STEM Lessons for College Students

JEE Main Problems | QUADRATIC EQUATIONS | 2013 to 2017 | Chapterwise Solutions

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In this video, we will discuss previous year JEE Main questions of the chapter QUADRATIC EQUATIONS (questions from year 2013 to 2017).


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JEE Main 2013 Online
If α and β are the root of the equation x2 + px + 3p/4 = 0 such that |α-β|=√10, then p belongs to the set
(1) {2, -5} (2) {-3, 2} (3) {-2, 5} (4) {3, -5}

JEE Main 2014
Let α and β be the roots of the equation px2 + qx + r = 0, p ≠ 0. If p, q, r are in AP, and 1/α + 1/β = 4, then the value of |α-β| is equal to

JEE Main 2014 B. Arch.
If the roots of the equation 1/(x+p)+1/(x+q)=1/r are equal in magnitude and opposite in sign, then the
product of the roots is
(1) -(p^2-q^2 )/2 (2) (p^2+q^2 ) (3) (p^2-q^2 )/2 (4) -(p^2+q^2 )/2

JEE Main 2014 Online
If α and β are the roots of the equation x2 – 4√2kx + 2 e^(4 ln⁡k )- 1 = 0, for some k. If α^2+β^2=66, then α^3+β^3 is equal to

JEE Main 2017 B. Arch.
If λ_1 and λ_2 are the two values of λ such that the roots α and β of the quadratic equation
λ(x^2-x)+x+5=0 satisfy: α/β + β/α + 4/5 =0, then λ_1/〖λ_2〗^2 + λ_2/〖λ_1〗^2 is equal to
(1) 488 (2) 536 (3) 512 (4) 504

JEE Main 2015
Let α and β be the roots of the equation x^2-6x-2=0. If a_n=α^n-β^n. Then (a_10-2a_8)/(2a_9 )=
(1) 6 (2) -6 (3) 3 (4) -3

JEE Main 2017 B. Arch. Paper 2
The sum of the abscissa of the points where the curves, y = kx2 + (5k + 3) x + 6k + 5, (k ∈ R), touch the x-axis, is equal to

JEE Main 2016 B. Arch. Paper 2
The number of integral values of m for which the equation, (1 + m2) x2 – 2(1 + 3m)x + (1 + 8m) = 0, has no real root, is:
(1) 1 (2) 2 (3) 3 (4) infinitely many

JEE Main 2013 Online
If p and q are non-zero real numbers and α^3+β^3=-p, αβ=q, then a quadratic equation whose roots are α^2/β and β^2/α is:

JEE Main 2014 Online
If 1/√α and 1/√β are the roots of the equation ax2 + bx + 1 = 0, (a ≠ 0, a, b ∈ R), then the equation
x(x + b3) + (a3 – 3abx) = 0 has roots:
(1) α^(3/2),β^(3/2) (2) αβ^(1/2),α^(1/2) β (3) √αβ,αβ (4) α^(-3/2),β^(-3/2)

JEE Main 2017
If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1)(x + 2) +…..+ (x + n – 1)(x + n) = 10n has two consecutive integral solutions, then n is equal to

JEE Main 2015 Online
If the two roots of the equation, (a – 1)(x4 + x2 + 1) + (a + 1)(x2 + x + 1)2 = 0 are real and distinct, then the set of all values of ‘a’ is:

JEE Main 2014 Online
The sum of the roots of the equation x^2+|2x-3|-4=0, is:
(1) 2 (2) -2 (3) √2 (4) -√2

JEE Main 2016
The sum of all real values of x satisfying this equation (x^2-5x+5)^((x^2+4x-60) )=1 is
(1) – 4 (2) 6 (3) 5 (4) 3

JEE Main 2015 Online
If 2 + 3i is one of the roots of the equation 2×3 – 9×2 + kx – 13 = 0, k ∈ R, then real root of this equation
(1) exists and is equal to 1 (2) exists and is equal to –1/2
(3) exists and is equal to 1/2 (4) does not exist

JEE Main 2014 Online
If the equations 2×2 + 3x + 4 = 0 and ax2 + bx + c = 0, a,b,c ∈ R, a ≠ 0, have a common root, then a : b : c is
(1) 1 : 2 : 3 (2) 2 : 3 : 4 (3) 4 : 3 : 2 (4) 3 : 2 : 1

JEE Main 2016 Online
If equations x2 + bx -1 = 0 and x2+ x+ b = 0 have a common root different from -1, then |b| is equal to
(1) √2 (2) 2 (3) √3 (4) 3

JEE Main 2013 Online
The value of “a” for which one root of the equation x2 – (a + 1)x + (a2 + a – 8) = 0 exceeds 2, and other is lesser than 2, are given by

JEE Main 2013 B. Arch. Paper 2
If the quadratic equation 3×2 + 2(a2 + 1)x + (a2 – 3a + 2) = 0 possess roots of opposite signs, then “a” lies in the interval

JEE Main 2015 B. Arch. Paper 2
The values of k for which each root of the equation, x2 – 6kx + 2 – 2k + 9k2 = 0 is greater than 3, always satisfy the inequality :

JEE Main 2014 Online
Let p(x) be a quadratic polynomial such that p(0) = 1. If p(x) leaves remainder 4 when divided by (x – 1) and it leaves remainder 6 when divided by (x + 1), then:
(1) p(2) = 11 (2) p(2) = 19 (3) p(-2) = 19 (4) p(-2) = 11

JEE Main 2014
If a ∈ R and the equation -3(x – [x])2 + 2(x – [x]) + a2 = 0 (where [x] denotes greatest integer less than equal to x) has no integral solution, then all the possible values of a lies in the interval
(1) (-2, -1) (2) (-∞, -2) ∪ (2, ∞) (3) (-1, 0) ∪ (0, 1) (4) (1, 2)

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