NDA Quadratic Equations and Inequations previous year questions with solutions

Answer: (d) 1a + 1c = 2b

Method. sum of coefficients zero ⇒ x = 1 is a root; equal roots ⇒ product = 1

Rearrange: a(b − c)x² + b(c − a)x + c(a − b) = 0.
The coefficients add to 0, so x = 1 is a root; equal roots ⇒ both are 1 ⇒ c(a − b)a(b − c) = 1.
ca − bc = ab − ac ⇒ 2ac = b(a + c).
Divide by abc: 2b = 1a + 1c.

Answer: (c) HP

Method. reciprocals in AP ⇒ HP

From the previous item, equal roots give 1a + 1c = 2b (since 2ac = b(a + c)).
So 1a, 1b, 1c are in AP.
Hence a, b, c are in HP.

Answer: (a) None

Method. a sum of squares vanishes only if each square vanishes

(x − 1)² + (x − 3)² + (x − 5)² = 0 for real x needs x = 1, x = 3 and x = 5 at once, impossible.
Check: the equation is 3x² - 18x + 35 = 0, D = 324 − 420 = -96 < 0.
So there is no real root.

Answer: (c) a², b², c² are in HP.

Method. sum of coefficients zero ⇒ x = 1 is a root; equal roots ⇒ product of roots = 1

Coefficients: A = a²(b² - c²), B = b²(c² - a²), C = c²(a² - b²); A + B + C = 0, so x = 1 is a root.
Roots equal ⇒ both roots are 1 ⇒ CA = 1.
c²(a² - b²) = a²(b² - c²) ⇒ 2a²c² = a²b² + b²c².
Divide by a²b²c²: 2b² = 1a² + 1c².
So a², b², c² are in HP.

Answer: (c) b²(c² - a²)2a²(c² - b²)

Method. equal roots: each root = -B2A

For equal roots each root is half the sum: x = -B2A = -b²(c² - a²)2a²(b² - c²) = b²(c² - a²)2a²(c² - b²).
This is option (c) (its value is 1, since x = 1 is a root).
The others are -BA = 2 (option a), BA = -2 (option b) and B2A = -1 (option d), none a root.

Answer: (c) a₀ = 6, a₃ = 0

Method. irrational roots of an integer-coefficient polynomial occur in conjugate pairs

The coefficients are integers, so with -⁠√2⁠ a root its conjugate ⁠√2⁠ is also a root; with ⁠√3⁠ a root, -⁠√3⁠ is also a root.
(Equivalently x² - 2 and x² - 3 both divide the quartic.)
The quartic is monic of degree 4, so it equals (x² - 2)(x² - 3) = x⁴ - 5x² + 6.
Thus a₀ = 6, a₁ = 0, a₂ = -5, a₃ = 0.
Only option (c), a₀ = 6 and a₃ = 0, holds.

Answer: (d) 1, 2 and 3

Method. b² = ac turns the equation into x² + rx + r² = 0

GP ⇒ b² = ac, with a, c > 0, so b ≠ 0.
1. D = b² - 4ac = ac − 4ac = -3ac < 0: imaginary roots. True.
2. Let r = ba; then c = b²a = ar², and dividing by a: x² + rx + r² = 0.
Since x³ - r³ = (x − r)(x² + rx + r²), the roots are rω and rω²; their ratio is ω : ω² = 1 : ω. True.
3. Product of roots = ca = b²a² (as ac = b²). True.
All three are correct.

Answer: (d) 1 − cosβ ≥ 0

Method. discriminant check, including the degenerate case cosβ = 1

For β∈[0, π]: sinβ ≥ 0 and 1 − cosβ ≥ 0.
If cosβ = 1 (β = 0) the equation becomes x = 0, which has the real root 0.
If cosβ ≠ 1: D = cos²β − 4(cosβ − 1)sinβ = cos²β + 4(1 − cosβ)sinβ ≥ 0, so the roots are real.
So the equation has a real root for every β∈[0, π], i.e. exactly when 1 − cosβ ≥ 0, which holds throughout.
(c) is only part of the answer: it leaves out β = 0, where the equation reduces to x = 0 and still has the real root 0; the full condition is (d).

Answer: (b) Real and unequal

Method. discriminant of a quadratic

With k = c2: D = (a + b + c)² - 4(a + b)·c2 = (a + b + c)² - 2c(a + b).
Expand: D = (a + b)² + 2c(a + b) + c² - 2c(a + b) = (a + b)² + c².
Since c ≠ 0, c² > 0, so D > 0.
Hence the roots are real and unequal.

Answer: (c) 1 and ca + b

Method. sum of coefficients zero ⇒ x = 1 is a root

With k = c, put x = 1: (a + b) − (a + b + c) + c = 0, so 1 is a root.
Product of roots = ca + b, so the other root is ca + b.
Roots: 1 and ca + b.

