Class 11 Mathematics Complex Numbers and Quadratic Equation MCQs Set 10

Question. If a, b, c are positive then both roots of the equation \( ax^2 + bx + c = 0 \)
(a) are real and negative
(b) are real and positive
(c) have negative real parts
(d) have postive real parts
Answer: (c) have negative real parts
Solution:
\( \alpha + \beta = -\frac{b}{a} < 0, \alpha\beta = \frac{c}{a} > 0 \)

Question. In a quadratic equation \( ax^2 + bx + c = 0 \) if 'a' and 'c' are of opposite signs and 'b' is real, then roots of the equation are
(a) real and distinct
(b) real and equal
(c) imaginary
(d) both roots positive
Answer: (a) real and distinct
Solution:
\( ac < 0 \)
\( \implies \) \( \Delta = b^2 - 4ac > 0 \)

Question. If one root of a quadratic equation is real and the other is imaginary, then the coefficients of the equation are
(a) real numbers
(b) rational numbers
(c) irrational numbers
(d) complex numbers
Answer: (d) complex numbers
Solution:
If 1 root is real and other is complex number then the coefficients will be complex numbers.

Question. If \( a \neq b \) the roots of the equation \( (x-a)(x-b) = b^2 \) are
(a) real and distinct
(b) real and equal
(c) real
(d) imaginary
Answer: (a) real and distinct
Solution:
\( \Delta = (a - b)^2 + 4b^2 > 0 \)

Question. If a, b, c are positive numbers in G.P. then the roots of the equation \( ax^2 + bx + c = 0 \)
(a) are real and negative
(b) have negative real parts
(c) are equal
(d) have negative imaginary parts
Answer: (b) have negative real parts
Solution:
\( b^2 = ac \)
\( \therefore b^2 - 4ac < 0 \)
\( \therefore \) roots are imaginary. a,b,c are +ve
\( \implies \) roots have -ve real parts.

Question. If the ratio of the roots of the equation \( ax^2 + bx + c = 0 \) is m : n then
(a) \( \frac{m}{n} + \frac{n}{m} = \frac{b^2}{ac} \)
(b) \( \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \frac{b}{\sqrt{ac}} = 0 \)
(c) \( \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{b^2}{ac} \)
(d) \( \frac{m}{n} + \frac{n}{m} = \frac{a^2}{b^2} \)
Answer: (b) \( \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \frac{b}{\sqrt{ac}} = 0 \)
Solution:
\( \frac{b^2}{ac} = \frac{(m+n)^2}{mn} \)
\( \implies \) \( \frac{m+n}{\sqrt{mn}} = \sqrt{\frac{b^2}{ac}} \)

Question. If \( \alpha, \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \), then the value of \( \frac{1}{a\alpha+b} + \frac{1}{a\beta+b} = \)
(a) \( \frac{a}{bc} \)
(b) \( \frac{b}{ac} \)
(c) \( \frac{c}{ab} \)
(d) \( \frac{ab}{c} \)
Answer: (b) \( \frac{b}{ac} \)
Solution:
\( a\alpha + b = -c/\alpha \); \( a\beta + b = -c/\beta \)

Question. If the roots of \( ax^2 + bx + c = 0 \) and \( px^2 + qx + r = 0 \) differ by the same quantity, then \( \frac{b^2 - 4ac}{q^2 - 4pr} = \)
(a) \( \left(\frac{p}{a}\right)^2 \)
(b) \( \left(\frac{c}{p}\right)^2 \)
(c) \( \left(\frac{a}{p}\right)^2 \)
(d) \( \left(\frac{p}{c}\right)^2 \)
Answer: (c) \( \left(\frac{a}{p}\right)^2 \)
Solution:
\( |\alpha - \beta| = |\alpha_1 - \beta_1| \)
\( \implies \) \( \sqrt{\frac{\Delta}{a^2}} = \sqrt{\frac{\Delta_1}{p^2}} \)
\( \implies \) \( \frac{b^2 - 4ac}{q^2 - 4pr} = \frac{a^2}{p^2} \)

Question. If one root of the equation \( ax^2+bx+c=0 \) is equal to the \( n^{\text{th}} \) power of the other, then \( (ac^n)^{1/(n+1)} + (a^n c)^{1/(n+1)} + b = \)
(a) 0
(b) 1
(c) -1
(d) 2
Answer: (a) 0
Solution:
roots are \( \alpha, \alpha^n \); \( \alpha + \alpha^n = \frac{-b}{a}, \alpha^{n+1} = \frac{c}{a} \).

