JEE Mathematics Quadratic Equations MCQs Set 09

Answer: (a) \( (-\infty, 2] \cup [6, \infty) \)

Explanation:
1. \( x^2 + 2x - 11 = 2y(x - 3) \) gives \( x^2 + (2 - 2y)x + (6y - 11) = 0 \).
2. For real \( x \), \( D \ge 0 \): \( (2 - 2y)^2 - 4(6y - 11) = 4y^2 - 32y + 48 = 4(y - 2)(y - 6) \ge 0 \).
3. So \( y \le 2 \) or \( y \ge 6 \). (No common factor in numerator and denominator, so no value is lost.)

Teacher's Note:
1. Put y equal to the function, form a quadratic in x and apply D ≥ 0.
2. Check x² + 2x - 11 has no root x = 3: 9 + 6 - 11 = 4 ≠ 0.

Answer: (d) \( \left[\dfrac{1}{7}, 7\right] \)

Explanation:
1. Both quadratics have \( D < 0 \) (denominator \( x^2 + 3x + 4 \): \( D = 9 - 16 < 0 \)), so the denominator never vanishes.
2. \( (y - 1)x^2 + 3(y + 1)x + 4(y - 1) = 0 \). For \( y \ne 1 \), \( D \ge 0 \): \( 9(y+1)^2 - 16(y-1)^2 \ge 0 \).
3. \( (3y + 3 - 4y + 4)(3y + 3 + 4y - 4) = (7 - y)(7y - 1) \ge 0 \), so \( \dfrac{1}{7} \le y \le 7 \). The value \( y = 1 \) is attained at \( x = 0 \).
4. Range: \( \left[\dfrac{1}{7}, 7\right] \).

Teacher's Note:
1. Form the quadratic in x and use D ≥ 0; check the case y = 1 separately because the coefficient of x² becomes 0.
2. Difference of squares helps to factorise 9(y+1)² - 16(y-1)².

Answer: (b) \( (-\infty, 1) \)

Explanation:
1. Write \( y = \dfrac{(x+1)^2 - 3}{(x+1)^2} = 1 - \dfrac{3}{(x+1)^2} \) with \( x \ne -1 \).
2. \( \dfrac{3}{(x+1)^2} \) takes every positive value, so \( y \) takes every value less than 1.
3. The range is \( (-\infty, 1) \).

Teacher's Note:
1. Complete the square in the numerator using the denominator (x + 1)².
2. The term 3/(x+1)² is strictly positive, so y never reaches 1.

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

Explanation:
1. The denominator \( x^2 + 2x + 3 \) has \( D = 4 - 12 < 0 \) and never vanishes.
2. \( (y - 1)x^2 + (2y - 14)x + (3y - 9) = 0 \). For \( y \ne 1 \), \( D \ge 0 \): \( (2y - 14)^2 - 4(y - 1)(3y - 9) \ge 0 \).
3. \( 4y^2 - 56y + 196 - 12y^2 + 48y - 36 = -8y^2 - 8y + 160 \ge 0 \), i.e. \( y^2 + y - 20 = (y + 5)(y - 4) \le 0 \), so \( -5 \le y \le 4 \).
4. \( y = 1 \) lies in this interval and is attained (\( -12x - 6 = 0 \), \( x = -\dfrac{1}{2} \)).

Teacher's Note:
1. The same method works whenever the denominator never vanishes.
2. Reduce the discriminant condition to a quadratic inequality in y.

Answer: (c) \( (-\infty, 1] \cup [2, \infty) \) and \( [0, \infty) \)

Explanation:
1. We need \( x^2 - 3x + 2 = (x-1)(x-2) \ge 0 \), i.e. \( x \le 1 \) or \( x \ge 2 \).
2. The inside takes every value in \( [0, \infty) \) on this domain, so the square root takes every value in \( [0, \infty) \).

Teacher's Note:
1. The expression under the root must be ≥ 0.
2. The quadratic is 0 at x = 1, 2 and unbounded above.