Answer: (d) 3

Method. non-real roots of a real quadratic occur in conjugate pairs

Taking a, b real, the other root is 2 + i⁠√3.⁠
Sum = 4 = -a, so a = -4.
Product = (2 − i⁠√3)(2⁠ + i⁠√3)⁠ = 4 + 3 = 7 = b.
a + b = -4 + 7 = 3.

Answer: (c) a + b − c = 0

Method. equal roots ⇔ discriminant zero; the discriminant is a perfect square

The equation is x² - (a + b + c)x + c(a + b) = 0.
D = (a + b + c)² - 4c(a + b) = [(a + b) − c]².
Equal roots: D = 0, so a + b − c = 0.

Answer: (d) k ≤ 1

Method. real roots ⇔ discriminant ≥ 0

D = 2² - 4.1·k = 4 − 4k.
Real roots: 4 − 4k ≥ 0, so k ≤ 1.

Answer: (c) 7

Method. real roots ⟺ q² ≥ 4p; count pairs

D = q² - 4p ≥ 0.
q = 1: 1 ≥ 4p, none. q = 2: 4 ≥ 4p ⟹ p = 1 (1 equation).
q = 3: 9 ≥ 4p ⟹ p = 1, 2 (2 equations). q = 4: 16 ≥ 4p ⟹ p = 1, 2, 3, 4 (4 equations).
Total = 0 + 1 + 2 + 4 = 7.

Answer: (c) a, b and c are in HP

Method. coefficients sum to zero, so x = 1 is a root; equal roots force the product to be 1

a(b − c) + b(c − a) + c(a − b) = 0, so x = 1 is a root; equal roots ⟹ both roots are 1.
Product of roots: c(a − b)a(b − c) = 1 ⟹ ca − bc = ab − ac ⟹ 2ac = ab + bc.
b = 2aca + c: a, b, c are in HP.

Answer: (c) m² ≥ 8

Method. real roots ⟺ D ≥ 0

D = m² - 4(1)(2) = m² - 8.
Real roots ⟺ m² - 8 ≥ 0 ⟹ m² ≥ 8.

Answer: (c) Both 1 and 2

Method. discriminant written as a sum of squares

Expand: 4x² - 4(p + q)x + 4pq − r² = 0.
D = 16(p + q)² - 16(4pq − r²) = 16[(p − q)² + r²].
D ≥ 0 for all real p, q, r, so the roots are always real: statement 1 is correct.
If p = q and r = 0, then D = 0 and the roots are equal (both equal p): statement 2 is correct.

Answer: (b) 1, 3, 4 and 5 only

Method. a rational expression is a quotient of two polynomials (a polynomial is one, with denominator 1)

1. x + x² - 1x = x³ + x² - 1x: rational.
2. Contains a square root of a non-square expression: not rational.
3. 3x² - 5x + ab is a polynomial = 3x² - 5x + ab1: rational.
4. 2x² - ax + b³: quotient of polynomials, rational.
5. 1x − 2x + 5 = 5 − xx(x + 5): rational.
Rational: 1, 3, 4 and 5.

Answer: (c) β ∈ (1, ∞)

Method. non-real roots of a real quadratic are conjugates; product = |z|²

α, β are real, so the non-real roots are conjugates: z = 1 + iy and 1 − iy with y ≠ 0 (Re z = 1).
Product of roots β = (1 + iy)(1 − iy) = 1 + y².
y ≠ 0 ⟹ y² > 0 ⟹ β > 1.
So β ∈ (1, ∞) is necessary.

Answer: (b) (p − q)/⁠(q − r), 1

Method. coefficients summing to zero give x = 1 as a root

Sum of coefficients: (q − r) + (r − p) + (p − q) = 0, so x = 1 is a root.
Product of roots = p − qq − r, so the other root is p − qq − r.
Roots: p − qq − r, 1.

Answer: (c) 181

Method. real roots ⟺ D ≥ 0

D = (-4)² - 4(1)(-log₃ P) = 16 + 4 log₃ P.
D ≥ 0 ⟹ log₃ P ≥ -4 ⟹ P ≥ 3^-4 = 181.
Minimum value of P is 181.

Answer: (c) 1

Method. shift to u = α + 1, v = β + 1 and use uv = 1 − q

x² - px − (p + q) = 0 ⇒ α + β = p, αβ = -(p + q).
Let u = α + 1, v = β + 1: uv = αβ + α + β + 1 = 1 − q.
α² + 2α + 1 = u², and α² + 2α + q = u² - (1 − q) = u² - uv = u(u − v).
So the first term is uu − v; likewise the second is vv − u.
Sum = u − vu − v = 1.

