Class 11 Mathematics Complex Numbers and Quadratic Equation MCQs Set 09

COMMON ROOTS

Question. If \(ax^2 + bx + c = 0\) and \(bx^2 + cx + a = 0\) have a common root and \(a \neq 0\) then \(\frac{a^3 + b^3 + c^3}{abc} =\)
(a) 1
(b) 2
(c) 3
(d) 9
Answer: (c) 3
Solution: Let \(\alpha\) be the common root. \(a\alpha^2 + b\alpha + c = 0\) and \(b\alpha^2 + c\alpha + a = 0\). Subtracting \((a-b)\alpha^2 + (b-c)\alpha + (c-a) = 0 \dots\) Another way: \((\alpha-1)[a\alpha + c] = 0 \Rightarrow \alpha=1 \therefore a+b+c = 0 \Rightarrow a^3+b^3+c^3 = 3abc\). So ratio is 3.

Question. The equations \(x^2 + 3x + 5 = 0\) and \(ax^2 + bx + c = 0\) have a common root. If \(a, b, c \in N\) then the least possible values of \(a + b + c\) is equal to
(a) 3
(b) 6
(c) 9
(d) 12
Answer: (c) 9
Solution: For \(x^2 + 3x + 5 = 0\), \(\Delta = 9 - 20 = -11 < 0\). The roots are imaginary and must occur in conjugate pairs. Thus, both roots are common. Therefore, \(\frac{a}{1} = \frac{b}{3} = \frac{c}{5} = \lambda \Rightarrow a = \lambda, b = 3\lambda, c = 5\lambda\). For least \(a,b,c \in N\), \(\lambda=1 \Rightarrow a+b+c = 1+3+5 = 9\).

Question. If the equations \(ax^2 + 2bx + 3c = 0\) and \(3x^2 + 8x + 15 = 0\) have a common root where a,b,c are the length of the sides of \(\Delta ABC\) then \(\sin^2 A + \sin^2 B + \sin^2 C\) is equal to
(a) 1
(b) \(\frac{3}{2}\)
(c) \(\sqrt{2}\)
(d) 2
Answer: (d) 2
Solution: For \(3x^2 + 8x + 15 = 0\), \(\Delta = 64 - 180 = -116 < 0\). Since roots are imaginary, both roots must be common. Hence \(\frac{a}{3} = \frac{2b}{8} = \frac{3c}{15} \Rightarrow \frac{a}{3} = \frac{b}{4} = \frac{c}{5} \Rightarrow a=3k, b=4k, c=5k\). This forms a right-angled triangle where C is \(90^\circ\). \(\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 A + \cos^2 A + \sin^2 90^\circ = 1 + 1 = 2\).

LOCATION OF ROOTS

Question. The set of values of p for which 6 lie between the roots of the equation \(x^2 + 2(p - 3)x + 9 = 0\) is
(a) \((\frac{3}{4}, \infty)\)
(b) \((-\infty, -\frac{3}{4})\)
(c) \((-\frac{3}{4}, \frac{3}{4})\)
(d) \([-\frac{3}{4}, \frac{3}{4}]\)
Answer: (b) \((-\infty, -\frac{3}{4})\)
Solution: \(af(k) < 0\) and \(\Delta > 0 \Rightarrow 1(36 + 12(p - 3) + 9) < 0 \Rightarrow 45 + 12p - 36 < 0 \Rightarrow 12p < -9 \Rightarrow p < -3/4\).

Question. All the values of m for which both roots of the equation \(x^2 - 2mx + m^2 - 1 = 0\) are greater than -2 but less than 4, lie in the interval
(a) \(-1 < m < 3\)
(b) \(1 < m < 4\)
(c) \(-2 < m < 0\)
(d) \(m > 3\)
Answer: (a) \(-1 < m < 3\)
Solution: Roots of \(x^2 - 2mx + (m^2 - 1) = 0\) are \(\frac{2m \pm \sqrt{4m^2 - 4(m^2 - 1)}}{2} = m \pm 1\). Given \(-2 < m - 1\) and \(m + 1 < 4 \Rightarrow m > -1\) and \(m < 3 \Rightarrow -1 < m < 3\).

