac<0{a^2}{k^2} + abk + ac < 0
Option D (a2k2+abk+ac<0{a^2}{k^2} + abk + ac < 0) is correct because it represents the condition where the quadratic function has different signs at the roots and at k.
127 Medium
127 of 362
Analyzing +4 | -1
Complete set of values of aa such that x2−x1−ax\frac{x^2-x}{1-a x} attains all real values, is
View answer and explanation
Correct option: A
[1,∞)[1, \infty)
Option A ([1, ∞)) is correct because this is the range of 'a' for which the given expression attains all real values, ensuring the discriminant is non-negative.
128 Medium
128 of 362
Application +4 | -1
If an unknown polynomial is divided by (x−1)(x-1) and (x−2)(x-2), we obtain the remainder 2 and 1 respectively, then the remainder resulting from the division of this polynomial by (x−1)(x−2)(x-1)(x-2) is
View answer and explanation
Correct option: C
3−x3-x
Option C (3−x3-x) is correct because it represents the remainder when the polynomial is divided by (x−1)(x−2)(x-1)(x-2), satisfying the given conditions.
129 Medium
129 of 362
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are the roots of the equation x3+px2+qx+r=0\mathrm{x}^3+\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0, then the cubic equation whose roots are α(β+γ),β(γ+α),γ(α+β)\alpha(\beta+\gamma), \beta(\gamma+\alpha), \gamma(\alpha+\beta) is :
View answer and explanation
Correct option: A
x3−2qx2+(q2+pr)x+r2−pqr=0\mathrm{x}^3-2 \mathrm{qx}^2+\left(\mathrm{q}^2+\mathrm{pr}\right) \mathrm{x}+\mathrm{r}^2-\mathrm{pqr}=0
Option A (x3−2qx2+(q2+pr)x+r2−pqr=0x^3-2 q x^2+\left(q^2+p r\right) x+r^2-p q r=0) is correct because by substituting y = alpha(beta + gamma) and using Vieta's formulas, we can derive the required cubic equation.
130 Medium
130 of 362
Application +4 | -1
If x2+px+q=0{x^2} + px + q = 0 has the roots α  and  β\alpha \;\mathrm{and}\;\beta, then the value of (α−β)2{\left( {\alpha - \beta } \right)^2} is equal to
View answer and explanation
Correct option: A
p2−4q{p^2} - 4q
Option A (p2−4q{p^2} - 4q) is correct because it is the simplified form of the expression (α−β)2(\alpha - \beta)^2 using the relationship between roots and coefficients of a quadratic equation.
131 Medium
131 of 362
Analyzing +4 | -1
Both the roots of given equation (x−a)(x−b)+(x-a)(x-b)+ (x−b)(x−c)+(x−c)(x−a)=0(x-b)(x-c)+(x-c)(x-a)=0 are always:
View answer and explanation
Correct option: C
Real
Option C (Real) is correct because the discriminant of the quadratic equation is always non-negative, ensuring that the roots are real.
132 Medium
132 of 362
Application +4 | -1
The value of 8+28+28+22  ,\sqrt {8 + 2\sqrt {8 + 2\sqrt {8 + 2\sqrt 2 } } } \;, is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the expression can be simplified by assuming the nested radical converges to a value x, and then solving the resulting quadratic equation.
133 Medium
133 of 362
Analyzing +4 | -1
For positive real number a,b,ca,b,c which of the following holds?
View answer and explanation
Correct option: A
a+b+c>3⇒a2+b2+c2>3a + b + c > 3 \Rightarrow {a^2} + {b^2} + {c^2} > 3
Option A (a+b+c>3⇒a2+b2+c2>3a + b + c > 3 \Rightarrow {a^2} + {b^2} + {c^2} > 3) is correct because it can be proven using the inequality 3(a2+b2+c2)≥(a+b+c)23\left( {{a^2} + {b^2} + {c^2}} \right) \ge {\left( {a + b + c} \right)^2}.
134 Medium
134 of 362
Application +4 | -1
12 If tan⁡A\tan A and tan⁡B\tan B are the roots of the quadratic equation x2−px+q=0x^2-p x+q=0, then value of sin⁡2(A+B)\sin ^2(A+B) is
View answer and explanation
Correct option: D
p2(1−q)2+p2\frac{p^2}{(1-q)^2+p^2}
Option D is correct because using the tangent addition formula and trigonometric identities, the expression simplifies to p2(1−q)2+p2\frac{p^2}{(1-q)^2+p^2}.
