JEE Mathematics Quadratic Equations MCQs Set 23

Answer: (d) 2

Explanation:
1. \(\alpha+\beta=\dfrac{p+1}2\), \(\alpha\beta=\dfrac{p-1}2\). \((\alpha-\beta)^2=(\alpha\beta)^2\) gives \(\dfrac{(p+1)^2}4-2(p-1)=\dfrac{(p-1)^2}4\).
2. \((p+1)^2-(p-1)^2=8(p-1)\) gives \(4p=8p-8\), so \(p=2\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) Both are true and Reason explains Assertion

Explanation:
1. \( f(0) + f(1) = a + b + 2c = 0 \) and \( f(-2) = 4a - 2b + c = 0 \) give \( c = -6a/5 \) and \( b = 7a/5 \).
2. Sum of roots = \( -b/a = -7/5 \), so the other root = \( -7/5 + 2 = 3/5 \), using the sum of zeroes from the reason.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) Assertion is false, Reason is true

Explanation:
1. \( D = 16 - 4(2 \lambda - 1)^2 \) >= 0 gives \( |2 \lambda - 1| \) <= 2, i.e. \( -1/2 \) <= \( \lambda \) <= 3/2.
2. The greatest integer is 1, not 2, so the assertion is false; the reason is true.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) 3 and 4

Explanation:
1. With \(t=2^x\): \(t^2-24t+128=0\) gives \(t=8\) or \(16\).
2. So \( x = 3 \) or 4.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(\dfrac{49}4\)

Explanation:
1. \(16+4p+12=0\) gives \(p=-7\).
2. Equal roots: \(49-4q=0\), so \(q=\dfrac{49}4\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) 3

Explanation:
1. Roots \(r,r^2\) with product \(r^3=1\). For real \(r=1\) the sum would be 2 giving \(p=-6<0\), rejected.
2. So \(r=\omega\): sum \(\omega+\omega^2=-1=-\dfrac p3\) gives \(p=3\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) 0

Explanation:
1. Let \(t=x+\dfrac1x\): \(t(t^2+1)=0\) gives \(t=0\) or \(t=\pm i\).
2. For real \(x\), \(|t|\ge2\), so none of these values is attainable: no real roots.

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) 1

Explanation:
1. Sum of squares \(=(a-2)^2+2(a+1)=a^2-2a+6=(a-1)^2+5\), least at \(a=1\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(a,\ b\)

Explanation:
1. \((x-\alpha)(x-\beta)=(x-a)(x-b)-5\).
2. So \((x-\alpha)(x-\beta)+5=(x-a)(x-b)\), which vanishes at \(a\) and \(b\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(2m=n+l\)

Explanation:
1. \(D=(n-l)^2-4(m-n)(l-m)=0\) expands to \((2m-n-l)^2=0\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) -1

Explanation:
1. \(\dfrac{3a+4}{a+1}=2\) gives \(a=-2\).
2. Sum \(=-\dfrac{2a+3}{a+1}=-\dfrac{-1}{-1}=-1\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) 64

Explanation:
1. \(2\alpha^2-35\alpha+2=0\) gives \(2\alpha-35=-\dfrac2\alpha\), similarly for \(\beta\).
2. \(\left(\dfrac{4}{\alpha\beta}\right)^3=4^3=64\) since \(\alpha\beta=1\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) \(2p^2=9q\)

Explanation:
1. \((\alpha+\beta)^2=9(\alpha-\beta)^2=9[(\alpha+\beta)^2-4\alpha\beta]\) gives \(p^2=9p^2-36q\).
2. So \(8p^2=36q\), i.e. \(2p^2=9q\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(rb^2=cq^2\)

Explanation:
1. For roots in ratio \(k\): \(\dfrac{b^2}c=k+\dfrac1k+2\), the same for both equations.
2. So \(\dfrac{b^2}c=\dfrac{q^2}r\), i.e. \(rb^2=cq^2\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (d) \(p=-1,\ q=-56\)

Explanation:
1. \(\alpha^2+\beta^2=25-32=-7\) and \(\dfrac{\alpha\beta}2=8\).
2. Sum \(=1=-p\), product \(=-56=q\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (b) \(-\dfrac15\)

Explanation:
1. \(4x^2-(5k+1)x+5k=0\): \((\alpha-\beta)^2=\dfrac{(5k+1)^2}{16}-5k=1\).
2. \(5k^2-14k-3=0\) gives \(k=3\) or \(-\dfrac15\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) 4

Explanation:
1. With \(t=|x-2|\): \(t^2+t-2=0\) gives \(t=1\) (as \(t\ge0\)).
2. \(x=1\) or \(x=3\), sum \(4\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (c) \(1+\sqrt2\)

Explanation:
1. Let \(x=2+\dfrac1x\): \(x^2-2x-1=0\), so \(x=1\pm\sqrt2\).
2. The value is positive (at least 2), so \(x=1+\sqrt2\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.

Answer: (a) \(x=3\)

Explanation:
1. \(x^2=6+x\) gives \((x-3)(x+2)=0\).
2. \(x>0\), so \(x=3\).

Teacher's Note:
1. Use the sum and product of roots, \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \), instead of finding the roots wherever possible.
2. The discriminant \( b^2 \) - 4ac decides whether the roots are real, equal or complex.