Class 11 Mathematics Complex Numbers and Quadratic Equation MCQs Set H

RELATION BETWEEN ROOTS AND COEFICIENTS

Question. If \(\alpha, \beta\) are the roots of \(ax^2 + bx + c = 0\) then the value \(\left(\frac{\alpha}{\beta} - \frac{\beta}{\alpha}\right)^2\) is
(a) \(\frac{b^2(b^2 - 4ac)}{c^2a^2}\)
(b) \(\frac{b^2(b^2 + 4ac)}{c^2a^2}\)
(c) \(\frac{b^2(b^2 - 2ac)}{c^2a^2}\)
(d) \(\frac{b^2(b^2 + 2ac)}{c^2a^2}\)
Answer: (a) \(\frac{b^2(b^2 - 4ac)}{c^2a^2}\)
Solution: \(\left(\frac{\alpha}{\beta} - \frac{\beta}{\alpha}\right)^2 = \frac{(\alpha^2 - \beta^2)^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2(\alpha - \beta)^2}{(\alpha\beta)^2} = \frac{(-b/a)^2(b^2 - 4ac)/a^2}{(c/a)^2} = \frac{b^2(b^2 - 4ac)}{c^2a^2}\)

Question. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
(a) \(x^2 + 18x + 16 = 0\)
(b) \(x^2 - 18x - 16 = 0\)
(c) \(x^2 + 18x - 16 = 0\)
(d) \(x^2 - 18x + 16 = 0\)
Answer: (d) \(x^2 - 18x + 16 = 0\)
Solution: Let roots be \(\alpha\) and \(\beta\). \(\frac{\alpha + \beta}{2} = 9 \Rightarrow \alpha + \beta = 18\); \(\sqrt{\alpha\beta} = 4 \Rightarrow \alpha\beta = 16\). \(\therefore\) The required equation is \(x^2 - 18x + 16 = 0\)

Question. If \(\alpha \neq \beta\) but \(\alpha^2 = 2\alpha - 3\); \(\beta^2 = 2\beta - 3\) then the equation whose roots are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is
(a) \(x^2 + 3x + 2 = 0\)
(b) \(3x^2 + 2x + 3 = 0\)
(c) \(x^2 - 3x + 2 = 0\)
(d) \(3x^2 - 2x + 3 = 0\)
Answer: (b) \(3x^2 + 2x + 3 = 0\)
Solution: Roots of \(x^2 - 2x + 3 = 0\) are \(\alpha, \beta\). The required quadratic equation is \(x^2 - \left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)x + \left(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}\right) = 0 \Rightarrow x^2 - \left(\frac{\alpha^2 + \beta^2}{\alpha\beta}\right)x + 1 = 0 \Rightarrow 3x^2 + 2x + 3 = 0\)

Question. If \(\alpha + \beta = -2\) and \(\alpha^3 + \beta^3 = -56\) then the quadratic equation whose roots are \(\alpha, \beta\), is
(a) \(x^2 + 2x - 16 = 0\)
(b) \(x^2 + 2x - 15 = 0\)
(c) \(x^2 + 2x - 12 = 0\)
(d) \(x^2 + 2x - 8 = 0\)
Answer: (d) \(x^2 + 2x - 8 = 0\)
Solution: \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \Rightarrow -56 = (-2)^3 - 3\alpha\beta(-2) \Rightarrow -56 = -8 + 6\alpha\beta \Rightarrow \alpha\beta = -8\). Equation is \(x^2 - (\alpha+\beta)x + \alpha\beta = 0 \Rightarrow x^2 + 2x - 8 = 0\)

Question. If the sum of the roots of the equation \(ax^2 + bx + c = 0\) is equal to sum of the squares of their reciprocals, then \(bc^2, ca^2, ab^2\) are in
(a) A.P
(b) G.P
(c) H.P
(d) A.G.P
Answer: (a) A.P
Solution: \(\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} \Rightarrow \alpha + \beta = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} \Rightarrow \alpha + \beta = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} \Rightarrow -\frac{b}{a} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} \Rightarrow -\frac{b}{a} = \frac{b^2 - 2ac}{c^2} \Rightarrow -bc^2 = ab^2 - 2a^2c \Rightarrow 2ca^2 = ab^2 + bc^2\). Hence, \(bc^2, ca^2, ab^2\) are in A.P.