135 Medium
135 of 362
Analyzing +4 | -1
If α\alpha and β\beta are roots of the quadratic equation, x2+xsin⁡θ−2sin⁡θ=0,θ∈(0,π2)x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right), then α12+β12(α−12+β−12)⋅(α−β)24\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right) \cdot(\alpha-\beta)^{24}} is equal to :
View answer and explanation
Correct option: C
212(sin⁡θ+8)12\frac{2^{12}}{(\sin \theta+8)^{12}}
Option C (212(sin⁡θ+8)12\frac{{2^{{12}}}}{{(\sin \theta+8)^{{12}}}}) is correct because it is obtained by simplifying the given expression using the relationships between the roots and coefficients of the quadratic equation.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
136 Medium
136 of 362
Analyzing +4 | -1
If x2−(a+b+c)x+(ab+bc+ca)=0x^2-(a+b+c) x+(a b+b c+c a)=0 has imaginary roots, where a,b,c∈R+a, b, c \in R^{+}, then a,b\sqrt{a}, \sqrt{b} and c\sqrt{c}
View answer and explanation
Correct option: A
can be sides of triangle
Option A (can be sides of triangle) is correct because the given condition implies that the square roots of a, b, and c satisfy the triangle inequality.
137 Medium
137 of 362
Analyzing +4 | -1
The absolute minimum value, of the function f(x)=∣x2−x+1∣+[x2−x+1]f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right], where [t][t] denotes the greatest integer function, in the interval [−1.2],[-1.2], is
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{3}{4}) is correct because the minimum value of the function occurs at x=1/2, and the greatest integer function evaluates to 0 at that point.
138 Medium
138 of 362
Analyzing +4 | -1
If roots of x2−(a−3)x+a=0x^2-(a-3) x+a=0 are such that both of them is greater than 2 , then
View answer and explanation
Correct option: C
a∈[9,10)a \in[9,10)
Option C (a∈[9,10)a \in[9,10)) is correct because it satisfies the conditions for both roots to be greater than 2, considering the discriminant, f(2) > 0, and -B/2A > 2.
139 Medium
139 of 362
Application +4 | -1
If one root of the equation is ix2−2(i+1)x+(2−i)=0i x^2-2(i+1) x+(2-i)=0 is 2−i2-i, then the other root is
View answer and explanation
Correct option: A
−i-i
The correct answer is A (-i) because the coefficients are not real, so we cannot assume conjugate roots, and we must use the product of roots formula.
140 Medium
140 of 362
Analyzing +4 | -1
If the roots of the equation ax2+x+b=0a x^2+x+b=0 are real, then the roots of the equation x2−4abx+1=0x^2-4 \sqrt{a b} x+1=0 will be
View answer and explanation
Correct option: D
imaginary
Option D is correct because the discriminant of the second equation is negative, implying imaginary roots.
141 Medium
141 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of ax2+bx+c=0a x^2+b x+c=0 and pn=αn+βnp_n=\alpha^n+\beta^n, then apn+1+bpn+cpn−1a p_{n+1}+b p_n+c p_{n-1} is equal to
View answer and explanation
Correct option: B
0
The correct answer is B (0) because substituting pnp_n and simplifying using the properties of roots results in zero.
142 Medium
142 of 362
Application +4 | -1
If x2+y2+z2=ax^2+y^2+z^2=a, then the value of xy+zy+zxx y+z y+z x lies in the interval
View answer and explanation
Correct option: C
[−a2,a]\left[\frac{-a}{2}, a\right]
The correct answer is C ([−a2,a\frac{-a}{2}, a]) because using (x+y+z)2≥0(x+y+z)^2 \geq 0 and (x−y)2+(y−z)2+(z−x)2≥0(x-y)^2+(y-z)^2+(z-x)^2 \geq 0 allows us to find the interval for xy+yz+zxxy+yz+zx.
