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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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301 Medium
301 of 914
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.
302 Medium
302 of 914
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.
303 Easy
303 of 914
Application +4 | -1
If k∈Rk \in \mathbf{R} lies between the roots of ax2+2bx+a x^2+2 b x+ c=0c=0, then
View answer and explanation
Correct option: B
a2k2+2abk+ac<0a^2 k^2+2 a b k+a c < 0
Option B (a2k2+2abk+ac<0a^2 k^2+2 a b k+a c < 0) is correct because if k lies between the roots, then a*f(k) < 0, where f(k) = ak2 + 2bk + c. Multiplying by 'a' gives a2k2 + 2abk + ac < 0.
304 Easy
304 of 914
Analyzing +4 | -1
If a root of the equation ax2+bx+c=0a{x^2} + bx + c = 0 be reciprocal of a root of the equation a′x2+b′x+c′=0a'{x^2} + b'x + c' = 0, then
View answer and explanation
Correct option: A
(cc′−aa′)2=(ba′−cb′)(ab′−bc′){\left( {cc' - aa'} \right)^2} = \left( {ba' - cb'} \right)\left( {ab' - bc'} \right)
Option A ((cc′−aa′)2=(ba′−cb′)(ab′−bc′){\left( {cc' - aa'} \right)^2} = \left( {ba' - cb'} \right)\left( {ab' - bc'} \right)) is correct because it represents the condition when one root of the first quadratic is the reciprocal of a root of the second quadratic.
305 Easy
305 of 914
Application +4 | -1
If the roots of the given equation (cos⁡p−1)x2+(cos⁡p)x+sin⁡p=0(\cos p - 1){x^2} + (\cos p)x + \sin p = 0 are real, then
View answer and explanation
Correct option: C
p∈(0  ,π)p \in \left( {0\;,\pi } \right)
Option C, p∈(0,π)p \in (0,\pi), is correct because it satisfies the condition for real roots of the quadratic equation, ensuring the discriminant is non-negative.
306 Easy
306 of 914
Application +4 | -1
If the roots of the equation x2+2ax+b=0x^2+2 a x+b=0 are real and distinct and differ by at most 2m2 m, then bb lies in the interval
View answer and explanation
Correct option: C
[a2−m2,a2)\left[a^2-m^2, a^2\right)
Option C ([a2-m2, a2)) is correct because the roots are real and distinct, and their difference is at most 2m, leading to the given interval for b.
307 Medium
307 of 914
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.
308 Medium
308 of 914
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.
309 Easy
309 of 914
Application +4 | -1
The least value of ∣a∣|a| for which sin⁡θ\sin \theta and cosec⁡θ\operatorname{cosec} \theta are the roots of the equation x2+ax+b=0x^{2}+a x+b=0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the minimum value of |sin(θ) + cosec(θ)| is 2, which is derived from AM-GM inequality.
310 Easy
310 of 914
Application +4 | -1
If x2−4x+log⁡1/2a=0x^{2}-4 x+\log _{1 / 2} a=0 does not have two distinct real roots, then maximum value of aa is
View answer and explanation
Correct option: B
116\frac{1}{16}
Option B (116\frac{{1}}{{16}}) is correct because the discriminant must be less than or equal to zero for the quadratic to not have two distinct real roots, leading to the inequality a≤116a \le \frac{{1}}{{16}}.
311 Medium
311 of 914
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.
312 Easy
312 of 914
Application +4 | -1
If x=1+ix=1+i is a root of x3−ix+1−i=0x^3-i x+1-i=0, then the quadratic equation whose roots are the remaining two roots of x3−ix+1−i=0x^3-i x+1-i=0 is
View answer and explanation
Correct option: B
x2+(1+i)x+i=0x^2+(1+i) x+i=0
Option B (x2+(1+i)x+i=0x^2+(1+i) x+i=0) is correct because it is the quadratic equation formed by the remaining two roots of the cubic equation after dividing by the factor corresponding to the given root.
313 Medium
313 of 914
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.
314 Medium
314 of 914
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.
