This set of Class 11 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) 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/(a2 + b2)
b) 0
c) 1
d) (c2 + a2)/(a2 + b2)
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
a2 cos2 θ = c2 + b2 sin2 θ – 2b c sinθ
a2 (1- sin2 θ) = c2 + b2 sin2 θ – 2b c sinθ
a2 – a2 sin2 θ = c2 + b2 sin2 θ – 2b c sinθ
Now rearranging the elements,
(a2 + b2) sin2 θ – 2b c sinθ +( c2 – a2) = θ
So, as sum of the roots are in the form –b/a if there is a quadratic equation ax2 + bx + c = 0
Now, we can conclude that
sinα + sinβ = 2bc/(a2 + b2).
2. If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα * sinβ ?
a) 2bc/(a2 + b2)
b) 0
c) 1
d) (c2 + a2)/(a2 + b2)
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
a2 cos2 θ = c2 + b2 sin2 θ – 2b c sinθ
a2 (1- sin2 θ) = c2 + b2 sin2 θ – 2b c sinθ
a2 – a2 sin2 θ = c2 + b2 sin2 θ – 2b c sinθ
Now rearranging the elements,
(a2 + b2) sin2 θ – 2b c sinθ +( c2 – a2) = θ
So, as sum of the roots are in the formc/a if there is a quadratic equation ax2 + bx + c = 0
Now , we can conclude that
sinα + sinβ = (c2 + a2)/(a2 + b2).
3. If x2 + ax + b = 0 and x2 + 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 x2 + ax + b = 0 to x2 + 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 2x2 – 3x – 6 = 0, then what is the equation whose roots are α2 + 2 and β2 + 2?
a) 4x2 + 49x + 118 =0
b) 4x2 – 49x + 118 =0
c) 4x2 – 49x – 118 =0
d) x2 – 49x + 118 =0
View Answer
Explanation: Let, y = x2 + 2
Then, 2x2 – 3x – 6 = 0
So, (3x)2 = (2x2 – 6)2
[2(y-2) – 6]2 = 9(y-2)
= 4x2 – 49x + 118 = 0.
5. If p and q are the roots of the equation x2 + 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 x2 + 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 x2 + px + 1 = 0 and (a – b)x2 + (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)x2 + (b – c)x + (c – a) = 0 and x2 + 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 x2 + bx + c = 0 and x2 + 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 |z1| = 4, |z2| = 3, then what is the value of |z1 + z2 + 3 + 4i|?
a) Less than 2
b) Less than 5
c) Less than 7
d) Less than 12
View Answer
Explanation: As, we know | z1 + z2 + …….. +zn| ≤ |z1| + |z2| + ………. + |zn|
So, |z1 + z2 + 3 + 4i| ≤ |z1| + |z2| + |3 + 4i|
Now, putting the given values in the equation, we get,
=> |z1 + z2 + 3 + 4i| ≤ 4 + 3 + √(9 + 16)
=> |z1 + z2 + 3 + 4i| ≤ 4 + 3 + 5
=> |z1 + z2 + 3 + 4i| ≤ 12.
9. According to De Moivre’s theorem what is the value of z1/n ?
a) r1/n [cos(2kπ + θ) + i sin(2kπ + θ)]
b) r1/n [cos(2kπ + θ)/n – i sin(2kπ + θ)/n]
c) r1/n [cos(2kπ + θ)/n + i sin(2kπ + θ)/n]
d) r1/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) = cosnθ – nC2 cosn-2θ sin2θ + nC4 cosn-4θ sin4θ + ……….
sin(nθ) = nC1 cosn-1θ sinθ – nC3 cosn-3θ sin3θ + ……….
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, z1/n = r1/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 x2 + bx + c = 0 and x2 + 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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