JEE Mathematics Quadratic Equations MCQs Set 10

Answer: (c) \( \alpha=\beta=\frac{\pi}{3} \)

Explanation:
1. \( 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) \). Put \( u=\cos(\alpha+\beta) \) and \( v=\cos(\alpha-\beta) \): the equation becomes \( u(u+v)=-\frac14 \), i.e. \( 4u^2+4uv+1=0 \).
2. Treat this as a quadratic in u. Real u needs \( 16v^2-16\ge0 \), i.e. \( v^2\ge1 \). Since \( |v|\le1 \), we get \( v=\pm1 \); with \( |\alpha-\beta|\lt\frac{\pi}{2} \) this means \( v=1 \) and \( \alpha=\beta \).
3. Then \( 4u^2+4u+1=0 \) gives \( u=-\frac12 \), so \( 2\alpha=\frac{2\pi}{3} \) and \( \alpha=\beta=\frac{\pi}{3} \).

Teacher's Note:
1. Writing the product of cosines as a sum turns the equation into a quadratic in cos(α + β).
2. Requiring a real root forces the discriminant condition and fixes α = β.

Answer: (a) \( a\lt2 \)

Explanation:
1. Let \( f(x)=x^2-2ax+a^2+a-3 \). Real roots: \( D=4a^2-4(a^2+a-3)\ge0\Rightarrow a\le3 \).
2. Both roots less than 3 needs \( f(3)\gt0 \): \( a^2-5a+6=(a-2)(a-3)\gt0\Rightarrow a\lt2\ \text{or}\ a\gt3 \), and the axis \( x=a\lt3 \).
3. Combining, a < 2.

Teacher's Note:
1. For both roots to be less than k we need D ≥ 0, f(k) > 0 and the axis to the left of k.
2. Intersect the three conditions.

Answer: (c) 3

Explanation:
1. Let the roots be r and r². Product: \( r^3=\frac33=1 \), so r = 1 or r is a complex cube root of unity.
2. If r = 1 the sum is 2, giving \( -\frac p3=2 \), p = −6, which is not positive.
3. So r = ω and the roots are ω, ω²: sum = −1 = \( -\frac p3 \), hence p = 3.

Teacher's Note:
1. Use the product of roots first: it gives r³ = 1.
2. Reject the case that makes p negative.

Answer: (b) \( \alpha\lt0\lt\beta\lt|\alpha| \)

Explanation:
1. Product \( \alpha\beta=c\lt0 \) so the roots have opposite signs: \( \alpha\lt0\lt\beta \).
2. Sum \( \alpha+\beta=-b\lt0 \), so the negative root has the larger magnitude: \( \beta\lt|\alpha| \).

Teacher's Note:
1. The sign of the product tells whether the roots have the same sign.
2. The sign of the sum tells which root is larger in magnitude.

Answer: (a) \( a\gt1 \)

Explanation:
1. Discriminant \( D=(a-b)^2-4(1-a-b)=b^2+(4-2a)b+(a^2+4a-4) \).
2. D > 0 for all b requires its own discriminant to be negative: \( (4-2a)^2-4(a^2+4a-4)=32-32a\lt0 \).
3. Hence a > 1.

Teacher's Note:
1. Treat D as a quadratic in b that must be positive for every b.
2. A quadratic with positive leading coefficient is always positive when its discriminant is negative.

Answer: (b) 2

Explanation:
1. Distinct real roots: \( D=64k^2-64(k^2-k+1)=64(k-1)\gt0\Rightarrow k\gt1 \).
2. Roots at least 4: \( f(4)=16-32k+16(k^2-k+1)=16(k-1)(k-2)\ge0 \) and axis \( 4k\ge4 \). With k > 1 this gives \( k\ge2 \).
3. The smallest value is 2.

Teacher's Note:
1. Use D > 0, f(4) ≥ 0 and axis ≥ 4 together.
2. The binding condition is f(4) ≥ 0, which gives k ≥ 2 once k > 1.

Answer: (b) \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \)

Explanation:
1. Let \( \alpha\beta=P \). Then \( q=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=-p^3+3pP \), so \( P=\frac{p^3+q}{3p} \).
2. Sum of new roots \( \frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{p^2-2P}{P}=\frac{p^2}{P}-2=\frac{3p^3}{p^3+q}-2=\frac{p^3-2q}{p^3+q} \); product = 1.
3. Equation: \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \).

Teacher's Note:
1. Find αβ from the sum of cubes identity.
2. Then compute the sum of α/β and β/α as (α² + β²)/αβ.

