Mathematics Questions and Answers – Quadratic Equations – 2

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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/(a2 + b2)
b) 0
c) 1
d) (c2 + a2)/(a2 + b2)
View Answer

Answer: a
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).
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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

Answer: d
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

Answer: c
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.
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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

Answer: b
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

Answer: a
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.
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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

Answer: b
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

Answer: c
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.
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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

Answer: d
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

Answer: c
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.
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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

Answer: d
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter