JEE Mathematics Quadratic Equations MCQs Set 07

Answer: (b) \( \frac{c}{a} > 0 \)

Explanation:
1. From the figure \( a < 0 \), \( c > 0 \), \( b > 0 \) and \( D > 0 \).
2. \( \frac{c}{a} < 0 \) because c and a have opposite signs, so \( \frac{c}{a} > 0 \) is false.
3. \( -\frac{b}{a} > 0 \) is the sum of roots, which is positive (vertex right of the y-axis).
4. \( -\frac{D}{4a} \) is the y-coordinate of the vertex, which lies above the x-axis, so it is positive.

Teacher's Note:
1. c/a is the product of the roots; the roots here lie on opposite sides of the origin, so the product is negative.
2. -b/a is the sum of the roots and -D/4a is the maximum value.

Answer: (d) \( [-3, -2] \)

Explanation:
1. Put \( t = \sin^2 x \in [0,1] \): \( t^2 - (k+2)t - (k+3) = 0 \).
2. Factorising, \( (t+1)(t-(k+3)) = 0 \), so \( t = -1 \) (impossible) or \( t = k+3 \).
3. We need \( 0 \le k+3 \le 1 \), i.e. \( -3 \le k \le -2 \).

Teacher's Note:
1. Replace sin²x by t and remember 0 ≤ t ≤ 1.
2. Check: k = -3 gives t = 0 (x = 0) and k = -2 gives t = 1 (x = π/2).

Answer: (d) real and distinct

Explanation:
1. Since \( a + b + c = 0 \): \( b + c - a = -2a \), \( c + a - b = -2b \) and \( a + b - c = -2c \).
2. The equation becomes \( -2(ax^2 + bx + c) = 0 \), i.e. \( ax^2 + bx + c = 0 \).
3. Its discriminant is \( b^2 - 4ac = (a + c)^2 - 4ac = (a - c)^2 > 0 \) because \( a \ne c \).
4. So the roots are real and distinct.

Teacher's Note:
1. Use a + b + c = 0 to replace each bracket by a single variable.
2. The discriminant simplifies to a perfect square, which is positive unless a = c.

Answer: (a) \( a = 4 \)

Explanation:
1. Equal roots need \( D = 0 \): \( 4(a-1)^2 - 12(2a-5) = 0 \).
2. \( a^2 - 2a + 1 - 6a + 15 = 0 \), i.e. \( a^2 - 8a + 16 = (a-4)^2 = 0 \).
3. So \( a = 4 \) (and then \( 2a - 5 = 3 \ne 0 \)).

Teacher's Note:
1. Equal roots mean zero discriminant.
2. Always check that the leading coefficient is not zero for the value found.

Answer: (d) \( m < -\frac{1}{2} \) or \( m > \frac{1}{2} \)

Explanation:
1. \( D = 1 - 4m^2 < 0 \), so \( m^2 > \frac{1}{4} \).
2. That means \( |m| > \frac{1}{2} \), i.e. \( m < -\frac{1}{2} \) or \( m > \frac{1}{2} \).

Teacher's Note:
1. Solve D < 0 as an inequality in m: m² > 1/4 gives two intervals.
2. Test m = 1: x² - x + 1 has no real root.

Answer: (b) positive

Explanation:
1. No real roots means \( f(x) = ax^2 + bx + c \) has the sign of \( a \) for every real \( x \).
2. So \( f(0) = c \) and \( f(1) = a + b + c \) both have the same sign as \( a \).
3. Hence \( c(a + b + c) > 0 \).

Teacher's Note:
1. f(0) = c and f(1) = a + b + c.
2. If D < 0, the parabola never crosses the x-axis, so f(0) and f(1) have the same sign.

Answer: (c) real

Explanation:
1. Expanding, \( 3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0 \).
2. \( D/4 = (a+b+c)^2 - 3(ab+bc+ca) = a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0 \).
3. So both roots are always real.

Teacher's Note:
1. Compute the discriminant and look for a sum of squares.
2. Equal roots occur only when a = b = c.

Answer: (a) 4

Explanation:
1. Put \( |x| = t \): \( t^2 - 3t + 2 = (t-1)(t-2) = 0 \), so \( t = 1 \) or \( t = 2 \).
2. \( |x| = 1 \) gives \( x = \pm 1 \) and \( |x| = 2 \) gives \( x = \pm 2 \).
3. There are 4 real solutions.

Teacher's Note:
1. Substitute |x| = t and solve the quadratic in t.
2. Each positive value of t gives two values of x.

Answer: (a) real and distinct

Explanation:
1. Let \( f(x) = (x-a)(x-c) + 2(x-b)(x-d) \), a quadratic with positive leading coefficient 3.
2. \( f(b) = (b-a)(b-c) < 0 \) because \( b > a \) and \( b < c \).
3. A quadratic that takes a negative value must cut the x-axis twice, so the roots are real and distinct.

Teacher's Note:
1. Find a point where the expression is negative.
2. With a positive leading coefficient, one negative value is enough to guarantee two real roots.