315 Easy
315 of 914
Application +4 | -1
The maximum value of the quadratic equation 1−x−3x2=0,x∈R1-x-3 x^2=0, x \in \mathbb{R} is
View answer and explanation
Correct option: D
13/1213 / 12
Option D (13/1213 / 12) is correct because it is the maximum value of the quadratic equation calculated using the formula 4AC−B24A\frac{{4AC-B^2}}{{4A}}.

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316 Hard
316 of 914
Application +4 | -1
The minimum value of P=bcx+cay+abz,  P = bcx + cay + abz,\; when   xyz=abc,  \;xyz = abc,\; is
View answer and explanation
Correct option: A
3abc3abc
Option A (3abc3abc) is correct because the minimum value of P is found using AM ≥ GM inequality, and given that xyz=abcxyz = abc.
317 Medium
317 of 914
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.
318 Easy
318 of 914
Application +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−2x+2=0x^2-\sqrt{2} x+2=0 then α16+β16=\alpha^{16}+\beta^{16}=
View answer and explanation
Correct option: B
−256-256
Option B (-256) is correct because the roots are complex numbers, and raising them to the 16th power simplifies to a real number.
319 Easy
319 of 914
Application +4 | -1
If tan⁡A\tan \mathrm{A} and tan⁡B\tan \mathrm{B} are roots of the quadratic equation x2−px+q=0x^2-p x+q=0, then the value of sin⁡2( A+B)\sin ^2(\mathrm{~A}+\mathrm{B}) is
View answer and explanation
Correct option: C
p2(1−q)2+p2\frac{p^2}{(1-q)^2+p^2}
Option C (p2(1−q)2+p2\frac{{p^2}}{{(1-q)^2+p^2}}) is correct. By using the tangent addition formula and trigonometric identities, we can express sin⁡2(A+B)\sin^2(A+B) in terms of p and q.
320 Medium
320 of 914
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.
321 Easy
321 of 914
Analyzing +4 | -1
The number of irrational solutions of the equation x2+x2+11+x2−x2+11=4,\sqrt {{x^2} + \sqrt {{x^2} + 11} } + \sqrt {{x^2} - \sqrt {{x^2} + 11} } = 4, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because after simplification and substitution, the equation yields two irrational solutions for x.
322 Medium
322 of 914
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.
323 Easy
323 of 914
Application +4 | -1
The minimum value of aa for which a2x2+(2a−1)x+1a^{2} x^{2}+(2 a-1) x+1 is non-negative for any real xx is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{{1}}{{4}}) is correct because the expression being non-negative implies that the discriminant is non-positive.
324 Easy
324 of 914
Analyzing +4 | -1
If each pair of the equation x2+ax+b=0,  x2+bx+c=0    and   x2+cx+a=0{x^2} + ax + b = 0,\;{x^2} + bx + c = 0\;\,\,\mathrm{and}\,\;{x^2} + cx + a = 0 has a common root, then product of all common roots is
View answer and explanation
Correct option: A
abc\sqrt {abc}
Option A is correct because the product of all common roots is abc\sqrt{abc}, which is derived from Vieta's formulas applied to the given equations.
325 Medium
325 of 914
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}.
326 Easy
326 of 914
Analyzing +4 | -1
Sum of the real roots of the equation x2+5∣x∣+6=0x^2+5|x|+6=0 is equal to
View answer and explanation
Correct option: D
Does not exist
Option D (Does not exist) is correct because the equation x2+5∣x∣+6=0x^2 + 5|x| + 6 = 0 has no real roots, as all terms are non-negative, and their sum cannot be zero.
327 Medium
327 of 914
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}.
328 Easy
328 of 914
Application +4 | -1
If one root of the equation 8x2−6x−a−3=08{x^2} - 6x - a - 3 = 0 is the square of the other, then the values of aa are
View answer and explanation
Correct option: D
−4,  24- 4,\;24
Option D (−4,  24- 4,\;24) is correct because these are the values of 'a' obtained by setting one root as the square of the other and solving the resulting equations.
329 Medium
329 of 914
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.
330 Easy
330 of 914
Application +4 | -1
α=sin⁡36∘\alpha = \sin 36^\circ is a root of which of the following equation?
View answer and explanation
Correct option: C
16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 0
Option C (16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 0) is correct because substituting α=sin⁡36∘\alpha = \sin 36^\circ into the equation yields a true statement.
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