Answer: (d) (p + q)/⁠2

Method. form the quadratic and set the x-coefficient to zero

r[(2x + q) + (2x + p)] = (2x + p)(2x + q) ⇒ r(4x + p + q) = 4x² + 2(p + q)x + pq.
⇒ 4x² + (2p + 2q − 4r)x + pq − r(p + q) = 0.
Sum of roots = 0 ⇒ 2p + 2q − 4r = 0 ⇒ r = p + q2.

Answer: (a) 0

Method. each root satisfies aα² + bα + c = 0

ap_n+1 + bp_n + cp_n-1 = α^n-1(aα² + bα + c) + β^n-1(aβ² + bβ + c).
Both brackets are 0, since α, β are roots.
So the value is 0.

Answer: (b) 4

Method. the root set {α, β} must equal {α², β²}

Need {α², β²} = {α, β} for x² + px + q = 0.
Case 1: α² = α and β² = β ⇒ α, β ∈ {0, 1}: roots (0, 0), (0, 1), (1, 1) ⇒ x² = 0, x² - x = 0, x² - 2x + 1 = 0.
Case 2: α² = β, β² = α, α ≠ β ⇒ α⁴ = α, α ∉ {0, 1} ⇒ α³ = 1, α ≠ 1 ⇒ roots ω, ω² ⇒ x² + x + 1 = 0.
Total: 4 equations.

Answer: (a) A, G

Method. sum and product of roots give A and G directly

α + β = 2b, αβ = c².
A = α + β2 = b, G = ⁠√(αβ)⁠ = c (c > 0).
x² - (b + c)x + bc = (x − b)(x − c), so its roots are b and c, i.e. A and G.

Answer: (b) 6

Method. (α − β)² = (α + β)² - 4αβ

α + β = k, αβ = k.
(α − β)² = k² - 4k = (2⁠√3)²⁠ = 12 ⇒ k² - 4k − 12 = 0 ⇒ (k − 6)(k + 2) = 0.
k = 6 or k = -2; among the options, k = 6.
Check: x² - 6x + 6 = 0 has roots 3 ± ⁠√3,⁠ which differ by 2⁠√3.⁠

Answer: (c) 1

Method. subtract the two root conditions and cancel (n − m)

n² + pn + m = 0 and m² + pm + n = 0.
Subtract: (n² - m²) + p(n − m) − (n − m) = 0 ⇒ (n − m)(n + m + p − 1) = 0.
Since m ≠ n, n + m + p − 1 = 0.
So p + m + n = 1.

Answer: (b) 2 only

Method. check whether each set of data fixes α to a single value

Statement 1: αβ = (α + β)² - (α² + β²)2 = -1, so the roots are 1 and -1. α could be either, so α is not determined: not sufficient.
Statement 2: a = 0 gives β = -α. Then αβ² = α³ = -1, so α = -1 (real), β = 1, b = -1, distinct real roots as required. α is determined: sufficient.
Only statement 2 is sufficient.

Answer: (b) 4

Method. HM = 2αβα + β from Vieta's relations

α + β = 4 + 2⁠√33⁠ + 2⁠√2,⁠ αβ = 8 + 4⁠√33⁠ + 2⁠√2.⁠
HM = 2αβα + β = 2(8 + 4⁠√3)4⁠ + 2⁠√3⁠ = 2 × 2(4 + 2⁠√3)4⁠ + 2⁠√3⁠ = 4.
(The discriminant (4 + 2⁠√3)²⁠ - 4(3 + 2⁠√2)(8⁠ + 4⁠√3)⁠ is negative, so the roots are complex conjugates; the HM is taken by the formula, and comes out real.)

Answer: (a) ⁠√2(√6⁠ - ⁠√3⁠ + ⁠√2⁠ − 1)

Method. GM = ⁠√(αβ)⁠; denest 8 + 4⁠√3⁠ and 3 + 2⁠√2⁠ as perfect squares

αβ = 8 + 4⁠√33⁠ + 2⁠√2.⁠
8 + 4⁠√3⁠ = 2(4 + 2⁠√3)⁠ = 2(⁠√3⁠ + 1)², and 3 + 2⁠√2⁠ = (⁠√2⁠ + 1)².
GM = ⁠√(αβ)⁠ = ⁠√2(√3⁠ + 1)⁠√2⁠ + 1 = ⁠√2(√3⁠ + 1)(⁠√2⁠ − 1).
(⁠√3⁠ + 1)(⁠√2⁠ − 1) = ⁠√6⁠ - ⁠√3⁠ + ⁠√2⁠ − 1, so GM = ⁠√2(√6⁠ - ⁠√3⁠ + ⁠√2⁠ − 1) ≈ 1.600.
(The roots are complex conjugates with αβ > 0, so ⁠√(αβ)⁠ is their common modulus.)

