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Test: Quadratic Equations- 2 - CAT MCQ


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10 Questions MCQ Test - Test: Quadratic Equations- 2

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Test: Quadratic Equations- 2 - Question 1

Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?

Detailed Solution for Test: Quadratic Equations- 2 - Question 1

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.

Test: Quadratic Equations- 2 - Question 2

Detailed Solution for Test: Quadratic Equations- 2 - Question 2

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Test: Quadratic Equations- 2 - Question 3

Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?

Detailed Solution for Test: Quadratic Equations- 2 - Question 3

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

Test: Quadratic Equations- 2 - Question 4

A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?

Detailed Solution for Test: Quadratic Equations- 2 - Question 4

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.

Test: Quadratic Equations- 2 - Question 5

I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?

Detailed Solution for Test: Quadratic Equations- 2 - Question 5

Explanation:

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


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

Test: Quadratic Equations- 2 - Question 6

Let f(x) be a quadratic ploynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(-2) is equal to

Detailed Solution for Test: Quadratic Equations- 2 - Question 6

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

Test: Quadratic Equations- 2 - Question 7

Let r and c be real numbers. If r and -r are roots of 5x3 + cx2 - 10x + 9 = 0, then c equals

Detailed Solution for Test: Quadratic Equations- 2 - Question 7

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).

Test: Quadratic Equations- 2 - Question 8

Suppose k is any integer such that the equation 2x2 + kx + 5 = 0 has no real roots and the equation x2 + (k - 5)x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is

Detailed Solution for Test: Quadratic Equations- 2 - Question 8

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).

Test: Quadratic Equations- 2 - Question 9

If (3 + 2√2) is a root of the equation ax2 + bx + c = 0, and (4 + 2√3) is a root of the equation ay2 + my + n = 0, where a, b, c, m and n are integers, then the value of 

Detailed Solution for Test: Quadratic Equations- 2 - Question 9

(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).

Test: Quadratic Equations- 2 - Question 10

If r is a constant such that |x2 – 4x - 13| = r has exactly three distinct real roots, then the value of r is?

Detailed Solution for Test: Quadratic Equations- 2 - Question 10

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).

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