## Complex Number & Quadratic Equations Quiz-2

A mathematical equation having a complex number comprises of the real and imaginary sections. Complex numbers are nothing but a combination of two numbers (real, imaginary)..

Q1. If the roots of the equation ax^2 + bx +c =0 are of the form (k+1)/k and (k+2)/(k+1) , then (a+b+c)^2 is equal to
•  2b^2-ac
•  ∑ a^2
•  b^2-4ac
•  b^2-2ac
Solution
(c)
We have,
(k + 1)/k + (k + 2)/(k + 1) = -b/a (1)
and (k + 1)/k + (k + 2)/(k + 1) = c/a
⇒ (k + 2)/k = c/a or 2/k = c/a - 1 = (c - a)/a or k = 2a/(c - a) (2)
Now, eliminate k. Putting the value of k in Eq. (1), we get
(c + a)/2a + 2c/(c + a) = -b/a
⇒(c + a )^2 + 4ac = - 2b( a + c)
⇒( a + c )^2 + 2b( a + c) = -4ac
Adding b^2 to both sides, we have
(a + b + c )^2 = b^2 - 4ac

Q2.Let r,s and t be the roots of the equation, 8x^3 + 1001x + 2008 = 0. The value of (r+s)^3 + (s+t)^3 + (t+r)^3 is
•  251
•  751
•  735
•  753
Solution
(d)
Equation 8x^3 + 1001x + 2008 = 0 has roots r,s and t
r + s + r = 0,rst = -2008/8 = -251
Now, let r + s = A, s + 1 = B, t + r = C,
∴ A + B + C = 2 (r + s + t) = 0
Hence,
A^3 + B^3 + C^3 = 3ABC
∴ (r + s)^3 + ( s + t)^3 + ( t + r)^3
=3(r + s )( s + t)( t + r)
=3(r + s + t - t)(s + t + r - r)(t + r + s - s)
=-3rst(as r + s + t = 0 )
=3(251) = 753

Q3.  If b>a , then the equation (x-a)(x-b) -1 = 0 has
•   Both roots in (a,b)
•  Both roots in (-∞,a)
•  Both roots in (b,+∞)
•   One root in (-∞,a) and the other in (b,+∞)
Solution
(d)
Given equation is
(x - a)(x - b) - 1 = 0
Let f(x) = (x - a)(x - b) - 1. Then,
f(a) = -1 and f(b) = -1
Also, graph of f(x) is concave upward; hence, a and b lie between the roots. Also, if b>a, then one root lies in (-∞,a) and the other root lies in (b,+∞)

Q4. If l,m,n are real l≠m, then the roots of the equation (l-m)x^2 - 5(l+m)x- 2(l-,m)=0 are
•  Real and equal
•  Complex
•  Real and unequal
•  None of these
Solution
(c)
l , m , n are real and l≠m. Given equation is
(l - m ) x^2 - 5(l + m )x - 2( l - m) = 0
D = 25( l + m)^2 + 8( l - m)^2 > 0,l,m∈R
Therefore, the roots are real and unequal
.

Q5. If the expression x^2 + 2(a+b+c)x + 3(bc+ca+ab) is a perfect square, then
•  a = b = c
•  a = ±b = ±c
•  a = b ≠ c
•  None of these
Solution
(a)
Given quadratic expression is x^2+2(a+b+c)x+3(bc+ca+ab), this quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero. Hence,
4(a + b + c)^2 - 4×3(bc + ca + ab) = 0
⇒(a + b + c)^2 - 3( bc + ca + ab) = 0
⇒a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3(bc + ca + ab ) = 0
⇒a^2 + b^2 + c^2 - ab - bc - ca = 0
⇒1/2 [ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ] = 0
⇒1/2 [(a^2 + b^2 - 2ab ) + ( b^2 + c^2 - 2bc ) + ( c^2 + a^2 - 2ca )] = 0
⇒1/2 [(a - b)^2 + (b -c )^2 + (c - a)^2 ] = 0
Which is possible only when
(a - b)^2 = 0,(b - c )^2 = 0 and ( c - a)^2 = 0, i.e., a = b = c

Q6. If |z| < √2 - 1 , then |z^2 + 2z cos⁡Î± |is
•  Less than 1
•  √2+1
• √2-1
•  None of these
Solution
(a)
|z^2 + 2z cos⁡Î± | ≤ | z^2 | + | 2z cos⁡Î± |
= |z|^2 + 2 | z || cos⁡Î± |
≤ |z|^2 + 2 | z |
< ( √2 - 1)^2 + 2(√2 - 1) = 1

Q7. if Ï‰ be a complex n^(th )root of unity, then
is equal to
•
•
•
•  None of these
Solution

Q8. If a,b∈R,a≠0 and the quadratic equation ax^2 -bx + 1 = 0 has imaginary roots then ( a + b + 1) is
•  Positive
•  TNegative
•  Zero
•  Dependent on the sign of b
Solution
(a) D = b^2 - 4a < 0 ⇒ a > 0
Therefore the graph is concave upwards
f(x) > 0 , ∀ x ∈R
⇒f(-1) > 0
⇒a + b + 1 > 0

Q9. Sum of the non-real roots of (x^2 + x - 2) ( x^2+ x - 3) = 12 is
•  -1
•  1
•  -6
•  6
Solution
(a)
Put x^2 + x = y, so that Eq.(1) becomes
(y - 2)(y - 3) = 12
⇒y^2 - 5y - 6 = 0
⇒(y - 6) ( y + 1)=0
⇒y = 6, -1
When y = 6, we get
x^2 + x - 6 = 0
⇒(x + 3)( x - 2) = 0 or x = -3 , 2
When y = -1, we get
Which has non-real roots and sum of roots is -1

Q10. Let z=cos⁡〖Î¸+i sin⁡Î¸ 〗. Then, the value of
is

•  1/sin⁡(2°)
•  1/(3 sin⁡(2°) )
•  1/(2 sin⁡(2° ) )
• 1/(4 sin⁡(2° ) )
Solution

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