JEE Mathematics Quadratic Equations MCQs Set 15

Answer: (a) \(\dfrac34\)

Explanation:
1. Let \(u=\sqrt{2x+1},\ v=\sqrt{2x-1}\). Then \(u-v=1\) and \(u^2-v^2=2\), so \(u+v=2\).
2. Hence \(u=\tfrac32,\ v=\tfrac12\) and \(\sqrt{4x^2-1}=uv=\tfrac34\).

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Answer: (c) \(\left(-\tfrac12,0\right)\cup\left(0,\tfrac12\right)\)

Explanation:
1. \(x^4+x^2+1=(x^2+x+1)(x^2-x+1)\) and \(x^2+x+1\ne0\) for real \(x\); dividing gives \((a-1)(x^2-x+1)+(a+1)(x^2+x+1)=0\).
2. This simplifies to \(ax^2+x+a=0\). Two real distinct roots need \(a\ne0\) and \(1-4a^2>0\).
3. So \(a\in\left(-\tfrac12,0\right)\cup\left(0,\tfrac12\right)\).

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Answer: (b) \(280\sqrt2\)

Explanation:
1. \(e^{4\ln k}=k^4\) (with \(k>0\)), so \(\alpha+\beta=4\sqrt2k\) and \(\alpha\beta=2k^4-1\).
2. \(\alpha^2+\beta^2=32k^2-2(2k^4-1)=66\Rightarrow k^4-8k^2+16=0\Rightarrow k=2\).
3. Then \(\alpha+\beta=8\sqrt2\), \(\alpha\beta=31\) and \(\alpha^3+\beta^3=(8\sqrt2)^3-3(31)(8\sqrt2)=1024\sqrt2-744\sqrt2=280\sqrt2\).

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Answer: (d) \(\left(-\infty,\tfrac23\right]\cup\left[\tfrac{10}3,\infty\right)\)

Explanation:
1. \(f(-1)=3a-2\) and \(f(2)=10-3a\). Exactly one root in \((-1,2)\) when \(f(-1)f(2) \lt 0\): \((3a-2)(10-3a) \lt 0\Rightarrow a \lt \tfrac23\) or \(a>\tfrac{10}3\).
2. At \(a=\tfrac23\): roots \(-1\) and \(-\tfrac23\), so exactly one root lies in \((-1,2)\): include. At \(a=\tfrac{10}3\): roots \(2\) and \(\tfrac53\): include.
3. A double root in the interval would count as two roots, so it is excluded. Hence \(a\le\tfrac23\) or \(a\ge\tfrac{10}3\).

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Answer: (d) 5

Explanation:
1. Each root satisfies \(\alpha-2=\dfrac1\alpha\), i.e. \(\alpha^2-2\alpha-1=0\).
2. As the roots are distinct, the equation is \(x^2-2x-1=0\): \(b=-2,\ c=-1\).
3. \(b^2+c^2=4+1=5\).

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Answer: (c) \(-(q+r)\)

Explanation:
1. \((\alpha-\gamma)(\alpha-\delta)=\alpha^2+p\alpha-r\), the second polynomial evaluated at \(\alpha\).
2. Since \(\alpha^2+p\alpha=-q\), the value is \(-q-r=-(q+r)\).

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Answer: (c) -1

Explanation:
1. Let \(f(x)=ax^2+bx+c\): \(f(0)=c>0\), \(f(1) \lt 0\), \(f(-1) \lt 0\).
2. Sign changes show one root in \((0,1)\) and one in \((-1,0)\).
3. \([\alpha]+[\beta]=0+(-1)=-1\).