Question. If \(a \in R\) and the roots of \(x^2 - 2x - a^2 + 1 = 0\) lies between the roots of \(x^2 - 2(a + 1)x + a(a - 1) = 0\) then ‘a’ belongs to
(a) \((1, \infty)\)
(b) \((-\frac{1}{4}, 1)\)
(c) \((-\infty, 0)\)
(d) \((-\infty, -\frac{1}{4})\)
Answer: (b) \((-\frac{1}{4}, 1)\)
Solution: Roots of \(x^2 - 2x - a^2 + 1 = 0\) are given by \((x-1)^2 = a^2 \Rightarrow x = 1 \pm a\). Let \(f(x) = x^2 - 2(a + 1)x + a(a - 1)\). The roots \(1 \pm a\) lie between roots of \(f(x)=0 \Rightarrow f(1+a) < 0\) and \(f(1-a) < 0\). \(f(1+a) < 0 \Rightarrow a \in (-1/4, 1)\).

SIGN OF THE EXPRESSIONS AND INEQUATIONS

Question. If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than \(\sqrt{5}\), then the set of possible values of a is
(a) \((-3, 3)\)
(b) \((-3, \infty)\)
(c) \((3, \infty)\)
(d) \((-\infty, -3)\)
Answer: (a) \((-3, 3)\)
Solution: \(|\alpha - \beta| < \sqrt{5} \Rightarrow \sqrt{a^2 - 4} < \sqrt{5} \Rightarrow a^2 - 4 < 5 \Rightarrow a^2 < 9 \Rightarrow -3 < a < 3\).

Question. The greatest negative integer satisfying \(x^2 + 4x - 77 < 0\) and \(x^2 > 4\) is
(a) -1
(b) -2
(c) -3
(d) -10
Answer: (c) -3
Solution: \((x+11)(x-7) < 0 \Rightarrow x \in (-11, 7)\). And \(x^2 > 4 \Rightarrow x \in (-\infty, -2) \cup (2, \infty)\). Intersection is \((-11, -2) \cup (2, 7)\). The greatest negative integer in this interval is -3.

Question. The solution set contained in R of the inequation \(3^x + 3^{1 - x} - 4 < 0\) is
(a) \((1, 3)\)
(b) \((0, 1)\)
(c) \((1, 2)\)
(d) \((0, 2)\)
Answer: (b) \((0, 1)\)
Solution: Let \(3^x = t\). \(t + 3/t - 4 < 0 \Rightarrow t^2 - 4t + 3 < 0 \Rightarrow (t - 1)(t - 3) < 0 \Rightarrow 1 < t < 3 \Rightarrow 1 < 3^x < 3 \Rightarrow x \in (0, 1)\).

Question. If the equation \(ax^2 + 2bx - 3c = 0\) has non-real roots and \(\frac{3c}{4} < a + b\) then c is always
(a) < 0
(b) > 0
(c) \(\geq 0\)
(d) zero
Answer: (a) < 0
Solution: Let \(f(x) = ax^2 + 2bx - 3c\). \(\frac{3c}{4} < a + b \Rightarrow 4a + 4b - 3c > 0 \Rightarrow f(2) > 0\). Since roots are non-real, \(f(x)\) has the same sign for all x. Since \(f(2) > 0\), \(f(x) > 0 \forall x \in R\). Thus \(f(0) > 0 \Rightarrow -3c > 0 \Rightarrow c < 0\).

Question. The equation \((\cos p - 1)x^2 + (\cos p)x + (\sin p) = 0\) in the variable x has real roots. Then p can take any value in the interval
(a) \((0, 2\pi)\)
(b) \((-\pi, 0)\)
(c) \((-\frac{\pi}{2}, \frac{\pi}{2})\)
(d) \((0, \pi)\)
Answer: (d) \((0, \pi)\)
Solution: For real roots, \(\Delta \geq 0 \Rightarrow \cos^2 p - 4(\cos p - 1)(\sin p) \geq 0 \Rightarrow (\cos p - 2\sin p)^2 + 4(1 - \sin p)\sin p \geq 0\). Since \((1 - \sin p) \geq 0\), we need \(\sin p \geq 0 \Rightarrow p \in (0, \pi)\).

MAXIMUM AND MINIMUM OF QUADRATIC EXPRESSION

Question. If a and \(b(\neq 0)\) are the roots of the equation \(x^2 + ax + b = 0\) then the least value of \(x^2 + ax + b (x \in R)\) is
(a) \(\frac{9}{4}\)
(b) \(-\frac{9}{4}\)
(c) \(-\frac{1}{4}\)
(d) \(\frac{1}{4}\)
Answer: (b) \(-\frac{9}{4}\)
Solution: Sum of roots \(a + b = -a \Rightarrow b = -2a\). Product of roots \(ab = b \Rightarrow a = 1\) (\(\because b \neq 0\)). Thus \(b = -2\). Expression is \(x^2 + x - 2\). Least value is \(\frac{-D}{4a} = \frac{4(-2) - 1^2}{4(1)} = \frac{-9}{4}\).