Answer: (d) Infinitely many

Method. sum = product reduces to one linear condition on the coefficients

For ax² + bx + c = 0: sum = -ba, product = ca.
Sum = product ⇔ c = -b.
Every equation ax² + bx − b = 0 (a ≠ 0, b any) qualifies, e.g. x² + 2x − 2 = 0, x² + 3x − 3 = 0, ….
So there are infinitely many such equations.

Answer: (a) 3⁠√c⁠

Method. (p − q)² = p² + q² - 2pq

pq = c.
p² + q² = 11pq, so (p − q)² = 11pq − 2pq = 9pq = 9c.
Since p > q and c > 0, p − q = 3⁠√c.⁠

Answer: (d) p³ + q³ - 3pq

Method. α³ + β³ = (α + β)³ - 3αβ(α + β)

α + β = -p, αβ = q.
α³ + β³ = (-p)³ - 3q(-p) = -p³ + 3pq = -m, so m = p³ - 3pq.
α³β³ = q³ = n.
m + n = p³ + q³ - 3pq.

Answer: (a) x² - αx − βx + αβ + c = 0

Method. sum and product of roots, then rebuild the equation

(x − a)(x − b) = c, so x² - (a + b)x + ab − c = 0 has roots α, β.
Hence α + β = a + b and αβ = ab − c, so a + b = α + β and ab = αβ + c.
The equation with roots a, b is x² - (a + b)x + ab = 0, that is x² - (α + β)x + αβ + c = 0, i.e. x² - αx − βx + αβ + c = 0.

Shortcut. (x − a)(x − b) − c = (x − α)(x − β) for every x, so (x − a)(x − b) = (x − α)(x − β) + c, and its zeros a, b solve x² - αx − βx + αβ + c = 0.

Answer: (d) 15

Method. α² - β² = (α + β)(α − β)

α + β = 8, so α² - β² = 8(α − β) = 16, giving α − β = 2.
α = 5, β = 3.
q = αβ = 15.

Answer: (d) No such value of k exists

Method. equal magnitude, opposite sign ⇒ sum of roots 0 and product negative

Roots r and -r (r ≠ 0) need sum = 0 and product = -r² < 0.
Sum = k² + 5k3 = 0 gives k = 0 or k = -5.
k = 0: the equation is 3x² = 0, roots 0, 0 — no roots of opposite sign.
k = -5: the equation is 3x² + 100 = 0, which has no real roots.
No value of k works (and options a–⁠c only offer positive k, for which the sum k² + 5k ≠ 0).

Answer: (b) One

Method. Vieta's relations with the coefficients themselves as roots

Sum: p + q = -p, so q = -2p.
Product: pq = q; since q ≠ 0, p = 1.
Then q = -2.
Check: x² + x − 2 = (x − 1)(x + 2) has roots 1 = p and -2 = q.
q has exactly one possible value.

Answer: (c) a² - b² + 2ac = 0

Method. square the sum of roots and use sin²θ + cos²θ = 1

sin θ + cos θ = -ba, sin θ cos θ = ca.
(sin θ + cos θ)² = 1 + 2 sin θ cos θ, so b²a² = 1 + 2ca.
Multiply by a²: b² = a² + 2ac.
So a² - b² + 2ac = 0.

Answer: (c) -15

Method. (α − β)² = (α + β)² - 4αβ

α + β = 5k + 14, αβ = 5k4.
(α − β)² = 1: (5k + 1)²16 − 5k = 1.
(5k + 1)² - 80k = 16 ⇒ 25k² - 70k − 15 = 0 ⇒ 5k² - 14k − 3 = 0.
(5k + 1)(k − 3) = 0, so k = 3 or k = -15.
Check k = -15: 4x² - 1 = 0, roots ±12, which differ by 1.
Of the options, k = -15.

Answer: (a) β = -2α² - 2α

Method. use α + β to write β, then replace the linear term using the equation α satisfies

α + β = -24 = -12, so β = -12 − α.
Since α is a root, 4α² + 2α − 1 = 0, i.e. 2α² = 12 − α.
Then -2α² - 2α = -(12 − α) − 2α = -12 − α = β.
Check the others with 4α² = 1 − 2α: 4α² - 3α = 1 − 5α, α² - 3α = 14 − 7α2, -2α² + 2α = -12 + 3α; none equals -12 − α.
So β = -2α² - 2α.

Answer: (c) 5

Method. reciprocal roots ⇒ product of roots = 1

If the roots are α and 1α, their product is 1.
Product = k5 = 1, so k = 5.
(Then 5x² + 26x + 5 = (5x + 1)(x + 5): roots -15 and -5, reciprocals.)

