JEE Mathematics Quadratic Equations MCQs Set 08

Answer: (a) \( (2, 3) \)

Explanation:
1. \( x^2 + x + 1 \) has \( D = -3 < 0 \) and is always positive.
2. So the sign is that of \( x^2 - 5x + 6 = (x-2)(x-3) \), which is negative for \( 2 < x < 3 \).

Teacher's Note:
1. A quadratic with D < 0 and a > 0 can be ignored in sign analysis.
2. Strict inequality: end points are excluded.

Answer: (d) \( (-1, 0) \cup (0, 1) \cup (1, 2) \)

Explanation:
1. The even powers \( (x-1)^2 \) and \( x^4 \) are positive except at \( x = 1 \) and \( x = 0 \). So the sign is that of \( \dfrac{x+1}{x-2} \).
2. \( \dfrac{x+1}{x-2} < 0 \) for \( -1 < x < 2 \).
3. Remove \( x = 0 \) (denominator zero) and \( x = 1 \) (numerator zero, value 0 not < 0).
4. Solution: \( (-1, 0) \cup (0, 1) \cup (1, 2) \).

Teacher's Note:
1. Replace each even power by 'positive but exclude its zero'.
2. Zeros of the numerator are excluded in a strict inequality.

Answer: (b) \( (1, 2) \cup (7, \infty) \)

Explanation:
1. \( \dfrac{2(x-4)(x-2) - (x-1)(x-7)}{(x-1)(x-7)(x-2)} \ge 0 \).
2. The numerator is \( 2x^2 - 12x + 16 - x^2 + 8x - 7 = x^2 - 4x + 9 = (x-2)^2 + 5 > 0 \).
3. So the sign is that of \( (x-1)(x-2)(x-7) \); we need it positive: \( x \in (1, 2) \cup (7, \infty) \).
4. The numerator is never zero, so no extra points are included.

Teacher's Note:
1. After combining the fractions, check whether the numerator quadratic has real roots.
2. If the numerator is always positive, only the denominator decides the sign.

Answer: (d) \( (-7, -4) \cup (-4, 1) \)

Explanation:
1. \( |x+4| > 0 \) for \( x \ne -4 \), so the sign is that of \( x^2 + 6x - 7 = (x+7)(x-1) \).
2. \( (x+7)(x-1) < 0 \) gives \( -7 < x < 1 \), and \( x = -4 \) must be excluded.
3. Solution: \( (-7, -4) \cup (-4, 1) \).

Teacher's Note:
1. A modulus in the denominator is positive wherever it is defined.
2. Exclude the point where the denominator vanishes.

Answer: (c) \( (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2}) \)

Explanation:
1. \( 2x^2 + 5x + 2 = (2x+1)(x+2) \). Moving everything to the left: \( \dfrac{2x(x+1) - (2x+1)(x+2)}{(2x+1)(x+2)(x+1)} > 0 \).
2. Numerator \( = 2x^2 + 2x - 2x^2 - 5x - 2 = -(3x + 2) \), so we need \( \dfrac{3x+2}{(2x+1)(x+2)(x+1)} < 0 \).
3. Critical points \( -2, -1, -\frac{2}{3}, -\frac{1}{2} \): the expression is positive for \( x > -\frac{1}{2} \), negative on \( (-\frac{2}{3}, -\frac{1}{2}) \), positive on \( (-1, -\frac{2}{3}) \), negative on \( (-2, -1) \), positive for \( x < -2 \).
4. Solution: \( (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2}) \).

Teacher's Note:
1. Factorise the quadratic in the denominator and put everything over a common denominator.
2. Order the critical points on the number line before assigning signs.

Answer: (b) \( (0, \frac{1}{3}) \)

Explanation:
1. The roots are \( 3a \) and \( a + 3 \). Since the expression is negative between its roots, \( [1, 3] \) must lie strictly between them.
2. Case \( 3a < a + 3 \) (i.e. \( a < \frac{3}{2} \)): need \( 3a < 1 \) and \( a + 3 > 3 \), so \( 0 < a < \frac{1}{3} \).
3. Case \( 3a > a + 3 \): need \( a + 3 < 1 \) and \( 3a > 3 \), which is impossible.
4. So \( a \in (0, \frac{1}{3}) \).

Teacher's Note:
1. For a quadratic with positive leading coefficient, f(x) < 0 only between the roots.
2. Test the end points x = 1 and x = 3: both must give a negative value.