Question. If \(x^2 + \frac{1}{x^2} = A\) and \(x - \frac{1}{x} = B\) \((|x|>1)\) then least value of \(\frac{A}{B}\) is
(a) 2
(b) \(\sqrt{2}\)
(c) \(-\sqrt{2}\)
(d) \(2\sqrt{2}\)
Answer: (d) \(2\sqrt{2}\)
Solution: \(\frac{A}{B} = \frac{x^2 + 1/x^2}{x - 1/x} = \frac{(x - 1/x)^2 + 2}{x - 1/x} = (x - 1/x) + \frac{2}{x - 1/x}\). Let \(p = x - 1/x > 0\). Expression is \(p + \frac{2}{p} \geq 2\sqrt{p \cdot 2/p} = 2\sqrt{2}\) (using AM \(\geq\) GM).

Question. If \(\alpha, \beta\) are the roots of \(x^2 - (a - 2)x - (a + 1) = 0\) where ‘a’ is a variable then the least value of \(\alpha^2 + \beta^2\) is
(a) 2
(b) 3
(c) 5
(d) 7
Answer: (c) 5
Solution: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (a - 2)^2 - 2(-(a + 1)) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6 = (a - 1)^2 + 5 \geq 5\).

MODULUS FUNCTIONS

Question. The minimum vlaue of \(|x| + |x + \frac{1}{2}| + |x - 3| + |x - \frac{5}{2}|\) is
(a) 2
(b) 4
(c) 6
(d) 4
Answer: (c) 6
Solution: The minimum of \(f(x)\) is achieved at the median of the critical points \(-1/2, 0, 5/2, 3\). Any x between \([0, 5/2]\) gives the minimum. \(f(0) = 0 + 1/2 + 3 + 5/2 = 6\).

Question. Sum of the roots of the equation \((x - 2)^2 - 2|x - 2| - 15 = 0\) is
(a) 4
(b) 0
(c) -4
(d) 8
Answer: (a) 4
Solution: \((|x - 2| - 5)(|x - 2| + 3) = 0 \Rightarrow |x - 2| = 5 \Rightarrow x - 2 = \pm 5 \Rightarrow x = 7, -3\). Sum = \(7 + (-3) = 4\).

Question. The real roots of the equation \(|x^2 + 4x + 3| + 2x + 5 = 0\) are
(a) \(4; -1 + \sqrt{3}\)
(b) \(-4; -1 - \sqrt{3}\)
(c) \(-6, -1\)
(d) \(6, -1\)
Answer: (b) \(-4; -1 - \sqrt{3}\)
Solution: Substitute options. x = -4 \(\Rightarrow |16 - 16 + 3| - 8 + 5 = 3 - 3 = 0\).

Question. If x satisfies \(|x - 1| + |x - 2| + |x - 3| \geq 6\) then
(a) \(0 \leq x \leq 4\)
(b) \(x \leq -2 (\text{or}) x \geq 4\)
(c) \(x \leq 0 (\text{or}) x \geq 4\)
(d) R
Answer: (c) \(x \leq 0 (\text{or}) x \geq 4\)
Solution: If \(x \geq 3\), \((x - 1) + (x - 2) + (x - 3) \geq 6 \Rightarrow 3x - 6 \geq 6 \Rightarrow x \geq 4\). If \(x \leq 1\), \(-(x - 1) - (x - 2) - (x - 3) \geq 6 \Rightarrow -3x + 6 \geq 6 \Rightarrow x \leq 0\). Thus \(x \leq 0\) or \(x \geq 4\).