Question. If \( ax^2 + 2bx + c = 0 \) and \( px^2 + 2qx + r = 0 \) have one and only one root in common and a, b, c, p, q, r being rational, then \( b^2 - ac \) and \( q^2 - pr \) are
(a) both are perfect squares
(b) \( b^2 - ac \) is a perfect square but \( q^2 - pr \) is not a perfect square
(c) \( q^2 - pr \) is a perfect square but \( b^2 - ac \) is not a perfect square
(d) both are not perfect squares
Answer: (a) both are perfect squares
Solution:
\( ax^2 + 2bx + c = 0 \) and \( px^2 + 2qx + r = 0 \) have one root common then it is rational other is also rational then \( b^2 - ac = \) perfect square and \( q^2 - pr = \) perfect square

Question. If both the roots of \( ax^2 + bx + c = 0 \) are positive then
(a) \( \Delta > 0, ab > 0, ac > 0 \)
(b) \( \Delta < 0, ab < 0, ac < 0 \)
(c) \( \Delta > 0, ab < 0, ac > 0 \)
(d) \( \Delta > 0, ab > 0, bc > 0 \)
Answer: (c) \( \Delta > 0, ab < 0, ac > 0 \)
Solution:
\( \Delta > 0 \), ab < 0 and ac > 0

Question. If both the roots of \( ax^2 + bx + c = 0 \) are negative then
(a) \( \Delta > 0, ab > 0, bc < 0 \)
(b) \( \Delta > 0, a, b, c, \) have the same signs
(c) \( \Delta < 0, ab > 0, ac < 0 \)
(d) \( \Delta < 0, ab > 0, bc > 0 \)
Answer: (b) \( \Delta > 0, a, b, c, \) have the same signs
Solution:
\( \Delta > 0 \) and ab > 0, ac > 0

Question. If both the roots of \( ax^2 + bx + c = 0 \) are negative and b < 0 then
(a) a < 0, c < 0
(b) a < 0, c > 0
(c) a > 0, c < 0
(d) a > 0, c > 0
Answer: (a) a < 0, c < 0
Solution:
a,b and c will have same sign

Question. If a > 0, then the expression \( ax^2 + bx + c \) is positive for all values of 'x' provided.
(a) \( b^2-4ac > 0 \)
(b) \( b^2-4ac < 0 \)
(c) \( b^2-4ac = 0 \)
(d) \( b^2-ac < 0 \)
Answer: (b) \( b^2-4ac < 0 \)
Solution:
roots are non real complex roots.

Question. For real x, the expression \( \frac{(x - a)(x - b)}{(x - c)} \) will assume real values provided
(a) a > b > c
(b) a < b < c
(c) c > a > b
(d) a < c < b
Answer: (d) a < c < b
Solution:
put \( y = \frac{(x-a)(x-b)}{(x-c)} \)
\( \implies \) \( x^2 - (a+b+y)x + ab + cy = 0 \)
\( x \in R \)
\( \implies \) \( \Delta = (a+b+y)^2 - 4(ab+cy) > 0 \)
\( \implies \) \( y^2 + 2(a+b-2c)y + (a-b)^2 > 0 \) true for all \( y \in R \)
\( \implies \) \( \Delta = 4(a+b-2c)^2 - 4(a-b)^2 < 0 \)
\( \implies \) \( (c-a)(c-b) < 0 \)

Question. If \( \alpha, \beta \) are the roots of \( ax^2 + bx + c = 0 \) and \( k \in R \) then the condition so that \( \alpha < k < \beta \) is
(a) ac > 0
(b) \( ak^2+bk+c > 0 \)
(c) ac < 0
(d) \( a^2k^2+abk+ac < 0 \)
Answer: (d) \( a^2k^2+abk+ac < 0 \)
Solution:
\( f(x) = ax^2 + bx + c \)
\( a f(k) < 0 \)
\( \implies \) \( a^2k^2 + abk + ac < 0 \).