Answer: (b) \( -i\sqrt3 \)

Explanation:
1. Subtract: \( (b-1)x-(1+b)=0 \), so the common root is \( x=\frac{b+1}{b-1} \).
2. Substitute in \( x^2+x+b=0 \): \( (b+1)^2+(b+1)(b-1)+b(b-1)^2=0 \), which simplifies to \( b^3+3b=0 \), i.e. \( b(b^2+3)=0 \).
3. b = 0 also works (common root −1) but is not among the options. Otherwise \( b^2=-3 \), so \( b=\pm i\sqrt3 \); the only option of this form is (b), b = −i√3.

Teacher's Note:
1. Subtracting the equations gives the common root directly.
2. Substitute that root back into either equation and solve for b.

Answer: (b) 1 : 2 : 3

Explanation:
1. The discriminant of \( x^2+2x+3 \) is \( 4-12=-8\lt0 \), so its roots are non-real conjugates.
2. A real quadratic with one non-real root has the conjugate as its other root, so the second equation has both roots in common with the first.
3. Hence the equations are proportional: a : b : c = 1 : 2 : 3.

Teacher's Note:
1. Non-real roots of a real quadratic always occur in conjugate pairs.
2. A common root therefore means both roots are common.

Answer: (a) does not exist

Explanation:
1. Let \( f(x)=2x^3+3x-k \). Then \( f'(x)=6x^2+3\gt0 \) for all real x.
2. So f is strictly increasing and crosses zero at most once.
3. Therefore no k gives two distinct real roots in [0, 1].

Teacher's Note:
1. Check monotonicity through the derivative.
2. A strictly increasing function has at most one zero.

Answer: (c) 1

Explanation:
1. Determinant of coefficients: \( (k+1)(k+3)-8k=k^2-4k+3=(k-1)(k-3) \). No solution is possible only for k = 1 or 3.
2. k = 1: the equations are \( 2x+8y=4 \) and \( x+4y=2 \), identical, so infinitely many solutions.
3. k = 3: \( 4x+8y=12 \) and \( 3x+6y=8 \) are parallel but inconsistent: no solution. So exactly 1 value.

Teacher's Note:
1. Set the determinant to zero, then test each value for consistency.
2. Parallel but distinct lines give no solution; coincident lines give infinitely many.

Answer: (a) \( \frac{2\sqrt{13}}{9} \)

Explanation:
1. \( \frac1\alpha+\frac1\beta=-\frac qr=4 \) gives q = −4r. A.P.: \( 2q=p+r \) so \( p=-9r \).
2. Equation: \( -9rx^2-4rx+r=0\Rightarrow 9x^2+4x-1=0 \).
3. \( |\alpha-\beta|=\frac{\sqrt{16+36}}{9}=\frac{2\sqrt{13}}{9} \).

Teacher's Note:
1. Express q and p in terms of r using the two given conditions.
2. Then |α − β| = √D / |a|.

Answer: (a) (i) and (iv)

Explanation:
1. Distinct real roots: \( 1-4\alpha^2\gt0\Rightarrow|\alpha|\lt\frac12 \).
2. \( |x_1-x_2|=\frac{\sqrt{1-4\alpha^2}}{|\alpha|}\lt1\Rightarrow1-4\alpha^2\lt\alpha^2\Rightarrow|\alpha|\gt\frac1{\sqrt5} \).
3. So \( S=\left(-\frac12,-\frac1{\sqrt5}\right)\cup\left(\frac1{\sqrt5},\frac12\right) \): intervals (i) and (iv) are subsets.

Teacher's Note:
1. Combine the condition for distinct roots with the condition on their difference.
2. Compare each interval with the resulting set S.

Answer: (c) \( -2\tan\theta \)

Explanation:
1. First equation: roots \( \sec\theta\pm|\tan\theta| \). Since θ is negative, \( |\tan\theta|=-\tan\theta \), so \( \alpha_1=\sec\theta-\tan\theta \).
2. Second equation: roots \( -\tan\theta\pm\sec\theta \), so \( \beta_2=-\tan\theta-\sec\theta \).
3. Sum: \( \alpha_1+\beta_2=-2\tan\theta \).

Teacher's Note:
1. Find the roots explicitly using the quadratic formula.
2. Pay attention to the sign of tan θ for θ in the fourth quadrant.

Answer: (d) 11

Explanation:
1. The sum of \( (x+k-1)(x+k) \) over k = 1 to n is \( nx^2+n^2x+\frac{n(n^2-1)}{3} \).
2. Setting equal to 10n and dividing by n: \( x^2+nx+\frac{n^2-1}{3}-10=0 \). Difference of roots \( =\sqrt{n^2-\tfrac{4(n^2-31)}{3}}=1 \) gives \( 3n^2-4n^2+124=3 \), so \( n^2=121 \) and n = 11.
3. Check: for n = 11 the roots are −5 and −6.