Answer: (a) 0

Explanation:
1. \( 5x^2 + 2x + 3 \) has minimum value \( 3 - \frac{1}{5} = 2.8 \) at \( x = -\frac{1}{5} \).
2. Since \( 2\sin x \le 2 < 2.8 \), the curves never meet.
3. The number of intersection points is 0.

Teacher's Note:
1. Compare the maximum of one curve with the minimum of the other.
2. Vertex value of ax² + bx + c is c - b²/4a.

Answer: (d) one root in \( (-\infty, a) \) and the other in \( (b, +\infty) \)

Explanation:
1. Let \( f(x) = (x-a)(x-b) - 1 \). Then \( f(a) = -1 < 0 \) and \( f(b) = -1 < 0 \).
2. Both \( a \) and \( b \) lie between the roots, because the parabola opens upward and is negative at these points.
3. So one root is less than \( a \) and the other is greater than \( b \).

Teacher's Note:
1. Evaluate f at a and b.
2. If f(a) < 0 and the parabola opens upward, a lies between the two roots.

Answer: (c) \( a > \frac{5}{3} \)

Explanation:
1. For all real x we need \( a - 1 > 0 \) and \( D < 0 \).
2. \( D = (a+1)^2 - 4(a-1)(a+1) = (a+1)(5 - 3a) < 0 \), so \( a < -1 \) or \( a > \frac{5}{3} \).
3. Combined with \( a > 1 \) this gives \( a > \frac{5}{3} \).

Teacher's Note:
1. Two conditions are needed: positive leading coefficient and negative discriminant.
2. Take the intersection of the solution sets.

Answer: (d) real and unequal

Explanation:
1. \( D = 25(l+m)^2 + 8(l-m)^2 \).
2. Since \( l \ne m \), \( 8(l-m)^2 > 0 \), so \( D > 0 \).
3. The roots are real and unequal.

Teacher's Note:
1. Both parts of the discriminant are non-negative squares.
2. Since the product of the roots is -2, the roots have opposite signs, which also shows they are real.

Answer: (a) non-negative

Explanation:
1. \( 2u = (x - 2y)^2 + (2y - 3z)^2 + (3z - x)^2 \).
2. A sum of squares is never negative, so \( u \ge 0 \) (and since x, y, z are different, in fact \( u > 0 \)).

Teacher's Note:
1. Express 2u as a sum of three squares.
2. Check by expanding: the cross terms give -4xy, -12yz and -6zx.

Answer: (b) \( x^2 - 4x + 1 = 0 \)

Explanation:
1. \( \tan 75^\circ = 2 + \sqrt{3} \).
2. For rational coefficients the other root is the conjugate \( 2 - \sqrt{3} \).
3. Sum \( = 4 \), product \( = 4 - 3 = 1 \), so the equation is \( x^2 - 4x + 1 = 0 \).

Teacher's Note:
1. Surd roots of a rational quadratic occur in conjugate pairs.
2. tan 75° = tan(45° + 30°) = (1 + 1/√3)/(1 - 1/√3) = 2 + √3.

Answer: (c) \( 4x^2 - 2x - 1 = 0 \)

Explanation:
1. \( \cos 36^\circ = \frac{1 + \sqrt{5}}{4} \), so the conjugate root is \( \frac{1 - \sqrt{5}}{4} \).
2. Sum \( = \frac{1}{2} \) and product \( = \frac{1 - 5}{16} = -\frac{1}{4} \).
3. The equation is \( x^2 - \frac{1}{2}x - \frac{1}{4} = 0 \), i.e. \( 4x^2 - 2x - 1 = 0 \).

Teacher's Note:
1. Use the known value cos 36° = (√5 + 1)/4.
2. The conjugate root equals -cos 72° = (1 - √5)/4.

Answer: (d) 6

Explanation:
1. From \( \alpha^2 = 2\alpha - 5 \) we get \( \alpha^3 = 2\alpha^2 - 5\alpha = -\alpha - 10 \).
2. \( \alpha^3 + \alpha^2 - \alpha + 21 = (-\alpha - 10) + (2\alpha - 5) - \alpha + 21 = 6 \).

Teacher's Note:
1. Reduce higher powers with the given equation.
2. Always substitute α² = 2α - 5 repeatedly until only constants remain.

Answer: (b) 7

Explanation:
1. \( \beta^2 = 2\beta - 5 \) and \( \beta^3 = -\beta - 10 \).
2. \( \beta^3 + 4\beta^2 - 7\beta + 37 = (-\beta - 10) + (8\beta - 20) - 7\beta + 37 = 7 \).

Teacher's Note:
1. Replace β² and β³ by linear expressions.
2. The coefficient of β cancels: -1 + 8 - 7 = 0.

Answer: (d) \( (-\infty, -6) \cup (-2, 0) \cup (3, \infty) \)

Explanation:
1. \( x^2 - x - 6 = (x-3)(x+2) \) and \( x^2 + 6x = x(x+6) \).
2. The inequality is \( (x+6)(x+2)x(x-3) > 0 \) with critical points -6, -2, 0, 3.
3. The product is positive for x > 3, negative on (0, 3), positive on (-2, 0), negative on (-6, -2) and positive for x < -6.
4. Solution: \( (-\infty, -6) \cup (-2, 0) \cup (3, \infty) \).