Answer: (d) \( (2x + y + 1)(x + y + 1) \)

Explanation:
1. The quadratic part factorises as \( 2x^2 + 3xy + y^2 = (2x + y)(x + y) \).
2. Write \( (2x + y + p)(x + y + q) \) and compare the coefficients of \( x \) and \( y \): \( 2q + p = 3 \) and \( q + p = 2 \), so \( q = 1 \), \( p = 1 \).
3. Check the constant term: \( pq = 1 \). The factors are \( (2x + y + 1)(x + y + 1) \).

Teacher's Note:
1. Factorise the second-degree terms first, then add unknown constants to each factor.
2. Compare coefficients of x, y and the constant.

Answer: (a) \( p^2 - 2q \) and \( p^3 - 3pq + 3r \)

Explanation:
1. \( \sum\alpha = p \), \( \sum\alpha\beta = q \), \( \alpha\beta\gamma = r \).
2. \( \sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = p^2 - 2q \).
3. \( \sum\alpha^3 - 3\alpha\beta\gamma = (\sum\alpha)(\sum\alpha^2 - \sum\alpha\beta) = p(p^2 - 3q) \), so \( \sum\alpha^3 = p^3 - 3pq + 3r \).

Teacher's Note:
1. Use the identity a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca).
2. Express everything with e1 = p, e2 = q, e3 = r.

Answer: (c) \( -4,\ \dfrac{3}{2},\ -\dfrac{3}{2} \)

Explanation:
1. Let the roots be \( \alpha, -\alpha, \gamma \). Sum of roots \( = \gamma = -\dfrac{16}{4} = -4 \).
2. Sum of products in pairs: \( -\alpha^2 + \gamma(\alpha - \alpha) = -\alpha^2 = -\dfrac{9}{4} \), so \( \alpha = \pm\dfrac{3}{2} \).
3. Roots: \( -4, \dfrac{3}{2}, -\dfrac{3}{2} \). Check: \( 4x^3 + 16x^2 - 9x - 36 = (x + 4)(2x - 3)(2x + 3) \).

Teacher's Note:
1. With roots α, -α, γ, the sum of roots gives γ immediately.
2. Use the sum of products of roots taken two at a time for α².

Answer: (d) 0

Explanation:
1. \( \dfrac{1 + t}{1 - t} = \dfrac{3(3t - t^3)}{1 - 3t^2} \). Cross-multiplying: \( (1 + t)(1 - 3t^2) = 3t(3 - t^2)(1 - t) \).
2. Expanding: \( 1 + t - 3t^2 - 3t^3 = 9t - 9t^2 - 3t^3 + 3t^4 \), i.e. \( 3t^4 - 6t^2 + 8t - 1 = 0 \).
3. This quartic has no \( t^3 \) term, so the sum of its roots, \( \tan\alpha + \tan\beta + \tan\gamma + \tan\delta \), is 0.

Teacher's Note:
1. Write everything in terms of t = tan x using the formula for tan 3x.
2. The sum of the roots is minus the coefficient of t³ divided by the leading coefficient.

Answer: (c) \( x^3 - 8x^2 + 19x - 15 = 0 \)

Explanation:
1. If \( y = x + 1 \) then \( x = y - 1 \); substitute in the given equation.
2. \( (y-1)^3 - 5(y-1)^2 + 6(y-1) - 3 = y^3 - 3y^2 + 3y - 1 - 5y^2 + 10y - 5 + 6y - 6 - 3 \).
3. \( = y^3 - 8y^2 + 19y - 15 \).
4. The required equation is \( x^3 - 8x^2 + 19x - 15 = 0 \).

Teacher's Note:
1. To increase every root by h, replace x by x - h.
2. Check: sum of roots increases by 3, from 5 to 8.

Answer: (c) 36

Explanation:
1. Factorise: \( x^3 - 12x^2 + 47x - 60 = (x - 3)(x - 4)(x - 5) \), so the sides are 3, 4, 5.
2. Semi-perimeter \( s = 6 \). By Heron's formula \( \Delta^2 = s(s-3)(s-4)(s-5) = 6 \cdot 3 \cdot 2 \cdot 1 = 36 \).
3. (Check: it is a right triangle with area \( \frac{1}{2} \cdot 3 \cdot 4 = 6 \).)