Question. If a > 0 and \( b^2-4ac < 0 \), then the graph of \( y=ax^2+bx+c \)
(a) lies entirely below the x-axis
(b) lies entirely above the x-axis
(c) cuts the x-axis
(d) touches the x-axis and lies below it
Answer: (b) lies entirely above the x-axis
Solution:
\( b^2-4ac < 0 \) and a > 0 then the graph lies entirely above the x-axis.

Question. If a > 0 and \( b^2-4ac=0 \), then the graph of \( y=ax^2+bx+c \)
(a) lies entirely above the x-axis
(b) touches the x-axis and lies above it
(c) touches the x-axis and lies below it
(d) cuts the x-axis
Answer: (b) touches the x-axis and lies above it
Solution:
\( b^2-4ac = 0 \) and a > 0 then the graph of y touches the x-axis and lies above x-axis.

Question. If a < 0 and \( b^2 -4ac < 0 \), then the graph of \( y=ax^2+bx+c \)
(a) lies entirely below the x-axis
(b) lies entirely above the x-axis
(c) cut the x-axis
(d) touches the x-axis
Answer: (a) lies entirely below the x-axis
Solution:
a < 0 and \( b^2 - 4ac < 0 \) the roots are imaginary then the graph lies entirely below the x - axis.

Question. If \( b^2-4ac > 0 \) then the graph of \( y=ax^2+bx+c \)
(a) cuts x-axis in two real points
(b) touches the x-axis
(c) lies entirely above the x-axis
(d) can not be determined
Answer: (a) cuts x-axis in two real points
Solution:
\( b^2 - 4ac > 0 \)
\( \implies \) roots are real and distinct.
\( \therefore \) graph cuts the x-axis in two distinct points.

Question. If a,b,c are positive real numbers, then the number of real roots of the equation \( ax^2 + b|x| + c = 0 \) is
(a) 2
(b) 4
(c) 0
(d) -1
Answer: (c) 0
Solution:
\( ax^2 + b|x| + c > 0, \forall x \)

Question. Product of real roots of the equation \( t^2x^2 + |x| + 9 = 0 \)
(a) is always positive
(b) is always negative
(c) does not exist
(d) \( \frac{9}{t^2} \)
Answer: (c) does not exist
Solution:
real roots does not exist.

Question. If \( \alpha \) and \( \beta \), (\( \alpha < \beta \)), are the roots of the equation \( x^2 + bx + c = 0 \), where \( c < 0 < b \), then
(a) \( 0 < \alpha < \beta \)
(b) \( \alpha < 0 < \beta < |\alpha| \)
(c) \( \alpha < \beta < 0 \)
(d) \( \alpha < 0 < |\alpha| < \beta \)
Answer: (b) \( \alpha < 0 < \beta < |\alpha| \)
Solution:
\( \alpha < \beta, c < 0 < b \) ; \( \alpha + \beta = -b, \alpha\beta = C \)
C is negative and b is positive
\( \alpha + \beta \) and \( \alpha\beta \) are negative
\( \therefore \) one root is positive and other root is negative
\( \alpha < 0 < \beta \)
\( \alpha + \beta = -b \)
\( \beta = -b - \alpha \)
\( \beta < |\alpha| \)