Question. \(\tan 22^\circ\) and \(\tan 23^\circ\) are roots of \(x^2 + ax + b = 0\) then
(a) \(a + b + 1 = 0\)
(b) \(a - b + 1 = 0\)
(c) \(b - a + 1 = 0\)
(d) \(a + b = 1\)
Answer: (b) \(a - b + 1 = 0\)
Solution: \(\tan 45^\circ = \frac{\tan 22^\circ + \tan 23^\circ}{1 - \tan 22^\circ \tan 23^\circ} \Rightarrow 1 = \frac{-a}{1 - b} \Rightarrow 1 - b = -a \Rightarrow a - b + 1 = 0\)

Question. If the difference of the squares of the roots of the equation \(x^2 - 6x + q = 0\) is 24, then the value of q is
(a) -7
(b) 8
(c) 5
(d) 4
Answer: (c) 5
Solution: \(|\alpha^2 - \beta^2| = 24 \Rightarrow |\alpha + \beta| \cdot |\alpha - \beta| = 24 \Rightarrow |-(-6)/1| \cdot \frac{\sqrt{b^2 - 4ac}}{|a|} = 24 \Rightarrow 6\sqrt{36 - 4q} = 24 \Rightarrow \sqrt{36 - 4q} = 4 \Rightarrow 36 - 4q = 16 \Rightarrow 4q = 20 \Rightarrow q = 5\)

Question. Let \(\alpha, \beta\) be the roots of the equation \((x - a)(x - b) = c, c \neq 0\), then the roots of the equation \((x - \alpha)(x - \beta) + c = 0\) are
(a) \(a, c\)
(b) \(b, c\)
(c) \(a, b\)
(d) \(a + b, b + c\)
Answer: (c) \(a, b\)
Solution: \((x - a)(x - b) - c = (x - \alpha)(x - \beta) \Rightarrow (x - a)(x - b) = (x - \alpha)(x - \beta) + c\). Therefore, roots of \((x - \alpha)(x - \beta) + c = 0\) are \(a, b\).

Question. If \(\alpha, \beta\) are the roots of the equation \(x^2 - 15x + 1 = 0\), then the value of \(\left( \frac{1}{\alpha} - 15 \right)^{-2} + \left( \frac{1}{\beta} - 15 \right)^{-2}\) is
(a) 225
(b) 900
(c) 223
(d) 0
Answer: (c) 223
Solution: \(\alpha + \beta = 15, \alpha\beta = 1 \Rightarrow \frac{1}{\alpha} = \beta, \frac{1}{\beta} = \alpha\). The given expression is \((\beta - 15)^{-2} + (\alpha - 15)^{-2}\). Also, \(\alpha - 15 = -1/\alpha\) and \(\beta - 15 = -1/\beta\). Therefore, \((-\alpha)^2 + (-\beta)^2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (15)^2 - 2(1) = 225 - 2 = 223\).

ROOTS ARE GIVEN WITH CONDITIONS

Question. The quadratic equation whose roots are the x and y intercepts of the line passing through \((1,1)\) and making a triangle of area A with the axes, may be
(a) \(x^2 + Ax + 2A = 0\)
(b) \(x^2 - 2Ax + 2A = 0\)
(c) \(x^2 - Ax + 2A = 0\)
(d) \(x^2 - 2Ax - 2A = 0\)
Answer: (b) \(x^2 - 2Ax + 2A = 0\)
Solution: Let equation of line be \(\frac{x}{a} + \frac{y}{b} = 1\) passing through \((1,1) \Rightarrow \frac{1}{a} + \frac{1}{b} = 1 \Rightarrow a + b = ab \dots (1)\). Area of the triangle \(A = \frac{1}{2}ab \Rightarrow ab = 2A\). From (1), \(a + b = 2A\). The quadratic equation whose roots are a and b is \(x^2 - (a + b)x + ab = 0 \Rightarrow x^2 - 2Ax + 2A = 0\).

