JEE Mathematics Quadratic Equations MCQs Set 14

Answer: (d) (i), (ii) and (iii)

Explanation:
1. If \(b^2\ge4ac\) the roots are real with sum \(-b/a \lt 0\) and product \(c/a>0\), so both are negative: (i) is correct (e.g. \(x^2+3x+2=0\), roots \(-1,-2\), which are also rational, so (iii) is correct).
2. If \(b^2 \lt 4ac\) the roots are complex with real part \(-b/(2a) \lt 0\); in the real case both roots are negative. So the real parts are always negative: (ii) is correct.
3. All three statements are correct.

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Answer: (c) (ii) and (iii) only

Explanation:
1. Let \(f(x)=(x-a)(x-c)+\lambda(x-b)(x-d)\). Then \(f(a)=\lambda(a-b)(a-d)\) has the sign of \(\lambda\), and \(f(c)=\lambda(c-b)(c-d)\) has the opposite sign, so there is a root between \(a\) and \(c\): (ii).
2. \(f(b)=(b-a)(b-c) \lt 0\) and \(f(d)=(d-a)(d-c)>0\), so there is a root between \(b\) and \(d\): (iii).
3. Since real roots exist, (i) is false; rationality is not guaranteed, so (iv) is not always true.

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Answer: (a) \(|\alpha|=|\beta|\)

Explanation:
1. Non-real roots of a real quadratic are complex conjugates, so \(|\alpha|=|\beta|\).
2. \(|\alpha|^2=\alpha\beta=\dfrac ca>1\), so \(|\alpha|\ne1\).

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Answer: (c) \(a=c\)

Explanation:
1. The roots are reciprocals, so their product \(\dfrac ca=1\), i.e. \(a=c\).
2. Non-real roots need \(b^2 \lt 4ac=4a^2\), i.e. \(|b| \lt 2|a|\), which does not force \(|b|\le a\), \(|b|\le|c|\) or \(b\le a\).
3. The statement that always holds is a = c.

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Answer: (a) \(b^2>4c^2\) and \(a^2+c^2>ab\)

Explanation:
1. Put \(y=\dfrac{x+a}{x^2+bx+c^2}\): \(yx^2+(by-1)x+(yc^2-a)=0\). Real \(x\) requires \((b^2-4c^2)y^2+(4a-2b)y+1\ge0\).
2. For a gap of excluded values \(y\), this quadratic in \(y\) must be negative on an interval: leading coefficient \(b^2-4c^2>0\) and discriminant \(>0\).
3. Discriminant: \((4a-2b)^2-4(b^2-4c^2)=16(a^2+c^2-ab)>0\Rightarrow a^2+c^2>ab\).

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Answer: (c) \(a\in[1,3]\)

Explanation:
1. Write \(y(a+x-2x^2)=ax^2+x-2\): \((a+2y)x^2+(1-y)x-(2+ay)=0\). For every real \(y\) a real \(x\) must exist.
2. \(D=(1-y)^2+4(a+2y)(2+ay)=(1+8a)y^2+(4a^2+14)y+(1+8a)\ge0\) for all \(y\).
3. This needs \(1+8a>0\) and \((4a^2+14)^2\le4(1+8a)^2\), i.e. \(4a^2+14\le2+16a\Rightarrow a^2-4a+3\le0\Rightarrow 1\le a\le3\).

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Answer: (b) 2

Explanation:
1. Let \(s=\sqrt{x^4+20}\ge0\); then \(x^4=s^2-20\) and \(s^2+s-42=0\), so \(s=6\) (the root \(-7\) is rejected).
2. \(x^4=16\Rightarrow x=\pm2\) (real), the other roots being imaginary.
3. There are 2 real roots.

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Answer: (c) Real and unequal roots

Explanation:
1. Since \(0 \lt \sin\alpha \lt \sin\beta \lt \sin\gamma \lt 1\), the left side is decreasing between consecutive poles, from \(+\infty\) to \(-\infty\).
2. So there is one root between the first two poles and another between the last two: two real, distinct roots.