Question. If \( 0 < a < b < c \), and the roots \( \alpha, \beta \) of the equation \( ax^2 + bx + c = 0 \) are non real complex roots, then
(a) \( |\alpha| = |\beta| \)
(b) \( |\alpha| > 1 \)
(c) \( |\beta| < 1 \)
(d) \( |\alpha| > 1, |\beta| < 1 \)
Answer: (a) \( |\alpha| = |\beta| \)
Solution:
\( 0 < a < b < c \), \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \)
For non real complex roots \( b^2 - 4ac < 0 \)
\( \implies \) \( \frac{b^2}{a^2} - \frac{4c}{a} < 0 \)
\( \implies \) \( (\alpha + \beta)^2 - 4\alpha\beta < 0 \)
\( \implies \) \( (\alpha - \beta)^2 < 0 \)
\( \because 0 < a < b < c \)
\( \therefore \) roots are conjugate, then \( |\alpha| = |\beta| \)

Question. If \( b^2 \ge 4ac \) for the equation \( ax^4 + bx^2 + c = 0 \), then all roots of the equation will be real, if
(a) \( b > 0, a > 0, c > 0 \)
(b) \( b < 0, a > 0, c > 0 \)
(c) \( b > 0, a > 0, c < 0 \)
(d) \( b < 0, a < 0, c < 0 \)
Answer: (b) \( b < 0, a > 0, c > 0 \)
Solution:
All roots of equation \( ax^4 + bx^2 + c = 0 \) will be real if boh roots of \( ay^2 + by + c = 0 \) will be + ve (let \( x^2 = y \)), ie, sum of roots \( = -\frac{b}{a} > 0 \) and product of roots \( = \frac{c}{a} > 0 \). Hence, a and b are of opposite sign, while a and c are of same sign.

LEVEL-I (C.W)

RELATION BETWEEN ROOTS AND COEFFICIENTS

Question. If \( \alpha, \beta \) are roots of \( ax^2 + bx + c = 0 \) then \( \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \)
(a) \( \frac{3abc - b^3}{a^3} \)
(b) \( \frac{3ab - b^3}{a^2c} \)
(c) \( \frac{3abc - b^3}{c^3} \)
(d) \( \frac{b^2 - 2ac}{ac} \)
Answer: (c) \( \frac{3abc - b^3}{c^3} \)
Solution:
\( \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^3} = \frac{(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)}{(\alpha\beta)^3} \)

Question. If \( \alpha, \beta \) are roots of the equation \( 2x^2 + 6x + b = 0 \ (b < 0) \) then \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \) is less than
(a) 2
(b) -2
(c) 18
(d) 0
Answer: (b) -2
Solution:
\( \alpha + \beta = -3, \alpha\beta = b/2 \)
\( D = 36 - 8b > 0 \ (\because b < 0) \)
\( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{(\alpha+\beta)^2}{\alpha\beta} - 2 = \frac{18}{b} - 2 < -2 \)

Question. If \( \alpha, \beta \) are the roots of \( x^2 - p(x+1) + C = 0 \) then \( (\alpha + 1)(\beta + 1) = \)
(a) 1-C
(b) 1+C
(c) C-1
(d) C
Answer: (b) 1+C
Solution:
\( \alpha\beta + \alpha + \beta + 1 = -p + c + p + 1 \)

ROOTS AND/OR CONDITIONS GIVEN

Question. If \( \omega \) is an imaginary cube root of unity. Then the equation whose roots are \( 2\omega + 3\omega^2, 2\omega^2 + 3\omega \) is
(a) \( x^2 + 5x + 7 = 0 \)
(b) \( x^2 + 5x - 7 = 0 \)
(c) \( x^2 - 5x + 7 = 0 \)
(d) \( x^2 - 5x - 7 = 0 \)
Answer: (a) \( x^2 + 5x + 7 = 0 \)
Solution:
\( \alpha + \beta = 2(\omega + \omega^2) + 3(\omega^2 + \omega) = -5 \)
\( \alpha\beta = (2\omega + 3\omega^2)(2\omega^2 + 3\omega) = 7 \)

Question. If p and q are the roots of \( x^2 + px + q = 0 \) then
(a) p = 1
(b) p = 1 or 0
(c) p = -2
(d) p = -2 or 0
Answer: (b) p = 1 or 0
Solution:
\( p + q = -p, pq = q \)
\( \implies \) \( q = 0, p = 1 \)
\( \implies \) \( p = 0, q = -2 \)