Question. If one root of \(x^2 + px + 12 = 0\) is 4 while the equation \(x^2 + px + q = 0\) has equal roots then q=
(a) -7
(b) 4
(c) 42
(d) \(\frac{49}{4}\)
Answer: (d) \(\frac{49}{4}\)
Solution: Substitute x = 4 in \(x^2 + px + 12 = 0 \Rightarrow 16 + 4p + 12 = 0 \Rightarrow p = -7\). Equation \(x^2 - 7x + q = 0\) has equal roots \(\Rightarrow \Delta = 0 \Rightarrow (-7)^2 - 4(1)(q) = 0 \Rightarrow q = \frac{49}{4}\).

Question. If \(k > 0\) and the product of the roots of the equation \(x^2 - 3kx + 2e^{2\log k} - 1 = 0\) is 7 then the sum of the roots is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (c) 6
Solution: Product of roots = \(2e^{2\log k} - 1 = 7 \Rightarrow 2k^2 - 1 = 7 \Rightarrow 2k^2 = 8 \Rightarrow k^2 = 4 \Rightarrow k = 2\) (\(\because k > 0\)). Sum of roots = \(3k = 3(2) = 6\).

Question. The value of ‘a’ for which one root of the quadratic equation \((\alpha^2 - 5a + 3)x^2 - (3a - 1)x + 2 = 0\) is twice as large as other is
(a) -2/3
(b) 1/3
(c) -1/3
(d) 2/3
Answer: (d) 2/3
Solution: Let roots be \(\alpha, 2\alpha\). For one root being twice the other (\(m:n = 1:2\)), condition is \(\frac{(m+n)^2}{mn} = \frac{b^2}{ac} \Rightarrow \frac{(1+2)^2}{(1)(2)} = \frac{(3a-1)^2}{2(a^2-5a+3)} \Rightarrow \frac{9}{2} = \frac{(3a-1)^2}{2(a^2-5a+3)} \Rightarrow 9a^2 - 45a + 27 = 9a^2 - 6a + 1 \Rightarrow 39a = 26 \Rightarrow a = \frac{2}{3}\).

Question. Two students while solving a quadratic equation in x, one copied the constant term incorrectly and got the roots as 3 and 2. The other copied the constant term and coefficient of \(x^2\) as -6 and 1 respectively. The correct roots are
(a) 3, -2
(b) -3, 2
(c) -6, -1
(d) 6, -1
Answer: (d) 6, -1
Solution: Let original equation be \(x^2 + ax + b = 0\). First student got sum of roots correctly: \(-a = 3 + 2 = 5 \Rightarrow a = -5\). Second student got product of roots correctly: \(b/1 = -6 \Rightarrow b = -6\). Correct equation is \(x^2 - 5x - 6 = 0\). Roots are \(6, -1\).

Question. Let A, G and H be the A.M, G.M and H.M of two positive numbers a and b. The quadratic equation whose roots are A and H is
(a) \(Ax^2 - (A^2 + G^2)x + AG^2 = 0\)
(b) \(Ax^2 - (A^2 + H^2)x + AH^2 = 0\)
(c) \(Hx^2 + (H^2 + G^2)x + HG^2 = 0\)
(d) \(Gx^2 - (H^2 + G^2)x + GH^2 = 0\)
Answer: (a) \(Ax^2 - (A^2 + G^2)x + AG^2 = 0\)
Solution: \(AH = G^2 \Rightarrow H = \frac{G^2}{A}\). Equation with roots A and H is \(x^2 - (A + H)x + AH = 0 \Rightarrow x^2 - \left(A + \frac{G^2}{A}\right)x + G^2 = 0 \Rightarrow Ax^2 - (A^2 + G^2)x + AG^2 = 0\).