Teacher's Note:
1. Factorise both quadratics completely before using the wavy-curve method.
2. The sign is positive to the right of the largest root when the leading coefficient is positive.

Answer: (a) \( (-\frac{7}{3}, -1) \cup (2, 3) \)

Explanation:
1. Critical points: \( -\frac{7}{3}, -1, 2, 3 \); all coefficients of x are positive.
2. The product is positive for x > 3, negative on (2, 3), positive on (-1, 2), negative on \( (-\frac{7}{3}, -1) \), positive for \( x < -\frac{7}{3} \).
3. The required (negative) regions give \( (-\frac{7}{3}, -1) \cup (2, 3) \).

Teacher's Note:
1. Start from the right with a plus sign and alternate at every simple root.
2. Convert 3x + 7 to 3(x + 7/3); the positive factor 3 does not change the signs.

Answer: (d) \( [-2, 1] \)

Explanation:
1. Multiply by -1 and reverse the sign: \( x^2 + x - 2 \le 0 \).
2. \( (x+2)(x-1) \le 0 \), so \( -2 \le x \le 1 \).

Teacher's Note:
1. Multiplying an inequality by a negative number reverses its sign.
2. A quadratic with positive leading coefficient is ≤ 0 between its roots.

Answer: (b) \( (-\infty, 1] \cup [\frac{4}{3}, \infty) \)

Explanation:
1. \( 3x^2 - 7x + 4 = (x-1)(3x-4) \), with roots 1 and \( \frac{4}{3} \).
2. Positive leading coefficient: the expression is \( \ge 0 \) outside the roots (including them).
3. Solution: \( (-\infty, 1] \cup [\frac{4}{3}, \infty) \).

Teacher's Note:
1. Factorise first: 3x² - 7x + 4 = (x - 1)(3x - 4).
2. ≥ 0 with a > 0 means outside the roots.

Answer: (c) \( (-6, -2) \cup (-2, 0) \cup (3, \infty) \)

Explanation:
1. The factor \( (x+2)^2 \) is positive except at \( x = -2 \), so we solve \( x(x+6)(x-3) > 0 \) and exclude \( x = -2 \).
2. For \( x(x+6)(x-3) \): positive for x > 3, negative on (0, 3), positive on (-6, 0), negative for x < -6.
3. So \( x \in (-6, 0) \cup (3, \infty) \) with \( x \ne -2 \), i.e. \( (-6, -2) \cup (-2, 0) \cup (3, \infty) \).

Teacher's Note:
1. Even power: drop it but remember to exclude its zero.
2. Do not forget the excluded point when it lies inside a solution interval.

Answer: (a) \( (-1, 1) \cup (1, 4) \)

Explanation:
1. \( (x-1)^2 > 0 \) for \( x \ne 1 \), and \( (x+1)^3 \) has the same sign as \( (x+1) \).
2. So we need \( (x+1)(x-4) < 0 \), i.e. \( -1 < x < 4 \), with \( x \ne 1 \).
3. Solution: \( (-1, 1) \cup (1, 4) \).

Teacher's Note:
1. Odd powers behave like linear factors; even powers only exclude their zeros.
2. x = 1 makes the product zero, so it must be removed.

Answer: (c) \( (-\infty, \frac{3}{2}] \cup (\frac{7}{3}, \infty) \)

Explanation:
1. Critical points: \( \frac{3}{2} \) (numerator zero, included) and \( \frac{7}{3} \) (denominator zero, excluded).
2. The fraction is positive for \( x > \frac{7}{3} \), negative between the two points and positive for \( x < \frac{3}{2} \).
3. Solution: \( (-\infty, \frac{3}{2}] \cup (\frac{7}{3}, \infty) \).

Teacher's Note:
1. Include zeros of the numerator, exclude zeros of the denominator.
2. Check x = 0: (-3)/(-7) > 0, which lies in the first interval.

Answer: (b) \( [0, \frac{8}{3}) \cup (\frac{8}{3}, 3) \cup \{4\} \)

Explanation:
1. Even powers: \( (2x-3)^2 \), \( (x-4)^6 \) and \( (3x-8)^4 \) never change the sign; odd powers \( x^3 \) and \( (x-3)^3 \) behave like \( x \) and \( (x-3) \).
2. The sign of the fraction is that of \( \dfrac{x}{x-3} \), which is \( \le 0 \) for \( 0 \le x < 3 \).
3. Exclude \( x = \frac{8}{3} \) (denominator zero). The zeros \( x = \frac{3}{2} \) and \( x = 4 \) of the numerator give value 0 and are allowed; \( x = \frac{3}{2} \) already lies in [0, 3) while \( x = 4 \) must be added separately.
4. Solution: \( [0, \frac{8}{3}) \cup (\frac{8}{3}, 3) \cup \{4\} \).

Teacher's Note:
1. Cancel the effect of even powers on the sign, but keep their zeros as special points.
2. Numerator zeros are included in a 'less than or equal to zero' inequality; denominator zeros never are.