NATURE OF THE ROOTS AND PROPERTIES

Question. If one root of \( x^2 - (3+2i)x + (1+3i) = 0 \) is 1+i then the other root is
(a) 1-i
(b) 2+i
(c) 3+i
(d) 1+3i
Answer: (b) 2+i
Solution:
\( \alpha + \beta = 3 + 2i \)

Question. If the expression \( x^2 - (5m - 2)x + (4m^2 + 10m + 25) \) can be expressed as a perfect square, then m =
(a) 8/3 or 4
(b) -8/3 or 4
(c) 4/3 or 8
(d) -4/3 or 8
Answer: (d) -4/3 or 8
Solution:
\( \Delta = 0 \)

Question. If the roots of the quadratic equation \( x^2 - 4x - \log_3 a = 0 \) are real, then the least value of a is
(a) 81
(b) 1/81
(c) 1/64
(d) 9
Answer: (b) 1/81
Solution:
\( D \geq 0 \)
\( \implies \) \( \log_3 a \geq -4 \)
\( \implies \) \( a \geq 3^{-4} \)
\( \implies \) \( a \geq \frac{1}{81} \)

Question. Roots of the equation \( 2x^2 - 5x + 1 = 0 \) and \( x^2 + 5x + 2 = 0 \) are
(a) Reciprocal and of the same sign
(b) Reciprocal and of opposite sign
(c) Equal in magnitude
(d) Imaginary
Answer: (b) Reciprocal and of opposite sign
Solution:
\( f(x) = 2x^2 - 5x + 1 = 0 \)
\( f(-\frac{1}{x}) = 0 \)
\( \implies \) \( x^2 + 5x + 2 = 0 \)

SOLVING EQUATIONS

Question. If \( 3^{1+x} + 3^{1-x} = 10 \) then the values of x are
(a) 1, -1
(b) 1, 0
(c) 1, 2
(d) -1, -2
Answer: (a) 1, -1
Solution:
Put \( 3^x = t \)

Question. If \( (x+1) \) is a factor of \( x^4 + (p-3)x^3 - (3p-5)x^2 + (2p-9)x + 6 \) then the value of p is
(a) -4
(b) 0
(c) 4
(d) 2
Answer: (c) 4
Solution:
Since x+1 is a factor of f(x), -1 must be a root of f(x)=0
\( \therefore f(-1) = 0 \)
\( \implies \) \( -6p + 24 = 0 \)
\( \implies \) \( p = 4 \)

Question. The number of real solutions of the equation \( \sin(e^x) = 5^x + 5^{-x} \) is (are)
(a) 0
(b) 1
(c) 2
(d) Infinitely many
Answer: (a) 0
Solution:
\( 5^x + 5^{-x} \geq 2 \) but \( \sin(e^x) \leq 1 \)

Question. The roots of the equation \( x^{\sqrt{x}} = (\sqrt{x})^x \) are
(a) 0 and 1
(b) 0 and 4
(c) 1 and 4
(d) 0,1 and 4
Answer: (c) 1 and 4

COMMON ROOTS

Question. If \( (x - 2) \) is a common factor of the expressions \( x^2 + ax + b \) and \( x^2 + cx + d \), then \( \frac{b - d}{c - a} = \)
(a) -2
(b) -1
(c) 1
(d) 2
Answer: (d) 2
Solution:
\( x - 2 \) is a common factor
\( \implies \) \( x = 2 \) is common root
\( \therefore 4 + 2a + b = 0 \) and \( 4 + 2c + d = 0 \)
\( \implies \) \( 2(a - c) + (b - d) = 0 \)
\( \implies \) \( \frac{b - d}{c - a} = 2 \)

SIGN OF THE EXPRESSIONS AND INEQUATIONS

Question. If x < 5, then the sign of the expression \( 2x + 7 - 5x^2 \) is
(a) positive
(b) negative
(c) non negative
(d) can not say
Answer: (d) can not say
Solution:
\( 5x^2 - 2x - 7 = 0 \);
\( 5x^2 + 5x - 7x - 7 = 0 \)
\( (x+1)(5x-7) = 0 \)