MISCELLANEOUS

Question. If \(a = \cos \frac{2\pi}{7} + i\sin \frac{2\pi}{7}\), \(\alpha = a + a^2 + a^4\) and \(\beta = a^3 + a^5 + a^6\) then \(\alpha, \beta\) are roots of The equation
(a) \(x^2 + x + 1 = 0\)
(b) \(x^2 + x + 2 = 0\)
(c) \(x^2 + 2x + 2 = 0\)
(d) \(x^2 + 2x + 3 = 0\)
Answer: (b) \(x^2 + x + 2 = 0\)
Solution: \(1, a, a^2, a^3, a^4, a^5, a^6\) are 7th roots of unity. \(1 + a + a^2 + a^3 + a^4 + a^5 + a^6 = 0 \Rightarrow \alpha + \beta = -1\). \(\alpha\beta = 3 + a + a^2 + a^3 + a^4 + a^5 + a^6 = 2\). The equation is \(x^2 - (\alpha+\beta)x + \alpha\beta = 0 \Rightarrow x^2 + x + 2 = 0\).

Question. The set of values of x for which the inequality \([x]^2 - 5[x] + 6 \leq 0\) (where [.] denote thegreatest integral function) hold good is
(a) \(2 \leq [x] < 3\)
(b) \(2 \leq x < 4\)
(c) \(2 \leq x < 3\)
(d) \(2 \leq x \leq 4\)
Answer: (b) \(2 \leq x < 4\)
Solution: \(([x] - 2)([x] - 3) \leq 0 \Rightarrow 2 \leq [x] \leq 3 \Rightarrow 2 \leq x < 4\).

Question. The number of quadratic equations which are unchanged by squaring their roots is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (b) 4
Solution: The possible root sets are \(\{0, 0\}, \{0, 1\}, \{1, 1\}, \{\omega, \omega^2\}\). Thus there are 4 equations: \(x^2 = 0\), \(x^2 - x = 0\), \(x^2 - 2x + 1 = 0\), \(x^2 + x + 1 = 0\).

Question. If \(\alpha\) is a root of the equation \(4x^2 + 2x - 1 = 0\) then the other root isgiven by
(a) \(-2\alpha, -1\)
(b) \(4\alpha^2 + \alpha - 1\)
(c) \(4\alpha^3 - 3\alpha\)
(d) \(4\alpha^2 - 3\alpha\)
Answer: (c) \(4\alpha^3 - 3\alpha\)
Solution: \(4x^2 + 2x - 1 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{5}}{4}\). Roots are \(\sin 18^\circ\) and \(-\sin 54^\circ\). Since \(\beta = -\sin 54^\circ = -(3\sin 18^\circ - 4\sin^3 18^\circ) = 4\alpha^3 - 3\alpha\).

Question. The least value of expression \(x^2 + 4y^2 + 9z^2 - 2x + 8y + 27z + 15\) is
(a) 15
(b) 5
(c) 0
(d) \(\frac{-41}{4}\)
Answer: (d) \(\frac{-41}{4}\)
Solution: \((x - 1)^2 + 4(y + 1)^2 + 9(z + 3/2)^2 - 41/4 \geq -41/4\).

Question. Let \(f(x) = ax^2 + bx + c\), if \(f(-1) < 1, f(1) > -1, f(3) < -4\) and \(a \neq 0\) then
(a) \(a > 0\)
(b) \(a > 1\)
(c) \(a < -\frac{1}{8}\)
(d) \(-\frac{1}{8} < a < 0\)
Answer: (c) \(a < -\frac{1}{8}\)
Solution: \(a - b + c < 1\), \(a + b + c > -1 \Rightarrow -a - b - c < 1\), \(9a + 3b + c < -4\). Adding equations yields \(12a + 4c < -1\) and \(6a - 2c < -1 \Rightarrow 24a < -3 \Rightarrow a < -1/8\).

Question. The number of distinct pairs \((x, y)\) of real numbers that satisfy the equation \(4x^2 + 4xy + 2y^2 - 2y + 1 = 0\)
(a) 0
(b) 1
(c) 4
(d) infinitely many
Answer: (b) 1
Solution: \((2x + y)^2 + (y - 1)^2 = 0 \Rightarrow 2x + y = 0\) and \(y - 1 = 0 \Rightarrow y = 1, x = -1/2\). Exactly 1 pair.

EQUATIONS REDUCIBLE TO QUADRATIC

Question. The number of solutions of the equation \(x^{\frac{2}{3}} + x^{\frac{1}{3}} - 2 = 0\) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
Solution: Let \(x^{1/3} = t \Rightarrow t^2 + t - 2 = 0 \Rightarrow (t+2)(t-1) = 0 \Rightarrow t=1, -2 \Rightarrow x=1, -8\).