Answer: (d) 4

Method. solve for the sum s = α + β and product p = αβ, each pair giving one equation x² - sx + p = 0

αβ = (αβ)² gives p = 0 or p = 1.
α² + β² = s² - 2p, so the first relation is s = s² - 2p.
p = 0: s² = s, so s = 0 or 1, giving x² = 0 and x² - x = 0.
p = 1: s² - s − 2 = 0, so s = 2 or -1, giving x² - 2x + 1 = 0 and x² + x + 1 = 0.
Each of the four satisfies both relations (x² + x + 1 = 0 has the complex roots ω, ω², and ω + ω² = ω² + ω⁴ = -1).
Number of such equations = 4.

The question does not say the roots are real or distinct, so x² + x + 1 = 0 (non-real roots) and the two equations with a repeated root all count. Restricting to real roots would give 3, option (c), which the question does not ask for.

Answer: (c) 32

Method. α² + β² = (α + β)² - 2αβ as a quadratic in k

α + β = k − 2, αβ = -k + 12.
α² + β² = (k − 2)² + (k + 1) = k² - 3k + 5 = (k − 32)² + 114.
Minimum at k = 32. (The roots are real there: D = 4(k − 2)² + 8(k + 1) = 21 > 0.)

Answer: (b) 7110

Method. p³ + q³ = (p + q)³ - 3pq(p + q)

p + q = 30, pq = 221.
p³ + q³ = 30³ - 3 × 221 × 30 = 27000 − 19890 = 7110.

Shortcut. x² - 30x + 221 = (x − 13)(x − 17); 13³ + 17³ = 2197 + 4913 = 7110.

Answer: (b) a < 0, b < 0

Method. signs of roots from sum and product

a + b = -qp < 0 and ab = rp > 0, since p, q, r > 0.
Positive product ⟹ same sign; negative sum ⟹ both negative.
a < 0, b < 0.

Answer: (d) Greatest value 94

Method. find α, β from sum and product, then complete the square

Sum: α + β = -α ⟹ β = -2α. Product: αβ = -β ⟹ α = -1 (as β ≠ 0). So β = 2.
-x² + αx + β = -x² - x + 2 = 94 − (x + 12)².
A downward parabola: no least value; greatest value 94 at x = -12.

Answer: (b) b²q²

Method. equal root ratios ⟺ b²/⁠(ac) = q²/⁠(pr)

Equal ratios of roots give (sum)²product equal: b²ac = q²pr = λ (say).
Then D₁ = b² - 4ac = ac(λ − 4) and D₂ = q² - 4pr = pr(λ − 4) (b, q ≠ 0, and the roots are not equal, so λ ≠ 0, 4; D₁/⁠D₂ is asked, so D₂ ≠ 0).
D₁D₂ = acpr = b²/⁠λq²/⁠λ = b²q².

Answer: (a) 3x² + 8x + 16 = 0

Method. sum and product of the new roots from α + β and αβ

α + β = -23, αβ = 13.
α + 1β = αβ + 1β, β + 1α = αβ + 1α.
Sum = (αβ + 1)(α + β)αβ = 43 · (-23)13 = -83.
Product = (αβ + 1)²αβ = 16913 = 163.
Equation: x² + 83x + 163 = 0 ⟹ 3x² + 8x + 16 = 0.

Answer: (a) (-3, -2] ∪ [2, 3)

Method. difference of roots ⁠√(D)⁠ for a monic quadratic

For x² + kx + 1 = 0: α + β = -k, αβ = 1.
(α − β)² = (α + β)² - 4αβ = k² - 4, so |α − β| = ⁠√(k²⁠ - 4).
⁠√(k²⁠ - 4) < ⁠√5⁠ ⟹ k² - 4 < 5 ⟹ k² < 9 ⟹ -3 < k < 3.
With |k| ≥ 2: k ∈ (-3, -2] ∪ [2, 3).

Answer: (a) p²m = l²q

Method. equal root ratios ⟺ equal value of (sum)²/⁠product

Let the roots of the first equation be α, β and of the second γ, δ, with α/⁠β = γ/⁠δ = t.
(α + β)²/⁠(αβ) = (t + 1)²/⁠t depends only on the ratio t, so it is the same for both equations.
First: (α + β)²/⁠(αβ) = p²/⁠q. Second: (γ + δ)²/⁠(γδ) = l²/⁠m.
p²/⁠q = l²/⁠m ⟹ p²m = l²q.

Answer: (c) H.P.

Method. symmetric functions of roots, then test the reciprocal sequence

α + β = -b, αβ = c.
1α² + 1β² = (α + β)² - 2αβ(αβ)² = b² - 2cc².
Given: -b = b² - 2cc² ⟹ 2c = b² + bc².
Divide by bc: 2b = bc + c, i.e. c, 1b, bc are in A.P.
These are the reciprocals of 1c, b, cb, so 1c, b, cb are in H.P.