Teacher's Note:
1. Try integer factors of 60 to find the roots.
2. Δ² = s(s-a)(s-b)(s-c); for a right triangle, area = half the product of the legs.

Answer: (a) \( d > \dfrac{11}{9} \)

Explanation:
1. Let \( f(x) = x^2 - 6dx + 9d^2 - 2d + 2 \). Conditions: \( D \ge 0 \), vertex \( 3d > 3 \) and \( f(3) > 0 \).
2. \( D = 36d^2 - 4(9d^2 - 2d + 2) = 8(d - 1) \ge 0 \) gives \( d \ge 1 \); the vertex condition gives \( d > 1 \).
3. \( f(3) = 9d^2 - 20d + 11 = (d - 1)(9d - 11) > 0 \) gives \( d < 1 \) or \( d > \dfrac{11}{9} \).
4. Intersection: \( d > \dfrac{11}{9} \).

Teacher's Note:
1. Both roots greater than k needs D ≥ 0, -b/2a > k and a·f(k) > 0.
2. Intersect all three conditions.

Answer: (b) \( -2 < k < 3 \)

Explanation:
1. The roots lie on either side of 2 when \( f(2) < 0 \) (leading coefficient positive).
2. \( f(2) = 4 - 2(k+1) + k^2 + k - 8 = k^2 - k - 6 = (k - 3)(k + 2) < 0 \).
3. So \( -2 < k < 3 \).

Teacher's Note:
1. A number lies between the roots of an upward parabola exactly when f(number) < 0.
2. Factorise k² - k - 6.

Answer: (b) empty (no real value of \( a \))

Explanation:
1. The leading coefficient \( a^2 + a + 1 \) is always positive, so we need \( f(3) < 0 \).
2. \( f(3) = 9(a^2 + a + 1) + 3(a - 1) + a^2 = 10a^2 + 12a + 6 \).
3. Its discriminant is \( 144 - 240 < 0 \) and the leading coefficient is positive, so \( f(3) > 0 \) for every real \( a \).
4. Hence no real \( a \) puts 3 between the zeros.

Teacher's Note:
1. Evaluate the polynomial at 3 and examine its sign for all a.
2. If f(3) is always positive, 3 can never lie between the zeros.

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

Explanation:
1. With \( A = a^2 - 5a + 6 \), we need \( A \cdot f(-1) < 0 \).
2. \( f(-1) = A + (a - 3) + 7 = a^2 - 4a + 10 > 0 \) always (its discriminant is negative).
3. So \( A < 0 \): \( (a - 2)(a - 3) < 0 \), i.e. \( 2 < a < 3 \).

Teacher's Note:
1. Roots on either side of a point: A·f(point) < 0.
2. Check whether f(-1) keeps a constant sign.

Answer: (c) \( \left[6, \dfrac{27}{4}\right) \)

Explanation:
1. Conditions: \( D \ge 0 \), \( f(-6) > 0 \), \( f(1) > 0 \) and \( -6 < -(k - 3) < 1 \).
2. \( D = 4(k-3)^2 - 36 \ge 0 \) gives \( k \le 0 \) or \( k \ge 6 \). \( f(-6) = 81 - 12k > 0 \) gives \( k < \dfrac{27}{4} \). \( f(1) = 2k + 4 > 0 \) gives \( k > -2 \).
3. Vertex: \( 2 < k < 9 \).
4. Combining all: \( 6 \le k < \dfrac{27}{4} \).

Teacher's Note:
1. Both roots in (d, e): D ≥ 0, f(d) > 0, f(e) > 0 and d < vertex < e.
2. Do not forget to intersect the D ≥ 0 branches with the vertex condition.

Answer: (b) \( (-6, -2) \cup (-2, 0) \)

Explanation:
1. Let \( f(x) = x^2 + mx + m^2 + 6m \). Then \( f(-2) = (m + 2)^2 \) and \( f(0) = m(m + 6) \).
2. Exactly one root in \( (-2, 0) \) needs \( f(-2)f(0) < 0 \): \( (m+2)^2 \, m(m+6) < 0 \), so \( -6 < m < 0 \) with \( m \ne -2 \).
3. Check the boundary cases: \( m = -2 \) gives roots \( -2, 4 \), none inside; \( m = 0 \) gives roots \( 0, 0 \) and \( m = -6 \) gives \( 0, 6 \), none inside.
4. Answer: \( (-6, -2) \cup (-2, 0) \).