143 Medium
143 of 362
Application +4 | -1
If α\alpha and β\beta are roots of ax2−2bx+c=0a x^2-2 b x+c=0, then α3β3+α2β3+α3β2\alpha^3 \beta^3+\alpha^2 \beta^3+\alpha^3 \beta^2 is
View answer and explanation
Correct option: A
c2(c+2b)a3\frac{c^2(c+2 b)}{a^3}
Option A (c2(c+2b)/a3c^2(c+2 b)/a^3) is correct because it results from substituting the sum and product of the roots into the simplified expression.
144 Medium
144 of 362
Application +4 | -1
If one root of a quadratic equation is 12+5\frac{1}{{2 + \sqrt 5 }}, then the equation is
View answer and explanation
Correct option: B
x2+4x−1=0x^2+4 x-1=0
Option B (x2+4x−1=0x^2+4 x-1=0) is correct because if one root is 12+5\frac{{1}}{{2 + \sqrt 5 }}, the other root is its conjugate, and we can form the quadratic equation using the sum and product of the roots.
145 Medium
145 of 362
Application +4 | -1
If p,q,rp, q, r are three positive real numbers are in APA P, then the roots of the quadratic equation px2+qx+r=0p x^2+q x+r=0 are all real for
View answer and explanation
Correct option: A
∣rp−7∣≥43\left|\frac{r}{p}-7\right| \geq 4 \sqrt{3}
Option A (∣rp−7∣≥43\left|\frac{{r}}{{p}}-7\right| \geq 4 \sqrt{{3}}) is correct because the condition for real roots of the quadratic equation, along with the arithmetic progression relationship, leads to this inequality.
146 Medium
146 of 362
Application +4 | -1
If tan⁡A\tan \mathrm{A} and tan⁡B\tan \mathrm{B} are the roots of x2+ax+b=0\mathrm{x}^2+\mathrm{ax}+\mathrm{b}=0, then the value of expression sin⁡2(A+B)+asin⁡(A+B)cos⁡(A+B)+bcos⁡2(A+B)\sin ^2(A+B)+a \sin (A+B) \cos (A+B)+b \cos ^2(A+B) is equal to -
View answer and explanation
Correct option: D
b\mathrm{b}
Option D (b) is correct because the expression simplifies to b after substituting the values of tan A + tan B and tan A * tan B and using trigonometric identities.
147 Medium
147 of 362
Application +4 | -1
If a+b+c=0a+b+c=0, then xa2/bc⋅xb2/ca⋅xc2/abx^{a^{2 / b c}} \cdot x^{b^2 / c a} \cdot x^{c^2 / a b} is equal to
View answer and explanation
Correct option: D
x3x^3
The correct answer is D (x3x^3) because when a+b+c=0a+b+c=0, a3+b3+c3=3abca^3+b^3+c^3=3abc, which simplifies the exponent to 3.
148 Medium
148 of 362
Analyzing +4 | -1
Consider two quadratic expressions f(x)=ax2+bx+cf(x)=a x^2+b x+c and g(x)=ax2+px+q(a,b,c,p,q∈R,b≠p)g(x)=a x^2+p x+q(a, b, c, p, q \in R, b \neq p) such that their discriminants are equal. If f(x)=g(x)f(x)=g(x) has a root x=αx=\alpha, then
View answer and explanation
Correct option: A
α\alpha will be AM of the roots of f(x)=0f(x)=0 and g(x)=0g(x)=0
The correct answer is A (α\alpha will be AM of the roots of f(x)=0f(x)=0 and g(x)=0g(x)=0) because it is derived from the given conditions and the properties of quadratic equations.
149 Medium
149 of 362
Application +4 | -1
The product of the roots of 8+x3+8−x3=1\sqrt[3]{8+x}+\sqrt[3]{8-x}=1 is
View answer and explanation
Correct option: A
−189-189
Option A (-189) is correct because it is the product of the solutions obtained after cubing and simplifying the given equation.
150 Medium
150 of 362
Application +4 | -1
If α\alpha and β\beta are roots of the equation x2+px+3p4=0x^{2}+p x+\frac{3 p}{4}=0, such that ∣α−β∣=10|\alpha-\beta|=\sqrt{10}, then pp belongs to the set :
View answer and explanation
Correct option: C
{−2,5}\{-2,5\}
Option C ({−2,5}\{-2,5\}) is correct because solving for p using the given condition ∣α−β∣=10|\alpha-\beta|=\sqrt{10} yields the set {−2,5}\{-2,5\}.
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