Quadratic Equations, Complex Numbers and Algebra — JEE Main Questions

47 JEE Main practice questions on Quadratic Equations, Complex Numbers and Algebra, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. The solution set of (x-1)(x-2)/((x-3)(x+4)) ≥ 0 is:

  1. (-∞, -4] ∪ [1, 2] ∪ [3, ∞)
  2. (-∞, -4) ∪ [1, 2] ∪ (3, ∞)
  3. (-∞, -4) ∪ (1, 2) ∪ (3, ∞)
  4. (-4, 1] ∪ [2, 3)

Answer: (-∞, -4) ∪ [1, 2] ∪ (3, ∞)

Critical points: -4, 1, 2, 3. Draw wavy curve starting from top right. Sign is positive on (-∞, -4), negative on (-4, 1), positive on (1, 2), negative on (2, 3), positive on (3, ∞). Since ≥ 0, include numerator zeros (1,2) but exclude denominator zeros (-4,3). So solution is (-∞, -4) ∪ [1, 2] ∪ (3, ∞).

2. If exactly one root of x² - (k+1)x + k = 0 lies in (0,3), then the set of possible values of k is:

  1. (-∞, 0] ∪ {3}
  2. (-∞, 0] ∪ [3, ∞)
  3. (-∞, 0) ∪ {3}
  4. (-∞, 0) ∪ (3, ∞)

Answer: (-∞, 0] ∪ [3, ∞)

The equation factors as (x-1)(x-k)=0, so roots are 1 and k. Exactly one root in (0,3) means k is not in (0,3). If k=0, root 0 is not in (0,3), so k=0 works. If k=3, root 3 is not in (0,3), so k=3 works. For k<0 or k>3, only root 1 lies in (0,3). Thus k ∈ (-∞,0] ∪ [3,∞).

3. If α and β are roots of x² - 5x + 6 = 0, find the equation whose roots are α+2 and β+2.

  1. x² - 9x + 14 = 0
  2. x² - 7x + 12 = 0
  3. x² - 5x + 6 = 0
  4. x² - 9x + 20 = 0

Answer: x² - 9x + 20 = 0

Using transformation: replace x by x-2 in original equation. (x-2)² - 5(x-2) + 6 = 0 simplifies to x² - 9x + 20 = 0. Alternatively, sum of new roots = (α+2)+(β+2) = (α+β)+4 = 5+4=9, product = (α+2)(β+2) = αβ + 2(α+β) + 4 = 6+10+4=20, so equation is x² - 9x + 20 = 0.

4. If 1, α, β are in AP and α, β are roots of x² - (k+1)x + k = 0, find k.

  1. k = 1
  2. k = 0
  3. k = 2
  4. k = 3

Answer: k = 1

For AP, 1 + β = 2α. From Vieta, α+β = k+1 and αβ = k. Substitute β = 2α - 1 into sum: α + (2α-1) = k+1 ⇒ 3α - 1 = k+1 ⇒ k = 3α - 2. Substitute into product: α(2α-1) = k = 3α - 2 ⇒ 2α² - α = 3α - 2 ⇒ 2α² - 4α + 2 = 0 ⇒ α² - 2α + 1 = 0 ⇒ α = 1. Then k = 3(1) - 2 = 1.

5. For what values of k does the equation x² + (k-1)x + (k+2) = 0 have both roots negative?

  1. k > 1
  2. k > -2
  3. k < -1
  4. k ≥ 7

Answer: k ≥ 7

For both roots negative, conditions: D ≥ 0, sum < 0, product > 0. Sum = -(k-1) < 0 ⇒ k > 1. Product = k+2 > 0 ⇒ k > -2. D = (k-1)² - 4(k+2) = k² - 6k - 7 ≥ 0 ⇒ (k-7)(k+1) ≥ 0 ⇒ k ≤ -1 or k ≥ 7. Intersection of k > 1, k > -2, and k ≥ 7 gives k ≥ 7.