Teacher's Note:
1. Exactly one root in (d, e) when f(d)f(e) < 0.
2. Test the end values of m where a product vanishes.

Answer: (a) all real values of \( \lambda \)

Explanation:
1. For \( \lambda > 0 \): \( f(b) = (b - a)(b - c) < 0 \) and the leading coefficient \( 1 + \lambda > 0 \), so the roots are real.
2. For \( \lambda < 0 \): \( f(a) = \lambda(a - b)(a - d) < 0 \) while \( f(c) = \lambda(c - b)(c - d) > 0 \), so there is a root between \( a \) and \( c \) (real); for \( \lambda = -1 \) the equation is linear and also has a real root.
3. For \( \lambda = 0 \) the roots are \( a \) and \( c \). So the roots are real for every real \( \lambda \).

Teacher's Note:
1. Look for two points where the polynomial has opposite signs.
2. A real quadratic with a real root has both roots real.

Answer: (a) \( a \ge -\dfrac{3}{5} \)

Explanation:
1. The square root requires \( 5a + 3 \ge 0 \), i.e. \( a \ge -\dfrac{3}{5} \).
2. The leading coefficient \( a^2 + 3 > 0 \), so we need \( D > 0 \): \( (5a + 3) + (a^2 + 3) > 0 \), i.e. \( (a + 2)(a + 3) > 0 \), which holds for \( a > -2 \) or \( a < -3 \).
3. For \( a \ge -\dfrac{3}{5} \) this always holds (even for \( a = -\dfrac{3}{5} \), the expression is \( (a^2 + 3)x^2 - \dfrac{1}{4} \), negative for small \( x \)).
4. So \( a \ge -\dfrac{3}{5} \).

Teacher's Note:
1. Find the domain of the expression first.
2. Then use D > 0 for a positive leading coefficient.

Answer: (d) \( m > 1 \)

Explanation:
1. The minimum of \( x^2 - 4x + 3m + 1 \) occurs at \( x = 2 \), which is positive, so the minimum over positive \( x \) is \( 4 - 8 + 3m + 1 = 3m - 3 \).
2. We need \( 3m - 3 > 0 \), so \( m > 1 \).

Teacher's Note:
1. Locate the vertex first: if it lies in the allowed region, the vertex value decides.
2. Positive for all x in the region means the minimum is positive.

Answer: (b) \( a < 4 \) or \( a > 12 \)

Explanation:
1. If \( a < 3 \): \( f(0) = a - 3 < 0 \), so \( f \) is negative for small negative \( x \).
2. If \( a \ge 3 \): the roots have sum \( -\dfrac{a}{4} < 0 \) and product \( \dfrac{a - 3}{4} \ge 0 \), so any real roots are non-positive; we need \( D = a^2 - 16a + 48 = (a - 4)(a - 12) > 0 \), i.e. \( 3 \le a < 4 \) or \( a > 12 \).
3. Together: \( a < 4 \) or \( a > 12 \). (At \( a = 4 \) and \( a = 12 \), \( D = 0 \) and \( f \ge 0 \).)

Teacher's Note:
1. Split into the cases f(0) < 0 and f(0) ≥ 0.
2. When f(0) ≥ 0 you need two negative roots (D > 0).

Answer: (c) \( a \le -1 \)

Explanation:
1. Real roots need \( D = 4(a-1)^2 - 4(a+5) = 4(a + 1)(a - 4) \ge 0 \), so \( a \le -1 \) or \( a \ge 4 \).
2. Sum of roots \( = -2(a - 1) \), product \( = a + 5 \).
3. For \( a \ge 4 \) the sum is negative and the product positive, so both roots are negative: rejected.
4. For \( a < -5 \) the product is negative (one positive root); for \( -5 \le a \le -1 \) the sum is positive and the product is non-negative, so a positive root exists.
5. So \( a \le -1 \).

Teacher's Note:
1. Check the sign of the sum and product of the roots in each interval.
2. Verify with a = 5: x² + 8x + 10 = 0 has both roots negative.