Question. If one root of the equation \(ax^2 + bx + c = 0\) is the square of the other, then
(a) \(b^2 + ac^2 + a^2c = 3abc\)
(b) \(b^3 + ac^2 + a^2c = 3abc\)
(c) \(b^2 + ac^2 + a^2c + 3abc = 0\)
(d) \(b^3 + ac^2 + a^2c + 3abc = 0\)
Answer: (b) \(b^3 + ac^2 + a^2c = 3abc\)
Solution: Roots are \(\alpha\) and \(\alpha^2\). \(\alpha + \alpha^2 = -b/a\) and \(\alpha \cdot \alpha^2 = \alpha^3 = c/a\). Cubing sum: \((\alpha + \alpha^2)^3 = (-b/a)^3 \Rightarrow \alpha^3 + \alpha^6 + 3\alpha^3(\alpha + \alpha^2) = -b^3/a^3 \Rightarrow \frac{c}{a} + \left(\frac{c}{a}\right)^2 + 3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right) = \frac{-b^3}{a^3} \Rightarrow \frac{c}{a} + \frac{c^2}{a^2} - \frac{3bc}{a^2} = \frac{-b^3}{a^3} \Rightarrow a^2c + ac^2 - 3abc = -b^3 \Rightarrow b^3 + ac^2 + a^2c = 3abc\).

Question. The roots of the equation \(ax^2 + bx + c = 0, a \in R^+\), are two consecutive odd positive integeres. Then
(a) \(|b| \leq 4a\)
(b) \(|b| \geq 4a\)
(c) \(|b| \geq 2a\)
(d) \(|b| \leq a\)
Answer: (b) \(|b| \geq 4a\)
Solution: Let \(\alpha = 2k-1, \beta = 2k+1, k \in N\). Sum of roots \(\alpha + \beta = 4k = -b/a\). Since \(k \geq 1\), \(4k \geq 4 \Rightarrow -b/a \geq 4 \Rightarrow -b \geq 4a \Rightarrow |b| \geq 4a\).

Question. The number of real solution of the equation \((9/10)^x = -3 + x - x^2\) is
(a) 2
(b) 0
(c) 1
(d) 3
Answer: (b) 0
Solution: LHS \((9/10)^x > 0\) for all real x. RHS \(x - x^2 - 3 = -(x^2 - x + 3) = -((x-1/2)^2 + 11/4) < 0\) for all real x. LHS is positive, RHS is negative, therefore 0 solutions.

Question. If the roots of \(x^2 - bx + c = 0\) are two consecutive integers then \(b^2 - 4c =\)
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
Solution: Roots are \(\alpha\) and \(\alpha+1\). Difference is 1. \(|\alpha - \beta| = 1 \Rightarrow \sqrt{b^2 - 4c} = 1 \Rightarrow b^2 - 4c = 1\).

Question. If the roots of the quadratic equation \(x^2 + px + q = 0\) are \(\tan 30^\circ\) and \(\tan 15^\circ\) respectively then the value of \(2 + q - p\) is
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (c) 3
Solution: \(\tan(30^\circ + 15^\circ) = \frac{\tan 30^\circ + \tan 15^\circ}{1 - \tan 30^\circ \tan 15^\circ} \Rightarrow \tan 45^\circ = 1 = \frac{-p}{1 - q} \Rightarrow 1 - q = -p \Rightarrow q - p = 1\). Thus, \(2 + q - p = 2 + 1 = 3\).

NATURE OF THE ROOTS AND PROPERTIES

Question. If \(a + b + c = 0\) then the equation \(3ax^2 + 2bx + c = 0\) has
(a) imaginary roots
(b) real and equal roots
(c) real and different roots
(d) rational roots
Answer: (c) real and different roots
Solution: \(\Delta = (2b)^2 - 4(3a)(c) = 4b^2 - 12ac = 4[(-a-c)^2 - 3ac] = 4[a^2 + c^2 + 2ac - 3ac] = 4[a^2 - ac + c^2] = 2[(a-c)^2 + a^2 + c^2] > 0\). Therefore, roots are real and different.

Question. If the roots of the equation \((a^2 + b^2)x^2 + 2(bc + ad)x + (c^2 + d^2) = 0\) are real then \(a^2, bd, c^2\) are in
(a) AP
(b) GP
(c) HP
(d) AGP
Answer: (b) GP
Solution: The equation can be written as \((ax+d)^2 + (bx+c)^2 = 0\). For real roots, both squares must be zero: \(ax+d = 0\) and \(bx+c = 0 \Rightarrow x = -d/a = -c/b \Rightarrow ad = bc \Rightarrow a^2c^2 = (bd)^2\). Hence \(a^2, bd, c^2\) are in GP.