Answer: (a) A.P.

Method. symmetric functions of roots

α + β = -1a, αβ = ca.
1α² + 1β² = (α + β)² - 2αβ(αβ)² = 1a² − 2cac²a² = 1 − 2acc².
Given: -1a = 1 − 2acc² ⟹ -c² = a − 2a²c ⟹ 2ca² = a + c².
So the middle term ca² is the mean of a and c²: a, ca², c² are in A.P.

Answer: (c) 3k²4

Method. complete the square

x² + kx + k² = (x + k2)² + k² - k²4 = (x + k2)² + 3k²4.
The square is least (0) at x = -k2.
Minimum value = 3k²4.

Answer: (c) 3

Method. f(1) = f(4) ⇒ axis of symmetry x = 52

f(1) = f(4) ⇒ a + b + c = 16a + 4b + c ⇒ 15a + 3b = 0 ⇒ b = -5a.
Sum of roots = -ba = 5; one root is 2, so the other is 3.
Check: f(x) = a(x − 2)(x − 3), f(1) = 2a = 2 ⇒ a = 1, and f(4) = 2.1 = 2. Consistent.

Shortcut. The roots are symmetric about x = 1 + 42 = 52, so the other root is 5 − 2 = 3.

Answer: (c) 2

Method. a + b + c = f(1)

a + b + c = f(1) = 2 (given).
(Indeed f(x) = x² - 5x + 6, so 1 − 5 + 6 = 2.)

Answer: (c) Both 1 and 2

Method. substitute convenient integers x = 0 and x = 1

x = 0: the value is n, so n must be an integer.
x = 1: the value is 1 + m + n, an integer; with n an integer, m must be an integer.
(Conversely, integers m, n clearly give integer values.)
Both statements are correct.

Answer: (b) 4

Method. roots 2 ± ⁠√(4⁠ −⁠ k): impose reality and the interval on each

Real roots: 16 − 4k ≥ 0, so k ≤ 4.
Roots are 2 - ⁠√(4⁠ −⁠ k) and 2 + ⁠√(4⁠ −⁠ k).
2 - ⁠√(4⁠ −⁠ k) > 0 ⇒ ⁠√(4⁠ −⁠ k) < 2 ⇒ k > 0.
2 + ⁠√(4⁠ −⁠ k) < 5 ⇒ ⁠√(4⁠ −⁠ k) < 3 ⇒ k > -5 (already implied).
So 0 < k ≤ 4: k = 1, 2, 3, 4 (k = 4 gives the double root 2, which lies in (0, 5)).
Number of integral values = 4.

Answer: (d) (2x − 3) (2x − 9) ≤ 0

Method. sign of a product of two linear factors between its zeros

1.5 ≤ x ≤ 4.5 means 3 ≤ 2x ≤ 9.
So 2x − 3 ≥ 0 and 2x − 9 ≤ 0.
A non-negative number times a non-positive number is ≤ 0: (2x − 3)(2x − 9) ≤ 0.
Strict < fails at the end points x = 1.5 and x = 4.5, where the product is 0.

Answer: (c) 0 < p < 2

Method. opposite signs means the product f(0)·f(1) is negative

f(0) = p, f(1) = 3 − 5 + p = p − 2.
Opposite signs: p(p − 2) < 0.
So 0 < p < 2.

Answer: (c) 1 < x < 2

Method. |E| > E exactly when E < 0

|E| > E holds iff E < 0.
x² - 3x + 2 = (x − 1)(x − 2) < 0 ⟹ 1 < x < 2.

Answer: (b) A = {x|x < a} and B = {x|x > b}

Method. a product of two factors is positive outside the roots

With a < b, (x − a)(x − b) > 0 ⟺ x < a or x > b.
So the set is {x | x < a} ∪ {x | x > b}. Of the four options, only (b) has A ∪ B equal to this set: (a) gives {x > a}, (c) gives {x < b}, and (d) gives all real x because a < b.

Answer: (b) g(x) > 0

Method. g is again a quadratic with a > 0; show its discriminant is negative

Let f(x) = ax² + bx + c with a > 0 and b² - 4ac < 0 (positive for all real x).
f'(x) = 2ax + b, f''(x) = 2a, so g(x) = ax² + (b + 2a)x + (c + b + 2a).
Discriminant of g: (b + 2a)² - 4a(c + b + 2a) = b² - 4ac − 4a² < 0.
Leading coefficient a > 0 and negative discriminant ⟹ g(x) > 0 for all real x.
Since the discriminant is negative, g(x) is never 0; so (d) g(x) ≥ 0, though not false, is the weaker statement, and (b) g(x) > 0 is the precise one.

