CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations Competitive Exam MCQs Set 02

Answer: (a) \( |z| = \frac{3}{2} \)

Explanation:
1. Group: 4z²(2iz + 3) + 9i(2iz + 3) = 8iz³ + 12z² − 18z + 27i, using 9i · 2iz = −18z.
2. So (2iz + 3)(4z² + 9i) = 0.
3. 2iz + 3 = 0 gives z = 3i/2, so |z| = 3/2.
4. 4z² = −9i gives |z|² = 9/4, so |z| = 3/2.

Teacher's Note:
1. Group the terms in pairs and look for a common factor such as 2iz + 3.
2. For z² = w you only need |z|² = |w| to find |z|; there is no need to find z itself.

Answer: (c) \( 2\sqrt{13} \)

Explanation:
1. By the triangle inequality, |z − 1 + 2i| + |4i − 3 − z| ≥ |(z − 1 + 2i) + (4i − 3 − z)|.
2. = |−4 + 6i| = \( \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \).
3. Equality holds when z lies on the segment joining 1 − 2i and −3 + 4i, so the minimum is \( 2\sqrt{13} \).

Teacher's Note:
1. Arrange the two moduli so that z cancels when they are added inside one modulus.
2. The sum of distances from z to two fixed points is least when z lies between them.

Answer: (a) purely real

Explanation:
1. Divide numerator and denominator by z₂ and let w = z₁/z₂ = x + iy: |w + i| = |w − i|.
2. x² + (y + 1)² = x² + (y − 1)² gives 4y = 0, so y = 0.
3. Hence z₁/z₂ is purely real.

Teacher's Note:
1. Dividing by z₂ turns a two-variable condition into one about the single number w = z₁/z₂.
2. |w + i| = |w − i| means w is equally far from i and −i, which happens only on the real axis.

Answer: (a) 4

Explanation:
1. |1 − z| ≤ |1| + |z| = 2.
2. |1 + z²| ≤ 1 + |z|² = 2.
3. So the sum ≤ 4; at z = −1 it equals 2 + 2 = 4, so the expression is always ≤ 4.

Teacher's Note:
1. Apply the triangle inequality to each modulus separately and add.
2. Check that the bound is reached (here at z = −1); smaller numbers in the options would then be false.

Answer: (c) \( \sqrt{5} - 2 \)

Explanation:
1. Let r = |z|. By the triangle inequality, |z² − 1| ≥ 1 − |z|².
2. So 4r ≥ 1 − r², i.e. r² + 4r − 1 ≥ 0, giving r ≥ −2 + √5 (r ≥ 0).
3. z = √5 − 2 (real) gives |z² − 1| = 1 − z² = 4√5 − 8 = 4|z|, so the minimum value of |z| is √5 − 2.

Teacher's Note:
1. Use |z² − 1| ≥ 1 − |z|² to get a lower bound on r = |z|, then solve the quadratic inequality in r.
2. For a real z between 0 and 1, |z² − 1| = 1 − z², which is how the bound is attained.

Answer: (d) infinitely many

Explanation:
1. Let z = x + iy: x² − y² + 2ixy + x² + y² = 0.
2. So 2x² + 2ixy = 0, giving x² = 0 and xy = 0, i.e. x = 0.
3. Every z = iy (y real) satisfies the equation, so there are infinitely many solutions.

Teacher's Note:
1. Separate real and imaginary parts and set each equal to zero.
2. All purely imaginary numbers satisfy z² = −|z|², which is why the count is infinite.

Answer: (a) max |2z₁ + z₂| = 4; (b) min |z₁ − z₂| = 1; (c) \( \left| z_2 + \frac{1}{z_1} \right| \le 3 \)

Explanation:
1. |2z₁ + z₂| ≤ 2|z₁| + |z₂| = 4, attained when z₂ = 2z₁; so (a) is true.
2. |z₁ − z₂| ≥ |z₂| − |z₁| = 1, attained when z₂ = 2z₁; so (b) is true.
3. \( \left| z_2 + \frac{1}{z_1} \right| \le |z_2| + \frac{1}{|z_1|} = 2 + 1 = 3 \); so (c) is true.

Teacher's Note:
1. The triangle inequality gives both the largest value |a| + |b| and the smallest value ||a| − |b|| of |a ± b|.
2. Equality needs the two numbers to point in the same (or opposite) direction, which is possible here.