Question. If x is real and \( 5x^2+2x+9 > 3x^2+10x+7 \), then x lies in the interval
(a) \( (2 - \sqrt{3}, 2 + \sqrt{3}) \)
(b) \( (-\infty, 2 - \sqrt{3}) \cup (2 + \sqrt{3}, \infty) \)
(c) \( (\sqrt{2} - 1, \sqrt{2} + 1) \)
(d) \( (2 + \sqrt{3}, \infty) \)
Answer: (b) \( (-\infty, 2 - \sqrt{3}) \cup (2 + \sqrt{3}, \infty) \)
Solution:
\( 2x^2 - 8x + 2 > 0 \); \( x^2 - 4x + 1 > 0 \)
root = \( \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} \)
\( = (-\infty, 2 - \sqrt{3}) \cup (2 + \sqrt{3}, \infty) \)

MAXIMUM AND MINIMUM

Question. Maximum value of \( -5x^2 + 2x + 3 \) is
(a) 14/5
(b) 13/5
(c) 12/5
(d) 16/5
Answer: (d) 16/5
Solution:
\( \frac{4ac - b^2}{4a} \)

Question. The maximum value of \( \frac{1}{4x^2 + 2x + 1} \) is
(a) 4/3
(b) 2/3
(c) 1
(d) 3/4
Answer: (a) 4/3
Solution:
Maximum value of \( \left(\frac{1}{4x^2 + 2x + 1}\right) = \frac{1}{\min \text{of } (4x^2 + 2x + 1)} = \frac{1}{\frac{4ac - b^2}{4a}} \)

Question. If \( x \in [2, 4] \) then for the expression \( x^2 - 6x + 5 \)
(a) the least value = -4
(b) the greatest value = 4
(c) the least value = 3
(d) the greatest value = -5
Answer: (a) the least value = -4
Solution:
\( x^2 - 6x + 5 = (x - 3)^2 - 4 \geq -4 \)

MODULUS FUNCTIONS

Question. Number of solutions of the equation \( |x|^2 - 3|x| + 2 = 0 \) is.........
(a) 4
(b) 2
(c) 0
(d) 1
Answer: (a) 4
Solution:
\( (|x| - 2)(|x| - 1) = 0 \)
\( \implies \) \( x = \pm 1, \pm 2 \)

Question. For the equation \( |x|^2 + |x| - 6 = 0 \) the roots are
(a) one and only one real number
(b) real with sum one
(c) real with sum zero
(d) real with product zero
Answer: (c) real with sum zero
Solution:
\( (|x| + 3)(|x| - 2) = 0 \)
\( \implies \) \( x = \pm 2 \)

TRANSFORMED EQUATIONS

Question. The equation formed by increasing each root of \( ax^2+bx+c=0 \) by 1 is \( 2x^2+8x+2=0 \) then
(a) a+b=0
(b) b+c=0
(c) b=c
(d) a=b
Answer: (c) b=c
Solution:
\( (\alpha + 1)(\beta + 1) = 1 \)
\( \implies \) \( \alpha\beta = -(\alpha + \beta) \)
\( \implies \) \( \frac{c}{a} = \frac{b}{a} \)
\( \implies \) \( b = c \)

QUADRATIC EXPRESSION IN TWO VARIABLES

Question. The condition for \( ax^2 + 2cxy + by^2 + 2bx + 2ay + c \) is resolvable into two linear factors is
(a) \( a^3+b^3+c^3=3abc \)
(b) \( a^3+b^3+c^3=abc \)
(c) \( a^3+b^3+c^3=ab+bc+ca \)
(d) \( a^3+b^3+c^3+abc=0 \)
Answer: (a) \( a^3+b^3+c^3=3abc \)
Solution:
\( \Delta \equiv abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \)

Question. If \( x^2 + 4xy + 4y^2 + 4x + cy + 3 \) can be written as the product of two linear factors, then c =
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (d) 8
Solution:
\( \Delta \equiv abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \)