This set of Mathematics Question Bank for Class 11 focuses on “Quadratic Equations – 2”.

1. If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα + sinβ?

a) 2bc/(a^{2} + b^{2})

b) 0

c) 1

d) (c^{2} + a^{2})/(a^{2} + b^{2})

View Answer

Explanation: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)

^{2}= (c – bsinθ)

^{2}

a

^{2}cos

^{2}θ = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

a

^{2}(1- sin

^{2}θ) = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

a

^{2}– a

^{2}sin

^{2}θ = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

Now rearranging the elements,

(a

^{2}+ b

^{2}) sin

^{2}θ – 2b c sinθ +( c

^{2}– a

^{2}) = θ

So, as sum of the roots are in the form –b/a if there is a quadratic equation ax

^{2}+ bx + c = 0

Now, we can conclude that

sinα + sinβ = 2bc/(a

^{2}+ b

^{2}).

2. If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα * sinβ ?

a) 2bc/(a^{2} + b^{2})

b) 0

c) 1

d) (c^{2} + a^{2})/(a^{2} + b^{2})

View Answer

Explanation: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)

^{2}= (c – bsinθ)

^{2}

a

^{2}cos

^{2}θ = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

a

^{2}(1- sin

^{2}θ) = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

a

^{2}– a

^{2}sin

^{2}θ = c

^{2}+ b

^{2}sin

^{2}θ – 2b c sinθ

Now rearranging the elements,

(a

^{2}+ b

^{2}) sin

^{2}θ – 2b c sinθ +( c

^{2}– a

^{2}) = θ

So, as sum of the roots are in the formc/a if there is a quadratic equation ax

^{2}+ bx + c = 0

Now , we can conclude that

sinα + sinβ = (c

^{2}+ a

^{2})/(a

^{2}+ b

^{2}).

3. If x^{2} + ax + b = 0 and x^{2} + bx + a = 0 have exactly 1 common root then what is the value of (a + b)?

a) 1

b) 0

c) -1

d) 3

View Answer

Explanation: Subtracting the equation x

^{2}+ ax + b = 0 to x

^{2}+ bx + a = 0 by solving the equation simultaneously, we get,

(a – b)x + (b – a) = 0

So, (a – b)x = (a – b)

Therefore, x = 1

Now, putting the value of x = 3 in any one of the equation, we get,

1 + a + b = 0

Therefore, a + b = -1.

4. If, α and β are the roots of the equation 2x^{2} – 3x – 6 = 0, then what is the equation whose roots are α^{2} + 2 and β^{2} + 2?

a) 4x^{2} + 49x + 118 =0

b) 4x^{2} – 49x + 118 =0

c) 4x^{2} – 49x – 118 =0

d) x^{2} – 49x + 118 =0

View Answer

Explanation: Let, y = x

^{2}+ 2

Then, 2x

^{2}– 3x – 6 = 0

So, (3x)

^{2}= (2x

^{2}– 6)

^{2}

[2(y-2) – 6]

^{2}= 9(y-2)

= 4x

^{2}– 49x + 118 = 0.

5. If p and q are the roots of the equation x^{2} + px + q =0 then, what are the values of p and q?

a) p = 1, q = -2

b) p = 0, q = 1

c) p = -2, q = 0

d) p = -2, q = 1

View Answer

Explanation: Since, p and q are the roots of the equation x

^{2}+ px + q =0

So, p + q = -p and pq = q

So, pq = q

And, q = 0 or p = 1

If, q = 0 then, p = 0 and if p = 1 then q = -2.

6. If x^{2} + px + 1 = 0 and (a – b)x^{2} + (b – c)x + (c – a) = 0 have both roots common, then what is the form of a, b, c?

a) a, b, c are in A.P

b) b, a, c are in A.P

c) b, a, c are in G.P

d) b, a, c are in H.P

View Answer

Explanation: Given, (a – b)x

^{2}+ (b – c)x + (c – a) = 0 and x

^{2}+ px + 1 =0

So, 1 / (a – b) = p / (b – c) = 1 / (c – a)

Equating the above equation, we get,

(b – c) = p(a – b) and

(b – c) = p(c – a)

So, p(a – b) = p(c – a)

=> a – b = c – a

So, 2a = b + c which means that b, a, c are in A.P.

7. What will be the sum of b + c if the equations x^{2} + bx + c = 0 and x^{2} + 3x + 3 = 0 have one common root?

a) 2

b) 4

c) 6

d) 8

View Answer

Explanation: Comparing the coefficients of the above equation we get,

1/1 = b/3 = c/3

This means b = 3 and c = 3

Therefore, b + c = 6.

8. If |z_{1}| = 4, |z_{2}| = 3, then what is the value of |z_{1} + z_{2} + 3 + 4i|?

a) Less than 2

b) Less than 5

c) Less than 7

d) Less than 12

View Answer

Explanation: As, we know | z

_{1}+ z

_{2}+ …….. +z

_{n}| ≤ |z

_{1}| + |z

_{2}| + ………. + |z

_{n}|

So, |z

_{1}+ z

_{2}+ 3 + 4i| ≤ |z

_{1}| + |z

_{2}| + |3 + 4i|

Now, putting the given values in the equation, we get,

=> |z

_{1}+ z

_{2}+ 3 + 4i| ≤ 4 + 3 + √(9 + 16)

=> |z

_{1}+ z

_{2}+ 3 + 4i| ≤ 4 + 3 + 5

=> |z

_{1}+ z

_{2}+ 3 + 4i| ≤ 12.

9. According to De Moivre’s theorem what is the value of z^{1/n} ?

a) r^{1/n} [cos(2kπ + θ) + i sin(2kπ + θ)]

b) r^{1/n} [cos(2kπ + θ)/n – i sin(2kπ + θ)/n]

c) r^{1/n} [cos(2kπ + θ)/n + i sin(2kπ + θ)/n]

d) r^{1/n} [cos(2kπ + θ) – i sin(2kπ + θ)]

View Answer

Explanation: If n is any integer, then (cosθ + isinθ)

^{n}= cos(nθ) + i sin(nθ).

Writing the binomial expansion of (cosθ + isinθ)

^{n}and equating real parts of cos(nθ) and the imaginary part to sin(nθ), we get,

cos(nθ) = cos

^{n}θ –

^{n}C

_{2}cos

^{n-2}θ sin

^{2}θ +

^{n}C

_{4}cos

^{n-4}θ sin

^{4}θ + ……….

sin(nθ) =

^{n}C

_{1}cos

^{n-1}θ sinθ –

^{n}C

_{3}cos

^{n-3}θ sin

^{3}θ + ……….

If, n is a rational number, then one of the value of (cosθ + isinθ)

^{n}= cos(nθ) + i sin(nθ).

If, n = p/q, where, p and q are integers (q>θ) and p, q have no common factor, then (cosθ + isinθ)

^{n}has q distinct values one of which is cos(nθ) + i sin(nθ)

If, z

^{1/n}= r

^{1/n}[cos(2kπ + θ)/n + i sin(2kπ + θ)/n], where k = 0, 1, 2, ……….., n – 1.

10. What will be the product of b * c if the equations x^{2} + bx + c = 0 and x^{2} + 3x + 3 = 0 have one common root?

a) 3

b) 4

c) 6

d) 9

View Answer

Explanation: Comparing the coefficients of the above equation we get,

1/1 = b/3 = c/3

This means b = 3 and c = 3

Therefore, b * c = 9.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 11**.

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