Answer: (a) \( \bar{z}_1, \bar{z}_2, \bar{z}_3, \bar{z}_4 \) are also roots of the equation; (b) z₁ is equal to at least one of \( \bar{z}_1, \bar{z}_2, \bar{z}_3, \bar{z}_4 \)

Explanation:
1. Taking the conjugate of a₀z_k⁴ + ... + a₄ = 0 with real coefficients gives a₀\( \bar{z}_k \)⁴ + ... + a₄ = 0, so (a) is true.
2. So \( \bar{z}_1 \) is a root, i.e. \( \bar{z}_1 = z_k \) for some k; then \( z_1 = \bar{z}_k \), so (b) is true.
3. (c) is false in general: z⁴ − z = 0 has the root 1, but −1 is not a root (1 + 1 = 2 ≠ 0).

Teacher's Note:
1. For a polynomial with real coefficients, non-real roots occur in conjugate pairs.
2. Replacing z by −z̄ changes the signs of odd-degree terms, so −z̄ is not a root in general.

Answer: (a) \( \alpha + \bar{\alpha} = 1 \); (c) \( \alpha + \bar{\alpha} = -1 \); (d) the absolute value of the real root is 1

Explanation:
1. Let x be the real root: αx² + x + \( \bar{\alpha} \) = 0. Conjugating, \( \bar{\alpha} \)x² + x + α = 0.
2. Subtracting: (α − \( \bar{\alpha} \))(x² − 1) = 0; α is non-real, so x² = 1, i.e. |x| = 1 and (d) holds.
3. Adding: (α + \( \bar{\alpha} \))(x² + 1) + 2x = 0, so α + \( \bar{\alpha} \) = −x = −1 (x = 1) or 1 (x = −1).
4. Hence (a), (c) and (d) can be true, but α + \( \bar{\alpha} \) = 0 is impossible.

Teacher's Note:
1. When an equation has a real root, take its conjugate and add or subtract the two equations.
2. Here α + ᾱ = 2Re(α), so Re(α) must be ±1/2.

Answer: (b) p² and −2

Explanation:
1. Split the middle term: x² + 2x − p²x − 2p² = 0.
2. Group: x(x + 2) − p²(x + 2) = 0, so (x + 2)(x − p²) = 0.
3. Hence the roots are p² and −2.

Teacher's Note:
1. Look for two numbers whose sum is the coefficient of x and whose product is the constant term; here 2 and −p² work.
2. Check quickly: sum of roots p² − 2 = −(2 − p²) and product −2p² match the equation.

Answer: (a) Rational and different

Explanation:
1. D = b² − 4ac = 81 − 56 = 25, which is positive and a perfect square.
2. So the roots are real, rational and unequal.
3. Indeed 7x² − 9x + 2 = (7x − 2)(x − 1), giving x = 1 and x = \( \frac{2}{7} \).

Teacher's Note:
1. With rational coefficients, D > 0 and a perfect square means rational, unequal roots.
2. If the sum of the coefficients is zero (7 − 9 + 2 = 0), x = 1 is always a root.

Answer: (a) Real and different

Explanation:
1. Here a = 1, b = \( -2\sqrt{2} \), c = 1.
2. D = 8 − 4 = 4 > 0, so the roots are real and different.
3. x = \( \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1 \), which are irrational, so (d) is wrong.

Teacher's Note:
1. D being a perfect square guarantees rational roots only when the coefficients themselves are rational.
2. Always find the actual roots when an option mentions rational or irrational.

Answer: (a) x² − 4x + 7 = 0

Explanation:
1. For real coefficients, complex roots occur in conjugate pairs, so the other root is \( 2 + i\sqrt{3} \).
2. Sum of roots = 4, product of roots = 4 − (i√3)² = 4 + 3 = 7.
3. Equation: x² − (sum)x + product = 0, i.e. x² − 4x + 7 = 0.

Teacher's Note:
1. The product (p + iq)(p − iq) = p² + q² is always real and positive.
2. Check: D = 16 − 28 = −12 < 0, so the equation indeed has non-real roots 2 ± i√3.

Answer: (b) x² − 2x + 2 = 0

Explanation:
1. With real coefficients, 1 − i must also be a root, so the degree is at least 2.
2. Sum = 2, product = (1 + i)(1 − i) = 1 + 1 = 2.
3. Required equation: x² − 2x + 2 = 0.

Teacher's Note:
1. A polynomial with real coefficients cannot have just one non-real root; its conjugate comes with it.
2. Verify by substitution: (1 + i)² − 2(1 + i) + 2 = 2i − 2 − 2i + 2 = 0.

Answer: (c) \( \frac{2c}{-b \pm \sqrt{b^2 - 4ac}} \)

Explanation:
1. The quadratic formula gives x = \( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), so (a) and (b) are wrong.
2. Multiply numerator and denominator by \( -b \mp \sqrt{b^2 - 4ac} \): the numerator becomes b² − (b² − 4ac) = 4ac.
3. So x = \( \frac{4ac}{2a(-b \mp \sqrt{b^2 - 4ac})} = \frac{2c}{-b \mp \sqrt{b^2 - 4ac}} \); taking both signs gives the same two roots as option (c).