Answer: (c) Both the roots are complex

Method. a parabola that never meets the x-axis has negative discriminant

If the graph lies entirely above the x-axis, the polynomial is never zero for real x.
So it has no real root: D < 0.
A real quadratic with D < 0 has two (conjugate) complex roots.
Both roots are complex.

Answer: (b) II only

Method. substitute t = ⁠√x⁠ ≥ 0

I. For x ≥ 0, ⁠√x⁠ + x + 1 ≥ 1 > 0, so the equation has no real root at all. False.
II. Put t = ⁠√x:⁠ t² - 5t + 4 = 0 ⇒ t = 1 or 4 ⇒ x = 1 or 16.
Check: 5 − 1 − 4 = 0, 20 − 16 − 4 = 0. Two rational roots. True.
Only II is correct.

Answer: (a) 0

Method. clear denominators: both equations share the numerator xy + 5

x + 5y = 4 ⇒ xy + 5 = 4y; y + 5x = -4 ⇒ xy + 5 = -4x.
So 4y = -4x ⇒ x + y = 0.
Check a solution exists: y = -x gives x² - 4x − 5 = 0, x = 5, y = -5: 5 − 1 = 4 and -5 + 1 = -4.
Hence x + y = 0.

Answer: (d) 30

Method. solve the quadratic in [x]; an integer x has [x] = x

Put t = [x]: t² - 30t + 221 = 0 ⇒ (t − 13)(t − 17) = 0 ⇒ t = 13 or 17.
For integer x, [x] = x, so the integer solutions are x = 13 and x = 17.
Sum = 13 + 17 = 30.

Answer: (b) g(x) = 0 has only one real root which is an integer

Method. substitute t = ⁠√x⁠ ≥ 0

g(x) = 9x − 8⁠√x⁠ − 1 = 0. Put t = ⁠√x⁠ ≥ 0: 9t² - 8t − 1 = 0 ⇒ (9t + 1)(t − 1) = 0.
t = -19 is rejected (t ≥ 0); t = 1 gives x = 1.
Check: 9 − 8 − 1 = 0.
So g(x) = 0 has only one real root, and it is an integer.

Answer: (b) 2

Method. substitute y = x − 3 to make the equation symmetric, then solve as a quadratic in y²

With y = x − 3: 1 − x = -(y + 2), 5 − x = -(y − 2).
(y + 2)⁴ + (y − 2)⁴ = 2(y⁴ + 24y² + 16) = 82.
y⁴ + 24y² - 25 = 0 ⇒ (y² + 25)(y² - 1) = 0.
y² = 1 gives y = ±1, i.e. x = 4, 2; y² = -25 gives no real y.
Number of real roots = 2.

Answer: (b) 12

Method. sum of all four roots of the quartic by Vieta (or from the substitution y = x − 3)

From the previous item the roots are y = ±1, ±5i with x = y + 3: x = 2, 4, 3 + 5i, 3 − 5i.
Sum = 2 + 4 + (3 + 5i) + (3 − 5i) = 12.
Check by expansion: (1 − x)⁴ + (5 − x)⁴ - 82 = 2x⁴ - 24x³ + …, so the sum of roots = 242 = 12.
(The sum of the real roots alone would be 6; the question asks for all the roots.)

Shortcut. The four roots pair off symmetrically about x = 3, so their sum is 4 × 3 = 12.

Answer: (b) Only two

Method. split |x − 4| + |x − 7| at x = 4 and x = 7

4 ≤ x ≤ 7: the sum is (x − 4) + (7 − x) = 3 ≠ 15, no solution.
x > 7: (x − 4) + (x − 7) = 2x − 11 = 15 ⇒ x = 13.
x < 4: (4 − x) + (7 − x) = 11 − 2x = 15 ⇒ x = -2.
Exactly two real numbers: -2 and 13.

Shortcut. The sum of distances from 4 and 7 is 15 > 7 − 4 = 3, so there is one point outside on each side.

Answer: (c) 8

Method. quadratic in x²

x⁴ - 10x² + 9 = (x² - 1)(x² - 9) = 0.
x² = 1 or 9, so x = ±1, ±3 (all real).
Sum of absolute values = 1 + 1 + 3 + 3 = 8.

