Test: Quadratic Equations- 2 - CAT MCQ

Explanation:

a/b + b/a = (a² + b²)/ab = (a² + b² + a + b)/ab 
= [(a + b)² - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)² - 2(4)]/4 = 56/4 = 14.

Explanation:

The quadratic equation whose roots are reciprocal of 2x² + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)² + 5(1/x) + 3 = 0
=> 3x² + 5x + 2 = 0

Explanation:

Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x 
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x² + 5x = 0
multiplying both sides by -1/10x
=> x² + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.

Explanation:

I.(a - 3)(a - 4) = 0
=> a = 3, 4

II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b

Let f(x) = ax² + bx + c
Since, f(x) ≥ 0, and f(2) = 0, it means the graph of the quadratic lies above x-axis and touches x-axis at x = 2.

Since x = 2 is the only root of the equation (i.e., graph is symmetric about x = 2),
⇒ f(2 + a) = f(2 – a)
⇒ f(2 + 2) = f(2 - 2)
⇒ f(4) = f(0)
⇒ f(0) = 6
⇒ c = 6

f(2) = 0
⇒ 4a + 2b + c = 0   
⇒ 4a + 2b = -6   …(2)

f(4) = 6
⇒ 16a + 4b + c = 6
⇒ 16a + 4b = 0   …(3)

Solving (2) and (4), we get
a = -3/2 and b = -6

⇒ f(-2) = 3/2 (-2)² – 6 × - 2 + 6 = 6 + 12 + 6 = 24

In a polynomial ax³ + bx² + cx + d = 0, whose roots are α, β and γ.
⇒ α + β + γ = -b/a
⇒ αβ + βγ + γα = c/a
⇒ αβγ = -d/a

If roots of 5x³ + cx² – 10x + 9 = 0, r, -r and γ
⇒ r – r + γ = -c/5   
⇒ γ = -c/5   …(1)

⇒ r × -r + r × γ – r × γ = (-10)/5 = -2
⇒ r² = 2   …(2)

⇒ r × - r × γ = -(9/5)
⇒ r² × γ = 9/5
⇒ γ = 9/10   …(3)

From (1) and (3)
⇒ -c/5 = 9/10
⇒ c = -9/2

Hence, option (b).

2x² + kx + 5 = 0 has no real roots ⇒ D < 0
⇒ k² – 4 × 2 × 5 < 0
⇒ k² < 40
⇒ -√40 < k < √40
∴ Possible integral values of k are -6, -5, -4, …, 0, …4, 5, 6   …(1)

Also, x² + (k - 5)x + 1 = 0 has two distinct roots ⇒ D > 0
⇒ (k - 5)² – 4 × 1 × 1 > 0
⇒ k² + 25 – 10k – 4 > 0
⇒ k² – 10k + 21 > 0
⇒ (k - 7)(k - 3) > 0
⇒ k ∈ (-∞, 3) ∪ (7, ∞)   …(2)

The integral value of k satisfying both (1) and (2) are
-6, -5, -4, -3, -2, -1, 0, 1, 2 i.e., 9 values.

Hence, option (c).

(3 + 2√2) is a root of the equation ax² + bx + c = 0 whose coefficients are integers
⇒ Since coefficients are rational the other root will be (3 - 2√2)

∴ Sum of roots = 6 = -b/a
⇒ b = -6a   …(1)

∴ product of roots = 1 = c/a
⇒ c = a   …(2)

(4 + 2√3) is a root of the equation ay² + my + n = 0 whose coefficients are integers
⇒ Since coefficients are rational the other root will be (4 - 2√3)

∴ Sum of roots = 8 = -m/a
⇒ m = -8a   …(3)

∴ product of roots = 4 = n/a
⇒ n = 4a   …(4)

Now,


Hence, option (c).

Given, |x² – 4x - 13| = r
∴ x² – 4x - 13 = ± r

We have two quadratic equations here but only three distinct roots it means one of the quadratic equations will have equal roots.

Case 1: x² – 4x - 13 = r has equal roots, i.e., Discriminant = 0
⇒ x² – 4x – 13 - r = 0 had D = 0

⇒ D = 16 – 4(-13 - r) = 0
⇒ 16 + 52 + 4r = 0
⇒ r = - 17 

Case 2: x² – 4x - 13 = - r has equal roots, i.e., Discriminant = 0
⇒ x² – 4x – 13 + r = 0 had D = 0

⇒ D = 16 – 4(-13 + r) = 0
⇒ 16 + 52 - 4r = 0
⇒ r = 17

Hence, option (c).