Teacher's Note:
1. Rationalising the numerator of the quadratic formula gives an equivalent form with 2c on top.
2. This form is handy when a is very small, since it avoids dividing by 2a.

Answer: (c) \( \frac{r-p}{p-q}, 1 \)

Explanation:
1. Sum of coefficients: (p − q) + (q − r) + (r − p) = 0, so x = 1 is a root.
2. Product of roots = \( \frac{r-p}{p-q} \).
3. Hence the other root is \( \frac{r-p}{p-q} \), and the roots are \( \frac{r-p}{p-q} \) and 1.

Teacher's Note:
1. Whenever a + b + c = 0 in ax² + bx + c = 0, one root is 1 and the other is c/a.
2. Spotting this saves you from using the long quadratic formula.

Answer: (d) None of these

Explanation:
1. Multiply by x (x ≠ 0): x² + 1 = 2x, i.e. x² − 2x + 1 = 0.
2. (x − 1)² = 0, so x = 1 (repeated).
3. x = 1 is not in options (a), (b) or (c), so the answer is (d).

Teacher's Note:
1. Clearing fractions turns the equation into a quadratic; remember x = 0 is not allowed.
2. D = 4 − 4 = 0 tells you at once that the two roots are equal.

Answer: (a) 1 ± i

Explanation:
1. Put t = x² + 2: t² − 6xt + 8x² = 0, i.e. (t − 2x)(t − 4x) = 0.
2. t = 4x gives x² − 4x + 2 = 0 with real roots 2 ± √2.
3. t = 2x gives x² − 2x + 2 = 0, so x = \( \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i \).
4. Hence the imaginary roots are 1 ± i.

Teacher's Note:
1. Treating x² + 2 as a single quantity turns the equation into a quadratic that factorises.
2. Check by substituting x = 1 + i: both sides equal 24i.

Answer: (b) Rational

Explanation:
1. D = 9 − 8 = 1, a positive perfect square.
2. 2x² + 3x + 1 = (2x + 1)(x + 1) = 0, so x = −1 or x = \( -\frac{1}{2} \).
3. Both roots are rational.

Teacher's Note:
1. For rational coefficients, a perfect-square discriminant always gives rational roots.
2. Factorising is the quickest way to confirm the nature of roots for small coefficients.

Answer: (c) Irrational and equal

Explanation:
1. D = (2√3)² − 4 × 3 = 12 − 12 = 0, so the roots are equal.
2. x² + 2√3x + 3 = (x + √3)², so both roots are −√3.
3. −√3 is irrational, so the roots are irrational and equal (the most precise option).

Teacher's Note:
1. D = 0 tells you the roots are equal; the value −b/2a then tells you whether they are rational.
2. Option (a) is true but less specific; always choose the option that describes the roots most exactly.

Answer: (b) (−4, 7)

Explanation:
1. p, q are real, so the other root is \( 2 - i\sqrt{3} \).
2. Sum of roots = 4 = −p, so p = −4.
3. Product of roots = 4 + 3 = 7 = q, so (p, q) = (−4, 7).

Teacher's Note:
1. Sum of roots is −p, not p; this sign is where most mistakes happen.
2. Product of conjugate roots (a + ib)(a − ib) equals a² + b².

Answer: (d) x² − 14x + 74 = 0

Explanation:
1. The other root is the conjugate 7 − 5i.
2. Sum = 14, product = 49 + 25 = 74.
3. Equation: x² − 14x + 74 = 0.

Teacher's Note:
1. The product of a complex number and its conjugate is the square of its modulus: |7 + 5i|² = 74.
2. The constant term must be positive here, which rules out options (a) and (c) at once.

Answer: (a) a = 25, b = −8

Explanation:
1. \( \frac{1}{4 - 3i} = \frac{4 + 3i}{16 + 9} = \frac{4 + 3i}{25} \); as a, b are real, the other root is \( \frac{4 - 3i}{25} \).
2. Sum of roots = \( \frac{8}{25} = -\frac{b}{a} \) and product = \( \frac{16 + 9}{625} = \frac{1}{25} = \frac{1}{a} \).
3. So a = 25 and b = −8.

Teacher's Note:
1. Write the root in the form p + iq first by multiplying by the conjugate of the denominator.
2. Here the product of roots is 1/a because the constant term is 1.

Answer: (d) 2

Explanation:
1. Taking a, b real, the other root is 1 + i.
2. Product of roots = (1 − i)(1 + i) = 1 + 1 = 2 = b.
3. Hence b = 2 (and the sum gives a = 2).

Teacher's Note:
1. In x² − ax + b = 0 the sum of roots is a and the product is b.
2. Remember i² = −1, so (1 − i)(1 + i) = 1 − i² = 2.