Question. If \(y = 2 + \frac{1}{4 + \frac{1}{4 + \dots \infty}}\) then
(a) \(y = 6\)
(b) \(y = 5\)
(c) \(y = \sqrt{6}\)
(d) \(y = \sqrt{5}\)
Answer: (d) \(y = \sqrt{5}\)
Solution: \(y - 2 = \frac{1}{4 + (y - 2)} \Rightarrow y - 2 = \frac{1}{y + 2} \Rightarrow y^2 - 4 = 1 \Rightarrow y^2 = 5 \Rightarrow y = \sqrt{5}\) (\(y > 0\)).

TRANSFORMED EQUATIONS

Question. If the equation \(ax^2 + bx + c = 0\) is not altered when each of the coefficient is increased by the same quantity then \(x^2 + x + 1 =\)
(a) 1
(b) 0
(c) 3
(d) 2
Answer: (b) 0
Solution: Equation is invariant \Rightarrow \frac{a+k}{a} = \frac{b+k}{b} = \frac{c+k}{c} \Rightarrow a=b=c. The equation reduces to \(x^2 + x + 1 = 0\).

Question. If \(\alpha\) and \(\beta\) are the roots of the equation \(ax^2 + bx + c = 0\) and if \(px^2 + qx + r = 0\) has roots \(\frac{1 - \alpha}{\alpha}\) and \(\frac{1 - \beta}{\beta}\) then r =
(a) \(a + 2b\)
(b) \(a + b + c\)
(c) \(ab + bc + ca\)
(d) abc
Answer: (b) \(a + b + c\)
Solution: Let \(y = \frac{1 - x}{x} \Rightarrow x = \frac{1}{y + 1}\). Substituting in \(ax^2 + bx + c = 0 \Rightarrow a\left(\frac{1}{y + 1}\right)^2 + b\left(\frac{1}{y + 1}\right) + c = 0 \Rightarrow cy^2 + (b + 2c)y + (a + b + c) = 0\). Comparing with \(px^2+qx+r=0\), the constant term \(r = a + b + c\).

RANGE

Question. Given that, for all real x, the expression \(\frac{x^2 - 2x + 4}{x^2 + 2x + 4}\) lies between \(\frac{1}{3}\) and 3. The values between which the expression \(\frac{9.3^{2x} + 6.3^x + 4}{9.3^{2x} - 6.3^x + 4}\) lies are
(a) \(\frac{1}{3}\) and 3
(b) -2 and 0
(c) -1 and 1
(d) 0 and 2
Answer: (a) \(\frac{1}{3}\) and 3
Solution: Let \(t = 3 \cdot 3^x > 0\). The expression becomes \(\frac{t^2 + 2t + 4}{t^2 - 2t + 4}\). The range of \(\frac{t^2 + 2t + 4}{t^2 - 2t + 4}\) for \(t \in (0, \infty)\) is [1/3, 3].

Question. Let \(f(x) = (1 + b^2)x^2 + 2bx + 1\) and \(m(b)\) be the minimum value of \(f(x)\). As b varies, the range of \(m(b)\) is
(a) \([0, 1]\)
(b) \([0, \frac{1}{2}]\)
(c) \([\frac{1}{2}, 1]\)
(d) \((0, 1]\)
Answer: (d) \((0, 1]\)
Solution: Minimum value of \(f(x)\) is \(m(b) = \frac{4(1 + b^2)(1) - (2b)^2}{4(1 + b^2)} = \frac{4}{4(1 + b^2)} = \frac{1}{1 + b^2}\). As \(b \in R, 1 + b^2 \geq 1\), so \(m(b) \in (0, 1]\).

INEQUALITIES

Question. If x is real and \(\frac{x - 1}{4x + 5} < \frac{x - 3}{4x - 3}\) then x lies in the interval
(a) \((-\frac{3}{4}, \frac{5}{4})\)
(b) \((-\frac{5}{4}, -\frac{3}{4})\)
(c) \((-\frac{5}{4}, \frac{3}{4})\)
(d) \((-\frac{3}{4}, \frac{7}{5})\)
Answer: (c) \((-\frac{5}{4}, \frac{3}{4})\)
Solution: \(\frac{x - 1}{4x + 5} - \frac{x - 3}{4x - 3} < 0 \Rightarrow \frac{(x - 1)(4x - 3) - (x - 3)(4x + 5)}{(4x + 5)(4x - 3)} < 0 \Rightarrow \frac{18}{(4x + 5)(4x - 3)} < 0 \Rightarrow (4x + 5)(4x - 3) < 0 \Rightarrow x \in (-5/4, 3/4)\).