Question. Both the roots of the equation \(\frac{(x - b)(x - c) + (x - c)(x - a) + (x - a)(x -b)}{(x-a)(x-b)} = 0\) are
(a) positive
(b) negative
(c) real
(d) imaginary
Answer: (c) real
Solution: Numerator is \(3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0\). \(\Delta = 4(a+b+c)^2 - 12(ab+bc+ca) = 4(a^2+b^2+c^2 - ab - bc - ca) = 2((a-b)^2 + (b-c)^2 + (c-a)^2) \geq 0\). Both roots are real.

Question. If a,b,c are in A.P. then the roots of the equation \(ax^2 + 2bx + c = 0\) are
(a) real and distinct
(b) real and equal
(c) real
(d) imaginary
Answer: (c) real
Solution: a,b,c in A.P \(\Rightarrow 2b = a + c\). \(\Delta = 4b^2 - 4ac = (a+c)^2 - 4ac = (a-c)^2 \geq 0\). The roots are real.

Question. If \(ax^2 - (2a + 3)x + (3 + 5a) = 0\) has no real roots, then ‘a’ lies in the interval
(a) \((-\frac{3}{4}, \frac{3}{4})\)
(b) \((-\infty, -\frac{3}{4})\)
(c) \((-\infty, \frac{3}{4})\)
(d) \((-\infty, -\frac{3}{4}) \cup (\frac{3}{4}, \infty)\)
Answer: (d) \((-\infty, -\frac{3}{4}) \cup (\frac{3}{4}, \infty)\)
Solution: \(\Delta < 0 \Rightarrow (2a+3)^2 - 4a(3+5a) < 0 \Rightarrow 4a^2 + 12a + 9 - 12a - 20a^2 < 0 \Rightarrow 9 - 16a^2 < 0 \Rightarrow 16a^2 > 9 \Rightarrow a \in (-\infty, -3/4) \cup (3/4, \infty)\).

Question. If the roots of \((x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0\) are equal then which of the following is not possible
(a) \(a + b + c = 0\)
(b) \(a = b = c\)
(c) \(a + bw + cw^2 = 0\)
(d) \(a + bw^2 + cw = 0\)
Answer: (a) \(a + b + c = 0\)
Solution: Equation is \(3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0\). For equal roots \(\Delta = 0 \Rightarrow a^2+b^2+c^2 - ab - bc - ca = 0 \Rightarrow \frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2) = 0 \Rightarrow a = b = c\). If \(a, b, c\) are non-zero equal values, \(a+b+c = 3a \neq 0\). So \(a+b+c=0\) is not possible.

Question. If the roots of \(ax^2 - bx - c = 0\) are changed by the same quantity then the expression in \(a, b, c\), that does not change is
(a) \(\frac{b^2 - 4ac}{a^2}\)
(b) \(\frac{b^2 - 4ac}{a}\)
(c) \(\frac{b^2 + 4ac}{a^2}\)
(d) \(\frac{b^2 + 4ac}{a}\)
Answer: (c) \(\frac{b^2 + 4ac}{a^2}\)
Solution: If the roots are changed by the same quantity, their difference remains unchanged. \(\Rightarrow (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (b/a)^2 - 4(-c/a) = \frac{b^2 + 4ac}{a^2}\) remains invariant.

Question. If the roots of the equation \(bx^2 + cx + a = 0\) be imaginary then for all real values of x, the expression \(3b^2x^2 + 6bcx + 2c^2\) is
(a) greater than 4ab
(b) less than 4ab
(c) greater than -4ab
(d) less than -4ab
Answer: (c) greater than -4ab
Solution: Roots of \(bx^2 + cx + a = 0\) are imaginary \(\Rightarrow c^2 - 4ab < 0 \Rightarrow c^2 < 4ab\). Expression \(3b^2x^2 + 6bcx + 2c^2 = 3(bx+c)^2 - c^2 \geq -c^2 > -4ab\).