Answer: (b) \( x = -2 \) and \( x = 1 \)

Explanation:
1. We need \( 5x + 16 \ge 0 \), i.e. \( x \ge -\frac{16}{5} \).
2. Case 1: \( 3x^2 + 12x + 6 = 5x + 16 \) gives \( 3x^2 + 7x - 10 = (3x + 10)(x - 1) = 0 \), so \( x = 1 \) or \( x = -\frac{10}{3} \); the latter is less than \( -\frac{16}{5} \) and is rejected.
3. Case 2: \( -(3x^2 + 12x + 6) = 5x + 16 \) gives \( 3x^2 + 17x + 22 = (3x + 11)(x + 2) = 0 \), so \( x = -2 \) or \( x = -\frac{11}{3} \); the latter is rejected.
4. Solutions: \( x = -2 \) and \( x = 1 \).

Teacher's Note:
1. The right-hand side of |A| = B must be non-negative.
2. Solve both cases A = B and A = -B and test each root against B ≥ 0.

Answer: (a) \( -5 - \sqrt{3} \)

Explanation:
1. Case \( x^2 + 4x + 3 \ge 0 \) (i.e. \( x \le -3 \) or \( x \ge -1 \)): \( x^2 + 6x + 8 = 0 \) gives \( x = -2, -4 \); only \( x = -4 \) satisfies the case condition.
2. Case \( x^2 + 4x + 3 < 0 \) (\( -3 < x < -1 \)): \( -(x^2 + 4x + 3) + 2x + 5 = 0 \) gives \( x^2 + 2x - 2 = 0 \), so \( x = -1 \pm \sqrt{3} \); only \( x = -1 - \sqrt{3} \approx -2.73 \) lies in \( (-3, -1) \).
3. The solutions are \( -4 \) and \( -1 - \sqrt{3} \), whose sum is \( -5 - \sqrt{3} \).

Teacher's Note:
1. Split the modulus into the two cases and check each root against its case condition.
2. Verify: x = -4 gives 3 - 8 + 5 = 0.

Answer: (a) 1210

Explanation:
1. From the first equation \( a + b = 10c \), so \( b = 10c - a \); from the second \( c + d = 10a \), so \( d = 10a - c \).
2. Since \( a \) satisfies the first equation, \( a^2 - 10ac - 11d = 0 \); since \( c \) satisfies the second, \( c^2 - 10ac - 11b = 0 \).
3. Substitute \( d \) and \( b \): \( a^2 - 10ac - 110a + 11c = 0 \) and \( c^2 - 10ac - 110c + 11a = 0 \).
4. Subtracting: \( (a^2 - c^2) - 121(a - c) = (a - c)(a + c - 121) = 0 \). As \( a \ne c \), \( a + c = 121 \).
5. \( a + b + c + d = 10c + 10a = 10(a + c) = 1210 \).

Teacher's Note:
1. Use sum of roots to express b and d in terms of a and c.
2. Plug a into its own equation and c into its own equation, then subtract to get a + c.

Answer: (d) \( \frac{2}{9}(2p - q)(2q - p) \)

Explanation:
1. \( \alpha + \beta = p \) and \( \frac{\alpha}{2} + 2\beta = q \).
2. Eliminating \( \alpha \): \( \frac{3}{2}\beta = q - \frac{p}{2} \), so \( \beta = \frac{2q - p}{3} \) and \( \alpha = p - \beta = \frac{4p - 2q}{3} \).
3. \( r = \alpha\beta = \frac{(4p - 2q)(2q - p)}{9} = \frac{2}{9}(2p - q)(2q - p) \).

Teacher's Note:
1. Solve the two linear equations for α and β.
2. The product of the roots of both equations equals r, so r = αβ.

Answer: (a) \( p^3 - (3p - 1)q + q^2 = 0 \)

Explanation:
1. Let the roots be \( \alpha \) and \( \alpha^2 \). Then \( \alpha + \alpha^2 = -p \) and \( \alpha^3 = q \).
2. Cube the first relation: \( (\alpha + \alpha^2)^3 = \alpha^3 + \alpha^6 + 3\alpha^3(\alpha + \alpha^2) = q + q^2 - 3pq \).
3. So \( -p^3 = q + q^2 - 3pq \), i.e. \( p^3 - 3pq + q + q^2 = p^3 - (3p - 1)q + q^2 = 0 \).

Teacher's Note:
1. Take the roots as α and α² and use sum and product.
2. Cubing the sum brings α³ and α⁶ into play, both expressible in q.

Answer: (d) \( \frac{5}{4} \)

Explanation:
1. \( x_1 + x_2 = 2m \) and \( x_1x_2 = m \).
2. \( x_1^3 + x_2^3 = 8m^3 - 6m^2 \) and \( x_1^2 + x_2^2 = 4m^2 - 2m \).
3. Equating: \( 8m^3 - 10m^2 + 2m = 2m(4m - 1)(m - 1) = 0 \), so \( m = 0, \frac{1}{4}, 1 \).
4. The question does not require real roots (for \( m = \frac{1}{4} \) the roots are a complex pair, but the sum and product relations used above still hold), so all three values are admitted.
5. Sum \( = 0 + \frac{1}{4} + 1 = \frac{5}{4} \).