Answer: (d) -2 ± ⁠√6⁠

Method. expand each 3 × 3 determinant along its first row

 x 0 2 2x 2 1 1 1 1  = x(2 − 1) − 0 + 2(2x − 2) = 5x − 4. 3x 0 2 x² 2 1 0 1 1  = 3x(2 − 1) − 0 + 2(x² - 0) = 2x² + 3x.
Sum: 2x² + 8x − 4 = 0 ⟹ x² + 4x − 2 = 0.
x = -4 ± ⁠√(16⁠ + 8)2 = -2 ± ⁠√6.⁠

Answer: (a) Zero

Method. a sum of non-negative terms plus a positive constant cannot vanish

For every real x, x² ≥ 0 and 3|x| ≥ 0.
So x² + 3|x| + 2 ≥ 2 > 0 for all real x.
The left side is never 0, so the equation has no real root.

Answer: (d) -2, 2, 4

Method. x + 2 ≥ 0 is forced; split by the sign of (x − 3)(x + 2)

RHS ≥ 0 ⟹ x ≥ -2. Note x² - x − 6 = (x − 3)(x + 2).
Where (x − 3)(x + 2) ≥ 0 (x = -2 or x ≥ 3): x² - x − 6 = x + 2 ⟹ x² - 2x − 8 = 0 ⟹ x = 4 or -2; both valid.
'… ⟹ x² = 4 ⟹ x = ±2; only x = 2 lies in -2 < x < 3 (x = -2 was already found above).'
Roots: -2, 2, 4.

Answer: (a) Zero

Method. every term of the left side is non-negative and 20 > 0

For real x: x² ≥ 0 and 9|x| ≥ 0, so x² + 9|x| + 20 ≥ 20 > 0.
The left side is never zero: no real roots.

Answer: (a) a rational root and an irrational root

Method. split the modulus at x = 1 and keep roots in their own range

Case x ≤ 1: 1 − x + x² = 5 ⟹ x² - x − 4 = 0 ⟹ x = 1 ± ⁠√172.⁠ Only 1 - ⁠√172⁠ (≈ -1.56) satisfies x ≤ 1.
Case x > 1: x − 1 + x² = 5 ⟹ x² + x − 6 = 0 ⟹ (x + 3)(x − 2) = 0. Only x = 2 satisfies x > 1.
Roots: 2 (rational) and 1 - ⁠√172⁠ (irrational).

Answer: (b) One

Method. fix [x] on each unit interval and solve the resulting quadratic

On [0, 1): [x] = 0, so x² - 4x = 0 ⟹ x = 0 or 4; only x = 0 lies in [0, 1).
On [1, 2): [x] = 1, so x² - 4x + 1 = 0 ⟹ x = 2 ± ⁠√3⁠ ≈ 0.27, 3.73; neither lies in [1, 2).
At x = 2: [x] = 2, and 4 − 8 + 2 = -2 ≠ 0.
Exactly one solution, x = 0.

Answer: (d) 1, 2 and 3

Method. multiplicativity and triangle inequality for absolute value

(1) |ab| = |a||b| holds for all real (hence integer) a, b.
(2) Triangle inequality: |a + b| ≤ |a| + |b|.
(3) Write a = (a − b) + b: |a| ≤ |a − b| + |b| ⟹ |a| − |b| ≤ |a − b|; swapping a and b gives |b| − |a| ≤ |a − b|. Hence ||a| − |b|| ≤ |a − b|.
All three are correct.

Answer: (d) 6

Method. substitute t = |x − 3| and reject the negative root

Put t = |x − 3| ≥ 0: t² + t − 2 = 0 ⟹ (t + 2)(t − 1) = 0.
t = -2 is impossible, so t = 1: |x − 3| = 1 ⟹ x = 4 or x = 2.
Sum of real roots = 4 + 2 = 6.

Answer: (c) both first quadrant and second quadrant only

Method. the region lies above both boundary lines, so y ≥ 1

x − y ≤ 3 ⇒ y ≥ x − 3; x + y ≥ 5 ⇒ y ≥ 5 − x.
Adding: 2y ≥ 2 ⇒ y ≥ 1, so no point of the region is in the third or fourth quadrant.
(4, 1) satisfies both, in the first quadrant; (-2, 8) gives -10 ≤ 3 and 6 ≥ 5, in the second quadrant.
So the solution lies in the first and second quadrants only.

Answer: (d) (-1, 2)

Method. test each point in all four strict inequalities

(0, 0): x < 0 fails.
(-2, 4): x + y = 2, and 2 < 2 fails.
(-1, 4): x + y = 3 < 2 fails.
(-1, 2): 5(-1) − 4(2) + 12 = -1 < 0 (holds); -1 + 2 = 1 < 2 (holds); x = -1 < 0 (holds); y = 2 > 0 (holds).
Only (-1, 2) lies in the common region.

Answer: (c) x ≤ -4, y = 0

Method. intersect the two systems of inequalities

All four conditions must hold together.
x ≤ 4 and x ≤ -4 ⟹ x ≤ -4.
y ≥ 0 and y ≤ 0 ⟹ y = 0.
Solution: x ≤ -4, y = 0.