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Answer: (b) \(a(a+|b|+c) \lt 0\)

Explanation:
1. Dividing by \(a\): the sign of \(a(a+|b|+c)\) is the sign of \(1+|\tfrac ba|+\tfrac ca=1+|\alpha+\beta|+\alpha\beta\).
2. If \(\alpha+\beta\ge0\): \(1+\alpha+\beta+\alpha\beta=(1+\alpha)(1+\beta) \lt 0\) as \(\alpha \lt -3,\beta>2\). If \(\alpha+\beta \lt 0\): \((1-\alpha)(1-\beta) \lt 0\).
3. So \(a(a+|b|+c) \lt 0\) always. Options (c), (d) fail: \(f(-3)\) and \(f(2)\) have the same sign (opposite to \(a\)).

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Answer: (c) 5

Explanation:
1. The leading coefficient \(a^2+a+2>0\) always, so \(1\) lies between the roots iff \(f(1) \lt 0\).
2. \(f(1)=a^2+a+2-a-4-7=a^2-9 \lt 0\Rightarrow-3 \lt a \lt 3\).
3. Integers: \(-2,-1,0,1,2\): \(5\) values.

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Answer: (c) 5

Explanation:
1. It must have two distinct real roots: \(D=4P^2-4(3P+4)>0\Rightarrow P^2-3P-4>0\).
2. \((P-4)(P+1)>0\Rightarrow P>4\) or \(P \lt -1\).
3. The smallest positive integer is \(5\).

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Answer: (c) 3

Explanation:
1. We need \((\lambda^2+\lambda-2)x^2+(\lambda+2)x-1 \lt 0\) for all \(x\). If \(\lambda=-2\) it reduces to \(-1 \lt 0\): valid. If \(\lambda=1\) it is \(3x-1 \lt 0\): not valid.
2. Otherwise we need \(\lambda^2+\lambda-2 \lt 0\) (\(-2 \lt \lambda \lt 1\)) and \(D=(\lambda+2)(5\lambda-2) \lt 0\Rightarrow\lambda \lt \tfrac25\). So \(-2 \lt \lambda \lt \tfrac25\).
3. Integral values: \(-2,-1,0\); \(k=-3\), \(|k|=3\).

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Answer: (c) 7

Explanation:
1. Let \(f(x)=x^2-(a+1)x+2a\): \(f(0)=2a\), \(f(3)=6-a\). Exactly one root in \((0,3)\) requires \(f(0)f(3) \lt 0\), i.e. \(a \lt 0\) or \(a>6\).
2. Check \(a=6\): roots \(3,4\), none inside. For \(a=1,\ldots,5\) the discriminant \(a^2-6a+1\) is negative (no real roots).
3. \(a=7\): \(x^2-8x+14\) has roots \(4\pm\sqrt2\), exactly one (\(\approx2.59\)) in \((0,3)\). So the smallest is \(7\).

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Answer: (b) 2

Explanation:
1. We need \(k-1>0\) and \(D \lt 0\): \((k+1)^2-4(k-1)(k+1) \lt 0\).
2. \((k+1)(5-3k) \lt 0\Rightarrow k>\tfrac53\) (since \(k>1\)).
3. The least integer is \(2\).

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Answer: (b) 6

Explanation:
1. Adding the first two at the common root \(r\): \(2r^2+(a+b)r+27=0\). The third gives \(r^2+(a+b)r+36=0\). Subtracting: \(r^2=9\Rightarrow r=3\).
2. First equation: \(9+3a+12=0\Rightarrow a=-7\). Second: \(9+3b+15=0\Rightarrow b=-8\). (Third check: \(9-45+36=0\).)
3. \(b-2a=-8+14=6\).

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Answer: (b) 1

Explanation:
1. The roots are \(e^{\pm i\pi/3}\), so \(\alpha^{2020}+\beta^{2020}=2\cos\dfrac{2020\pi}3\).
2. \(2020=6\cdot336+4\), so the angle reduces to \(\dfrac{4\pi}3\) and \(2\cos\dfrac{4\pi}3=-1\).
3. Hence the modulus is \(1\).