6. The number of real solutions of |x^2 - 3x + 2| = 1 is

  1. 0
  2. 4
  3. 2
  4. 3

Answer: 2

The equation |x^2 - 3x + 2| = 1 gives x^2 - 3x + 2 = 1 or x^2 - 3x + 2 = -1. The first gives x^2 - 3x + 1 = 0, with discriminant 5, so two real roots. The second gives x^2 - 3x + 3 = 0, discriminant -3, no real roots. Hence total 2 real solutions.

7. Solve (x+1)(x+2)(x+3)(x+4) = 120 for real x.

  1. x = 1, -6, -2, -3
  2. x = 1, -6
  3. x = 1, -6, -2
  4. x = 1, -6, -3

Answer: x = 1, -6

Pair (x+1)(x+4)=x²+5x+4 and (x+2)(x+3)=x²+5x+6. Let t=x²+5x. Then (t+4)(t+6)=120 ⇒ t²+10t-96=0 ⇒ t=6 or t=-16. Back-substitute: x²+5x=6 ⇒ x=1,-6; x²+5x=-16 has discriminant 25-64=-39<0, no real roots. So only real solutions are x=1 and x=-6.

8. The range of f(x) = (x² + x + 1)/(x² - x + 1) for real x is:

  1. [1/3, 3]
  2. (1/3, 3)
  3. [1/3, 3)
  4. (1/3, 3]

Answer: [1/3, 3]

Set y = f(x) and cross-multiply: (y-1)x² - (y+1)x + (y-1)=0. For y=1, x=0 works. For y≠1, discriminant ≥0 gives -3y²+10y-3≥0, i.e., (3y-1)(y-3)≤0, so y∈[1/3,3]. Both endpoints are attained: y=3 at x=1, y=1/3 at x=-1. Hence range is [1/3,3].

9. Let A: x² - 3x + 2 = 0 and B: x² - (m+1)x + m = 0. For what m does B have both roots in (0,3)?

  1. m > 0
  2. 0 < m < 3
  3. m < 3
  4. 0 ≤ m ≤ 3

Answer: 0 < m < 3

B factors as (x-1)(x-m), so roots are 1 and m. Both in (0,3) requires 0<1<3 (true) and 0<m<3. Also check vertex condition: vertex (m+1)/2 must lie in (0,3), which gives -1<m<5, automatically satisfied for m∈(0,3). So m∈(0,3).

10. What is the minimum value of the quadratic expression 3x^2 - 6x + 7?

  1. 3
  2. 4
  3. 5
  4. 7

Answer: 4

For a quadratic ax^2+bx+c with a>0, the minimum value is -D/(4a). Here D = (-6)^2 - 4*3*7 = 36 - 84 = -48. So minimum = -(-48)/(4*3) = 48/12 = 4. Alternatively, completing the square gives 3(x-1)^2+4, so minimum is 4 at x=1.

11. Find the range of y = (x² - 3x + 4)/(x² + 3x + 4).

  1. [1/7, 7]
  2. [1/7, 7)
  3. (1/7, 7]
  4. (1/7, 7)

Answer: [1/7, 7]

Cross-multiply: y(x²+3x+4)=x²-3x+4 ⇒ (y-1)x²+3(y+1)x+4(y-1)=0. For real x, D≥0: 9(y+1)²-16(y-1)²≥0 ⇒ -7y²+50y-7≥0 ⇒ 7y²-50y+7≤0 ⇒ (7y-1)(y-7)≤0 ⇒ y∈[1/7,7]. Endpoints attained: y=7 at x=1, y=1/7 at x=-1. So range is [1/7,7].

12. Which of the following is a step in solving 2x^2 - 4x - 3 = 0 by completing the square?

  1. Divide by 2 to get x^2 - 4x = 3/2
  2. Divide by 2 to get x^2 - 2x = -3/2
  3. Divide by 2 to get x^2 - 2x = 3/2
  4. Divide by 2 to get x^2 - 4x = -3/2

Answer: Divide by 2 to get x^2 - 2x = 3/2

To complete the square, first divide the equation by the coefficient of x^2 (which is 2) to make the coefficient of x^2 equal to 1. This gives x^2 - 2x - 3/2 = 0, then move the constant to the right: x^2 - 2x = 3/2.

More Mathematics topics

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