Teacher's Note:
1. Express both sides in terms of the sum and product of roots.
2. Factorise the resulting cubic in m.

Answer: (b) \( 2\alpha + \dfrac{1}{\beta},\ 2\beta + \dfrac{1}{\alpha} \)

Explanation:
1. The equation is \( ax^2 + 2bx - a = 0 \), so \( \alpha\beta = -1 \), i.e. \( \dfrac{1}{\beta} = -\alpha \) and \( \dfrac{1}{\alpha} = -\beta \).
2. Then \( 2\alpha + \dfrac{1}{\beta} = 2\alpha - \alpha = \alpha \) and \( 2\beta + \dfrac{1}{\alpha} = 2\beta - \beta = \beta \).
3. So the pair in option (b) is simply \( \alpha, \beta \), the roots themselves.
4. The other pairs do not reduce to the roots (for example \( \alpha + \dfrac{1}{\beta} = 0 \)).

Teacher's Note:
1. Product of roots is -a/a = -1, so 1/β = -α.
2. Substitute this relation into each option.

Answer: (d) harmonic progression

Explanation:
1. The coefficients add up to \( p(q-r) + q(r-p) + r(p-q) = 0 \), so \( x = 1 \) is a root. Equal roots mean both roots are 1.
2. Product of roots: \( \dfrac{r(p-q)}{p(q-r)} = 1 \), so \( rp - rq = pq - pr \), i.e. \( 2pr = q(p + r) \).
3. Dividing by \( pqr \): \( \dfrac{2}{q} = \dfrac{1}{p} + \dfrac{1}{r} \), which is the condition for an HP.

Teacher's Note:
1. If the coefficients add to zero, 1 is a root.
2. Use the product of roots to get a relation among p, q, r.

Answer: (b) \( \alpha^2\beta,\ \alpha\beta^2 \)

Explanation:
1. Put \( x = \dfrac{c}{a}t \): \( a^3 \cdot \dfrac{c^2}{a^2}t^2 + abc \cdot \dfrac{c}{a}t + c^3 = c^2(at^2 + bt + c) = 0 \).
2. So \( t = \alpha \) or \( \beta \) and \( x = \dfrac{c}{a}t = \alpha\beta\,t \).
3. The roots are \( \alpha^2\beta \) and \( \alpha\beta^2 \).

Teacher's Note:
1. Look for a substitution that turns the new equation into the original one.
2. c/a = αβ.

Answer: (c) \( (q - s)^2 + (p - r)(ps - qr) \)

Explanation:
1. \( (\alpha - \gamma)(\beta - \gamma) = \gamma^2 + p\gamma + q \) and likewise for \( \delta \), so the product is \( (\gamma^2 + p\gamma + q)(\delta^2 + p\delta + q) \).
2. Using \( \gamma^2 = -r\gamma - s \): \( \gamma^2 + p\gamma + q = (p - r)\gamma + (q - s) \), similarly for \( \delta \).
3. Product \( = (p-r)^2\gamma\delta + (p-r)(q-s)(\gamma + \delta) + (q-s)^2 = (p-r)^2 s - r(p-r)(q-s) + (q-s)^2 \).
4. \( (p-r)[(p-r)s - r(q-s)] = (p-r)(ps - qr) \), so the value is \( (q-s)^2 + (p-r)(ps - qr) \).

Teacher's Note:
1. Replace (α-γ)(β-γ) by the polynomial value at γ.
2. Reduce γ² and δ² using the second equation.

Answer: (d) \( p = 1,\ q = 2 \)

Explanation:
1. Both roots are infinite when \( a = 0 \), \( b = 0 \) and \( c \ne 0 \).
2. \( 2p - q = 0 \) and \( p - 1 = 0 \) give \( p = 1 \) and \( q = 2 \); \( c = 5 \ne 0 \).

Teacher's Note:
1. Both roots infinite means the equation reduces to a non-zero constant = 0.
2. Set a = 0 and b = 0.

Answer: (b) (i) and (ii) only

Explanation:
1. A function is symmetric if \( f(\alpha, \beta) = f(\beta, \alpha) \).
2. (i) \( \beta^2\alpha + \beta\alpha^2 \) is the same, so symmetric.
3. (ii) \( \cos(\beta - \alpha) = \cos(\alpha - \beta) \), so symmetric.
4. (iii) \( \sin(\beta - \alpha) = -\sin(\alpha - \beta) \), not symmetric.
5. (iv) \( \beta^2 - \alpha \ne \alpha^2 - \beta \) in general, not symmetric.