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Answer: (a) \(-4,\ -1-\sqrt3\)

Explanation:
1. We need \(|x^2+4x+3|=-(2x+5)\ge0\), so \(x\le-\tfrac52\).
2. For \(x\le-3\): \(x^2+4x+3+2x+5=0\Rightarrow x^2+6x+8=0\Rightarrow x=-4\) (since \(-2>-3\) is rejected).
3. For \(-3 \lt x\le-\tfrac52\): \(-(x^2+4x+3)+2x+5=0\Rightarrow x^2+2x-2=0\Rightarrow x=-1-\sqrt3\approx-2.73\) (valid); \(-1+\sqrt3\) is rejected.

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Answer: (b) \(x\in(-\infty,-5)\cup(-3,3)\cup(5,\infty)\)

Explanation:
1. If \(|x| \lt 3\), the left side is negative, so the inequality holds: \(x\in(-3,3)\).
2. If \(|x|>3\), then \(|x|-3>0\) and the inequality becomes \(|x|-3>2\Rightarrow|x|>5\).
3. Union: \((-\infty,-5)\cup(-3,3)\cup(5,\infty)\).

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Answer: (c) \(x\in\left[0,\tfrac85\right]\cup\left[\tfrac52,\infty\right)\)

Explanation:
1. The inequality is \(-1\le\dfrac{x^2-5x+4}{x^2-4}\le1\), with \(x\ne\pm2\).
2. \(\dfrac{x^2-5x+4}{x^2-4}\le1\iff\dfrac{-5x+8}{x^2-4}\le0\); \(\dfrac{x^2-5x+4}{x^2-4}\ge-1\iff\dfrac{2x^2-5x}{x^2-4}\ge0\).
3. Solving the two sign conditions and intersecting gives \(0\le x\le\tfrac85\) or \(x\ge\tfrac52\) (the points \(\pm2\) are not in these sets).

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Answer: (c) \(pq-3r\)

Explanation:
1. \(pq=(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)=\sum\alpha^2\beta+3\alpha\beta\gamma\).
2. Hence \(\sum\alpha^2\beta=pq-3r\).

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Answer: (c) \(-10,\ -3\)

Explanation:
1. The constant term was correct: \(q=(-2)(-15)=30\).
2. Original equation: \(x^2+13x+30=0=(x+3)(x+10)\), so the roots are \(-3,-10\).

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Answer: (d) \((0,\pi)\)

Explanation:
1. For \(p\in(0,\pi)\): \(\cos p-1 \lt 0\) and \(\sin p>0\), so \(D=\cos^2p-4(\cos p-1)\sin p=\cos^2p+4(1-\cos p)\sin p>0\): real roots.
2. For other intervals the discriminant becomes negative for some \(p\) (e.g. \(p=-1\), \(p=4\)), so they fail.

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Answer: (b) An even integer

Explanation:
1. \(\alpha\) satisfies the second equation; with \(p=-(\alpha+\beta)\), \(q=\alpha\beta\), dividing by \(\alpha^n\) gives \(\alpha^n+(-1)^n(\alpha+\beta)^n+\beta^n=0\).
2. \(\dfrac\alpha\beta\) satisfies the third equation; multiplying by \(\beta^n\) gives \(\alpha^n+\beta^n+(\alpha+\beta)^n=0\).
3. Both hold only if \((-1)^n=1\), so \(n\) is even (for example \(n=2\) works with \(\alpha^2+\alpha\beta+\beta^2=0\)).

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Answer: (a) 2500

Explanation:
1. \(\dfrac{x^3+y^3}{x^2+y^2-xy}=x+y\), so the condition is \(\log_{10}(x+y)\le2\), i.e. \(x+y\le100\).
2. By AM-GM, \(xy\le\left(\dfrac{x+y}2\right)^2\le2500\), with equality at \(x=y=50\).