Question. If \((\log_5 x)^2 + \log_5 x < 2\) then x belongs to the interval
(a) \((\frac{1}{25}, 5)\)
(b) \((\frac{1}{5}, \frac{1}{\sqrt{5}})\)
(c) \((1, \infty)\)
(d) \((5, 25)\)
Answer: (a) \((\frac{1}{25}, 5)\)
Solution: Let \(\log_5 x = t\). \(t^2 + t - 2 < 0 \Rightarrow (t + 2)(t - 1) < 0 \Rightarrow -2 < t < 1 \Rightarrow 5^{-2} < x < 5^1 \Rightarrow 1/25 < x < 5\).

Question. If \((x - 1)(x - 2)(x + 5) < 0\) then
(a) \(x < -3; 0 < x < 2\)
(b) \(x < -5; 1 < x < 2\)
(c) \(x > 2; -5 < x < 1\)
(d) \(x < 1; -5 < x < 2\)
Answer: (b) \(x < -5; 1 < x < 2\)
Solution: Wavy curve method gives \(x \in (-\infty, -5) \cup (1, 2)\).

OTHER MODELS

Question. If \(x^3 + ax + 1 = 0\) and \(x^4 + ax^2 + 1 = 0\) have common root then the exhaustive set of values of a is
(a) \((-\infty, -2)\)
(b) \([-2, \infty)\)
(c) \(\{-2\}\)
(d) \([-2, 2]\)
Answer: (c) \(\{-2\}\)
Solution: Let \(\alpha\) be common root: \(\alpha^3 + a\alpha + 1 = 0 \Rightarrow \alpha^4 + a\alpha^2 + \alpha = 0\). Subtracting \(\alpha^4 + a\alpha^2 + 1 = 0\) gives \(\alpha - 1 = 0 \Rightarrow \alpha = 1\). Sub into equation \(\Rightarrow 1 + a + 1 = 0 \Rightarrow a = -2\).

Question. If x is real then the value of \(\frac{x^2 - 3x + 4}{x^2 + 3x + 4}\) lies in the interval
(a) \([\frac{1}{3}, 3]\)
(b) \([\frac{1}{5}, 5]\)
(c) \([\frac{1}{6}, 6]\)
(d) \([\frac{1}{7}, 7]\)
Answer: (d) \([\frac{1}{7}, 7]\)
Solution: Let \(y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4} \Rightarrow x^2(y - 1) + 3x(y + 1) + 4(y - 1) = 0\). Since x is real, \(\Delta \geq 0 \Rightarrow 9(y + 1)^2 - 16(y - 1)^2 \geq 0 \Rightarrow (7y - 1)(-y + 7) \geq 0 \Rightarrow y \in [1/7, 7]\).

Question. In \(\Delta PQR\), \(\angle R = \frac{\pi}{4}\), \(\tan(\frac{P}{3})\), \(\tan(\frac{Q}{3})\) are the roots of the equation \(ax^2 + bx + c = 0\), then
(a) \(a + b = c\)
(b) \(b + c = 0\)
(c) \(a + c = 0\)
(d) \(b = c\)
Answer: (a) \(a + b = c\)
Solution: \(P+Q = \pi - \pi/4 = 3\pi/4 \Rightarrow \frac{P+Q}{3} = \frac{\pi}{4}\). \(\tan\left(\frac{P}{3} + \frac{Q}{3}\right) = 1 \Rightarrow \frac{\tan(P/3) + \tan(Q/3)}{1 - \tan(P/3)\tan(Q/3)} = 1 \Rightarrow \frac{-b/a}{1 - c/a} = 1 \Rightarrow -b = a - c \Rightarrow a + b = c\).

Question. If the harmonic mean between the roots of \((5 + \sqrt{2})x^2 - bx + (8 + 2\sqrt{5}) = 0\) is 4, then the value of b is 
(a) 2
(b) 3
(c) \(4 - \sqrt{5}\)
(d) \(4 + \sqrt{5}\)
Answer: (d) \(4 + \sqrt{5}\)
Solution: \(H.M = \frac{2\alpha\beta}{\alpha + \beta} = \frac{2(c/a)}{b/a} = \frac{2c}{b'}\) (where b' is coeff of x). \(4 = \frac{2(8 + 2\sqrt{5})}{b} \Rightarrow b = \frac{8 + 2\sqrt{5}}{2} = 4 + \sqrt{5}\).