Teacher's Note:
1. Sum the products using the standard sums of k and k².
2. Consecutive integral roots means the root difference equals 1.

Answer: (a) 12

Explanation:
1. With \( \alpha=\frac{1+\sqrt5}2,\ \beta=\frac{1-\sqrt5}2 \): \( \alpha^4=\frac{7+3\sqrt5}2,\ \beta^4=\frac{7-3\sqrt5}2 \).
2. \( a_4=\frac{7(p+q)}2+\frac{3\sqrt5(p-q)}2=28 \).
3. Rational and irrational parts: p + q = 8 and p = q, so p = q = 4 and \( p+2q=12 \).

Teacher's Note:
1. Compute α⁴ and β⁴ explicitly.
2. Use the fact that a + b√5 = 0 with a, b rational forces a = b = 0.

Answer: (d) \( a_{11}+a_{10} \)

Explanation:
1. Each root satisfies \( t^2=t+1 \). Multiply by \( t^{10} \): \( t^{12}=t^{11}+t^{10} \).
2. Multiply by p for α and q for β and add: \( a_{12}=a_{11}+a_{10} \).

Teacher's Note:
1. Use t² = t + 1 to get a recurrence for the sequence.
2. Apply it to both roots with their coefficients and add.

Answer: (d) 1

Explanation:
1. The roots are \( e^{\pm i\pi/3} \), so \( \alpha^6=\beta^6=1 \).
2. \( \alpha^{101}=\alpha^{5}=e^{5i\pi/3} \) and \( \beta^{107}=\beta^{5}=e^{-5i\pi/3} \).
3. Sum \( =2\cos\frac{5\pi}3=1 \).

Teacher's Note:
1. Find the order of the roots: they are 6th roots of unity.
2. Reduce each exponent modulo 6.

Answer: (d) contains exactly two elements

Explanation:
1. Put \( s=\sqrt x\ge0 \). If s ≥ 3: \( 2s-6+s^2-6s+6=s^2-4s=0 \) gives s = 4 (s = 0 rejected).
2. If s < 3: \( 6-2s+s^2-6s+6=s^2-8s+12=0 \) gives s = 2 (s = 6 rejected).
3. So x = 16 and x = 4: two elements.

Teacher's Note:
1. Substitute s = √x and split the modulus at s = 3.
2. Reject roots that violate the case condition.

Answer: (d) \( \frac{2^{12}}{(\sin\theta+8)^{12}} \)

Explanation:
1. \( \alpha^{-12}+\beta^{-12}=\frac{\alpha^{12}+\beta^{12}}{(\alpha\beta)^{12}} \), so the expression equals \( \frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}=\left[\frac{\alpha\beta}{(\alpha-\beta)^2}\right]^{12} \).
2. \( \alpha\beta=-2\sin\theta,\ (\alpha-\beta)^2=\sin^2\theta+8\sin\theta \).
3. Ratio inside = \( \frac{-2\sin\theta}{\sin\theta(\sin\theta+8)}=\frac{-2}{\sin\theta+8} \); raised to 12: \( \frac{2^{12}}{(\sin\theta+8)^{12}} \).

Teacher's Note:
1. Rewrite the reciprocal powers in terms of αβ.
2. Use αβ and (α − β)² from the coefficients.

Answer: (d) \( 8\sqrt5 \)

Explanation:
1. Sum of roots \( =\frac3{m^2+1} \) is greatest when m = 0.
2. Then the equation is \( x^2-3x+1=0 \): \( \alpha+\beta=3,\ \alpha\beta=1,\ |\alpha-\beta|=\sqrt5 \).
3. \( |\alpha^3-\beta^3|=|\alpha-\beta|(\alpha^2+\alpha\beta+\beta^2)=\sqrt5\,(9-1)=8\sqrt5 \).

Teacher's Note:
1. The sum is largest when the denominator is smallest.
2. Use α³ − β³ = (α − β)(α² + αβ + β²).

Answer: (b) \( p^2-4q-12=0 \)

Explanation:
1. Real coefficients: the other root is \( 2+\sqrt3 \).
2. Sum = 4 = −p so p = −4; product = 1 = q.
3. Check: \( p^2-4q-12=16-4-12=0 \).

Teacher's Note:
1. Irrational roots of a real-coefficient quadratic come in conjugate pairs here.
2. Substitute p = −4, q = 1 into each option.