Teacher's Note:
1. Interchange α and β and compare.
2. Only symmetric functions can be written in terms of α + β and αβ.

Answer: (d) \( \dfrac{1}{8} \)

Explanation:
1. \( 16y^2 = -x^2 + 3x - 2 = (x - 1)(2 - x) \), which needs \( x \in [1, 2] \).
2. On \( [1, 2] \), \( (x-1)(2-x) \) has maximum \( \dfrac{1}{4} \) at \( x = \dfrac{3}{2} \).
3. So \( 16y^2 \le \dfrac{1}{4} \), \( |y| \le \dfrac{1}{8} \) and the maximum of \( y \) is \( \dfrac{1}{8} \) (with \( x \in [1, 2] \)).

Teacher's Note:
1. Treat the equation as a quadratic in x: real x needs D ≥ 0 which bounds y.
2. Discriminant: 9 - 4(16y² + 2) ≥ 0 gives y² ≤ 1/64.

Answer: (a) \( k = -\dfrac{5}{4} \) or \( k = -\dfrac{7}{4} \)

Explanation:
1. The first equation has roots \( x = 1 \) and \( x = 2 \).
2. If \( x = 1 \) is common: \( 3 + 4k + 2 = 0 \), so \( k = -\dfrac{5}{4} \).
3. If \( x = 2 \) is common: \( 12 + 8k + 2 = 0 \), so \( k = -\dfrac{7}{4} \).

Teacher's Note:
1. When one equation factorises, test each root in the other equation.
2. Both choices give a common root, so both values of k are valid.

Answer: (d) \( (-4, 5) \)

Explanation:
1. The roots of \( x^2 - 4x + 5 = 0 \) are \( 2 \pm i \), which are non-real.
2. A real quadratic with one non-real root also has its conjugate as a root, so both roots are common.
3. Hence \( x^2 + ax + b \) is the same equation: \( a = -4 \), \( b = 5 \).

Teacher's Note:
1. Non-real roots of a real polynomial occur in conjugate pairs.
2. One common non-real root forces both roots to be common.

Answer: (b) \( \dfrac{\pi}{6} \) and \( \dfrac{5\pi}{6} \)

Explanation:
1. The coefficients of the second equation add up to \( a^2b^2 - a^2c^2 + b^2c^2 - a^2b^2 + a^2c^2 - b^2c^2 = 0 \), so \( x = 1 \) is a root. Equal roots mean both roots are 1.
2. The common root must be 1: \( 4\sin^2\theta - 4\sin\theta + 1 = (2\sin\theta - 1)^2 = 0 \).
3. So \( \sin\theta = \dfrac{1}{2} \), which gives \( \theta = \dfrac{\pi}{6} \) or \( \dfrac{5\pi}{6} \) in \( (0, \pi) \).

Teacher's Note:
1. Check whether the coefficients of the second equation add to zero.
2. If it has equal roots and 1 is a root, then 1 is the only root.

Answer: (c) \( -\dfrac{1}{8} \)

Explanation:
1. The minimum occurs at \( x = \dfrac{3}{4} \).
2. \( y = 2\cdot\dfrac{9}{16} - \dfrac{9}{4} + 1 = \dfrac{9}{8} - \dfrac{18}{8} + \dfrac{8}{8} = -\dfrac{1}{8} \).

Teacher's Note:
1. Minimum value of ax² + bx + c (a > 0) is -D/4a.
2. Here D = 9 - 8 = 1 and a = 2, so -D/4a = -1/8.

Answer: (a) \( \dfrac{81}{8} \)

Explanation:
1. The maximum occurs at \( x = \dfrac{5}{4} \).
2. \( y = 7 + \dfrac{25}{4} - \dfrac{25}{8} = \dfrac{56 + 50 - 25}{8} = \dfrac{81}{8} \).

Teacher's Note:
1. For a < 0 the maximum value is -D/4a.
2. D = 25 + 56 = 81 and a = -2, so -D/4a = 81/8.

Answer: (c) \( R - \{2\} \)

Explanation:
1. \( y = 2 + \dfrac{1}{x + 1} \), and \( \dfrac{1}{x+1} \) takes every real value except 0.
2. So \( y \) takes every real value except 2.

Teacher's Note:
1. For (ax + b)/(cx + d) the excluded value is a/c.
2. Here a/c = 2/1 = 2.

Answer: (b) \( R - \{1\} \)

Explanation:
1. For \( x \ne 1 \) the function simplifies to \( y = x \).
2. Since \( x = 1 \) is not in the domain, \( y = 1 \) is not attained.
3. The range is \( R - \{1\} \).

Teacher's Note:
1. Cancel the common factor but keep the excluded point in mind.
2. The hole in the graph at x = 1 removes y = 1.