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Answer: (c) 5

Explanation:
1. Let the common ratio be \(r\): \(\alpha(1+r)=4\) and \(\alpha r^2(1+r)=36\), so \(r^2=9\), \(r=3\), \(\alpha=1\).
2. The terms are \(1,3,9,27\): \(A=\alpha\beta=3\), \(B=\gamma\delta=243=3^5\).
3. \(A^t=B\Rightarrow3^t=3^5\Rightarrow t=5\).

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Answer: (b) 2

Explanation:
1. \(3x^2-17x+10=(3x-2)(x-5)\), so the roots are \(5\) and \(\tfrac23\).
2. Common root \(5\): \(25-25+m=0\Rightarrow m=0\). Common root \(\tfrac23\): \(\tfrac49-\tfrac{10}3+m=0\Rightarrow m=\tfrac{26}9\).
3. \(A=\tfrac{26}9\approx2.89\), so \([A]=2\).

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Answer: (c) 1

Explanation:
1. \(rs=(r+s)^2\Rightarrow r^2+rs+s^2=0\Rightarrow\left(\dfrac rs\right)^2+\dfrac rs+1=0\).
2. So \(\dfrac rs\) is a primitive cube root of unity \(\omega\), and \(\omega^3=1\).

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57. o
58. f
59.
60. u
61. n
62. i
63. t
64. y
65. .

Answer: (c) 2

Explanation:
1. The second factor has discriminant \(k^2+4>0\) for every \(k\), so it always has two distinct real roots.
2. The first factor may have none (e.g. \(k=0\)). So the minimum number of distinct real roots is \(2\).

Teacher's Note:
1. O
2. n
3. e
4.
5. f
6. a
7. c
8. t
9. o
10. r
11.
12. a
13. l
14. w
15. a
16. y
17. s
18.
19. c
20. o
21. n
22. t
23. r
24. i
25. b
26. u
27. t
28. e
29. s
30.
31. t
32. w
33. o
34.
35. r
36. e
37. a
38. l
39.
40. r
41. o
42. o
43. t
44. s
45. .

Answer: (b) 3

Explanation:
1. Let \(f(x)=4x^2-16x+\lambda\) (upward parabola). The conditions are \(f(1)>0,\ f(2) \lt 0,\ f(3)>0\).
2. \(f(1)=\lambda-12>0\), \(f(2)=\lambda-16 \lt 0\), \(f(3)=\lambda-12>0\), so \(12 \lt \lambda \lt 16\).
3. Integers: \(13,14,15\): \(3\) values.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. s
10. i
11. g
12. n
13.
14. o
15. f
16.
17. t
18. h
19. e
20.
21. q
22. u
23. a
24. d
25. r
26. a
27. t
28. i
29. c
30.
31. a
32. t
33.
34. t
35. h
36. e
37.
38. p
39. o
40. i
41. n
42. t
43. s
44.
45. t
46. h
47. a
48. t
49.
50. s
51. e
52. p
53. a
54. r
55. a
56. t
57. e
58.
59. t
60. h
61. e
62.
63. r
64. o
65. o
66. t
67. s
68. .

Answer: (c) 9

Explanation:
1. \(2x^2+3x+4\) has discriminant \(9-32 \lt 0\): its roots are non-real conjugates.
2. A real-coefficient quadratic sharing one non-real root also has the conjugate, so the equations are proportional: \((a,b,c)=k(2,3,4)\).
3. The least natural numbers occur at \(k=1\): \(a+b+c=9\).

Teacher's Note:
1. A
2.
3. n
4. o
5. n
6. -
7. r
8. e
9. a
10. l
11.
12. r
13. o
14. o
15. t
16.
17. f
18. o
19. r
20. c
21. e
22. s
23.
24. b
25. o
26. t
27. h
28.
29. r
30. o
31. o
32. t
33. s
34.
35. t
36. o
37.
38. b
39. e
40.
41. c
42. o
43. m
44. m
45. o
46. n
47. .