Answer: (d) \( k < -1 \) or \( k > \dfrac{5}{4} \)

Explanation:
1. Let \( t = x^2 \ge 0 \): \( t^2 + (1 - 2k)t + k^2 - 1 = 0 \) with discriminant \( D = (1 - 2k)^2 - 4(k^2 - 1) = 5 - 4k \), sum of roots \( 2k - 1 \) and product \( k^2 - 1 \).
2. If \( k > \dfrac{5}{4} \), \( D < 0 \): no real \( t \), so no real \( x \).
3. If \( k < -1 \): \( D > 0 \), the product \( k^2 - 1 > 0 \) and the sum \( 2k - 1 < 0 \), so both values of \( t \) are negative: no real \( x \).
4. For \( k = -1 \), \( t = 0 \) is possible (see below), and for \( -1 < k \le \dfrac{5}{4} \) there is a non-negative \( t \).
5. So there is no real solution for \( k < -1 \) or \( k > \dfrac{5}{4} \).

Teacher's Note:
1. Put t = x² and analyse the quadratic in t: only non-negative values of t give real x.
2. Each positive t gives two values of x; t = 0 gives one.

Answer: (b) \( k = -1 \)

Explanation:
1. Let \( t = x^2 \ge 0 \): \( t^2 + (1 - 2k)t + k^2 - 1 = 0 \) with discriminant \( D = (1 - 2k)^2 - 4(k^2 - 1) = 5 - 4k \), sum of roots \( 2k - 1 \) and product \( k^2 - 1 \).
2. One real solution needs \( t = 0 \) to be a root with the other root negative.
3. \( t = 0 \) is a root when \( k^2 - 1 = 0 \), i.e. \( k = \pm 1 \).
4. For \( k = 1 \) the other root is \( t = 2k - 1 = 1 > 0 \) (three solutions); for \( k = -1 \) the equation is \( t^2 + 3t = 0 \), so \( t = 0 \) or \( t = -3 \).
5. Hence exactly one real solution (\( x = 0 \)) occurs only for \( k = -1 \).

Teacher's Note:
1. Put t = x² and analyse the quadratic in t: only non-negative values of t give real x.
2. Each positive t gives two values of x; t = 0 gives one.

Answer: (d) \( k = 1 \)

Explanation:
1. Let \( t = x^2 \ge 0 \): \( t^2 + (1 - 2k)t + k^2 - 1 = 0 \) with discriminant \( D = (1 - 2k)^2 - 4(k^2 - 1) = 5 - 4k \), sum of roots \( 2k - 1 \) and product \( k^2 - 1 \).
2. Three real solutions need \( t = 0 \) and another positive \( t \).
3. \( k^2 - 1 = 0 \) gives \( k = \pm 1 \). For \( k = 1 \): \( t^2 - t = 0 \), so \( t = 0 \) or \( t = 1 \), giving \( x = 0, \pm 1 \).
4. For \( k = -1 \) the other root is negative, so \( k = 1 \) is the only value.

Teacher's Note:
1. Put t = x² and analyse the quadratic in t: only non-negative values of t give real x.
2. Each positive t gives two values of x; t = 0 gives one.

Answer: (a) \( 1 < k < \dfrac{5}{4} \)

Explanation:
1. Let \( t = x^2 \ge 0 \): \( t^2 + (1 - 2k)t + k^2 - 1 = 0 \) with discriminant \( D = (1 - 2k)^2 - 4(k^2 - 1) = 5 - 4k \), sum of roots \( 2k - 1 \) and product \( k^2 - 1 \).
2. Four real solutions need two distinct positive values of \( t \).
3. This requires \( D > 0 \) (\( k < \dfrac{5}{4} \)), sum \( 2k - 1 > 0 \) (\( k > \dfrac{1}{2} \)) and product \( k^2 - 1 > 0 \) (\( k > 1 \) or \( k < -1 \)).
4. Intersection: \( 1 < k < \dfrac{5}{4} \).

Teacher's Note:
1. Put t = x² and analyse the quadratic in t: only non-negative values of t give real x.
2. Each positive t gives two values of x; t = 0 gives one.