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Answer: (a) \(\alpha^{3/2}\) and \(\beta^{3/2}\)

Explanation:
1. Let \(r_1=\alpha^{-1/2},\ r_2=\beta^{-1/2}\): \(r_1+r_2=-\dfrac ba\), \(r_1r_2=\dfrac1a\).
2. The proposed roots \(u=\alpha^{3/2}=r_1^{-3}\), \(v=r_2^{-3}\) have \(uv=a^3\) and \(u+v=\dfrac{r_1^3+r_2^3}{(r_1r_2)^3}=a^3\cdot\dfrac{3ab-b^3}{a^3}=3ab-b^3\).
3. The equation is \(x^2+(b^3-3ab)x+a^3=0\), whose sum of roots is \(3ab-b^3\) and product \(a^3\): it matches.

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Answer: (b) \(\sqrt2\)

Explanation:
1. For \(x\ge\tfrac32\): \(x^2+2x-7=0\Rightarrow x=-1+2\sqrt2\approx1.83\) (valid).
2. For \(x \lt \tfrac32\): \(x^2-2x-1=0\Rightarrow x=1-\sqrt2\approx-0.41\) (valid); \(1+\sqrt2\) is rejected.
3. Sum \(=(-1+2\sqrt2)+(1-\sqrt2)=\sqrt2\).

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19. n
20. d
21.
22. c
23. h
24. e
25. c
26. k
27.
28. e
29. a
30. c
31. h
32.
33. r
34. o
35. o
36. t
37.
38. a
39. g
40. a
41. i
42. n
43. s
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45.
46. i
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48. s
49.
50. r
51. a
52. n
53. g
54. e
55. .

Answer: (a) no solution

Explanation:
1. We need \(x\ge3\). Squaring: \(3x^2+x+5=x^2-6x+9\Rightarrow2x^2+7x-4=0\Rightarrow x=\tfrac12\) or \(-4\).
2. Neither satisfies \(x\ge3\), so there is no solution.

Teacher's Note:
1. A
2. l
3. w
4. a
5. y
6. s
7.
8. c
9. h
10. e
11. c
12. k
13.
14. r
15. o
16. o
17. t
18. s
19.
20. a
21. g
22. a
23. i
24. n
25. s
26. t
27.
28. x
29.
30. -
31.
32. 3
33.
34. >
35. =
36.
37. 0
38. .

Answer: (c) \(-2 \lt a \lt 3\)

Explanation:
1. The roots lie on either side of \(2\) iff \(f(2) \lt 0\) (upward parabola).
2. \(f(2)=4-2(a+1)+a^2+a-8=a^2-a-6=(a-3)(a+2) \lt 0\Rightarrow-2 \lt a \lt 3\).

Teacher's Note:
1. P
2. l
3. a
4. c
5. e
6.
7. t
8. h
9. e
10.
11. n
12. u
13. m
14. b
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17.
18. b
19. e
20. t
21. w
22. e
23. e
24. n
25.
26. t
27. h
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30. r
31. o
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36. u
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47. i
48. g
49. n
50.
51. o
52. f
53.
54. f
55. (
56. 2
57. )
58. .

Answer: (c) \(\{-2,5\}\)

Explanation:
1. \((\alpha-\beta)^2=p^2-3p=10\Rightarrow p^2-3p-10=0\).
2. \((p-5)(p+2)=0\Rightarrow p=5\) or \(-2\).

Teacher's Note:
1. U
2. s
3. e
4.
5. (
6. a
7. l
8. p
9. h
10. a
11.
12. -
13.
14. b
15. e
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18. )
19. ^
20. 2
21.
22. =
23.
24. (
25. s
26. u
27. m
28. )
29. ^
30. 2
31.
32. -
33.
34. 4
35.
36. (
37. p
38. r
39. o
40. d
41. u
42. c
43. t
44. )
45. .