SOLVING EQUATIONS

Question. Let \(f(x) = x^2 - 3x + 4\), the value(s) of x which satisfies \(f(1) + f(x) = f(1)f(x)\) is
(a) 1
(b) 2
(c) 1 or 2
(d) 1 and 0
Answer: (c) 1 or 2
Solution: \(f(1) = 1 - 3 + 4 = 2\). The equation becomes \(2 + f(x) = 2f(x) \Rightarrow f(x) = 2 \Rightarrow x^2 - 3x + 4 = 2 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow x = 1, 2\).

Question. The number of non-zero solutions of the equation \(x^2 - 5x - (\text{sgn} \, x)6 = 0\) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Solution: CASE I: When \(x < 0\), \(\text{sgn}(x) = -1\). Equation is \(x^2 - 5x + 6 = 0 \Rightarrow x = 2, 3\). But \(x < 0\), so no solution. CASE II: When \(x > 0\), \(\text{sgn}(x) = 1\). Equation is \(x^2 - 5x - 6 = 0 \Rightarrow (x-6)(x+1) = 0 \Rightarrow x = 6, -1\). Since \(x > 0\), \(x=6\) is the only solution.

Question. A root of equation \(\frac{a + c}{x + a} + \frac{b + c}{x + b} = \frac{2(a + b + c)}{x + a + b}\) is
(a) a
(b) b
(c) c
(d) a+b+c
Answer: (c) c
Solution: Substituting \(x=c\) satisfies the equation (by options).

Question. The equation \(\log_2(3 - x) + \log_2(1 - x) = 3\) has
(a) One root
(b) Two roots
(c) Infinite roots
(d) No root
Answer: (a) One root
Solution: Domain: \(x < 1\). \(\log_2((3-x)(1-x)) = 3 \Rightarrow (3-x)(1-x) = 8 \Rightarrow x^2 - 4x - 5 = 0 \Rightarrow (x-5)(x+1) = 0 \Rightarrow x = 5, -1\). As \(x < 1\), \(x=5\) is rejected. \(\therefore x = -1\) is the only root.

Question. The number of real roots of \(\sin(2^x)\cos(2^x) = \frac{1}{4}(2^x + 2^{-x})\) is
(a) 1
(b) 2
(c) 3
(d) No solution
Answer: (d) No solution
Solution: \(\frac{1}{2}\sin(2 \cdot 2^x) = \frac{1}{4}(2^x + 2^{-x}) \Rightarrow \sin(2^{x+1}) = \frac{2^x + 2^{-x}}{2}\). The RHS is the AM of \(2^x\) and \(2^{-x}\), which is \(\geq 1\). Since \(2^x \neq 2^{-x}\) for \(x \neq 0\), equality to 1 only occurs if \(x=0\). At \(x=0\), LHS \(= \sin 2 < 1\), RHS \(= 1\). Therefore, no solution.

Question. The equation \(e^{\sin x} - e^{-\sin x} - 4 = 0\) has
(a) infinite number of real roots
(b) no real roots
(c) exactly one real root
(d) exactly four real roots
Answer: (b) no real roots
Solution: Let \(t = e^{\sin x}\), then \(t - 1/t - 4 = 0 \Rightarrow t^2 - 4t - 1 = 0 \Rightarrow t = 2 \pm \sqrt{5}\). But \(t = e^{\sin x}\), so \(1/e \leq t \leq e\). The values \(2 \pm \sqrt{5}\) are out of this range (approx 0.36 to 2.71). Therefore, no real roots.

Question. The real value of ‘a’ for which \(x^2 + i(a - 1)x + 5 = 0\) will have a pair of conjugate complex roots is
(a) values satisfying \(a^2 - 2a + 21 > 0\)
(b) 1
(c) all values
(d) no value
Answer: (b) 1
Solution: Let the roots be \(p+iq\) and \(p-iq\) (conjugate pair). The sum of the roots is \(2p\), which is purely real. Sum of roots = \(-i(a-1)\). Since this must be real, \(a-1 = 0 \Rightarrow a = 1\).