9 + Interview Questions in COMPLEX NUMBER IN GENERAL AWARENESS Page 1 InterviewSolution

1.	If α,β are roots of the equation (x–a)(x–b)+c = 0 (c ≠ 0), then then roots of the equation (x – c – α)(x – c – β) = c are(a) a and b + c(b) a + c and b(c) a + c and b + c(d) None of these
Answer» Correct option (c) a+c and b+c Explanation: x²– (a + b)x + ab + c = 0 α + β = a + b and α β = ab + c Now (x – c – α)(x – c – β ) = c ⇒ (x–c)² – (α + β)(x – c) + α β – c = 0 (x – c)²– (a + b)(x – c) + ab = 0 (x – c)²– a(x – c) – b(x – c) + ab = 0 ⇒ (x – c – a)(x – c – b) = 0 x = c + a and b + c

1.

If α,β are roots of the equation (x–a)(x–b)+c = 0 (c ≠ 0), then then roots of the equation (x – c – α)(x – c – β) = c are(a) a and b + c(b) a + c and b(c) a + c and b + c(d) None of these

Answer»

Correct option (c) a+c and b+c

Explanation:

x²– (a + b)x + ab + c = 0

α + β = a + b and α β = ab + c

Now (x – c – α)(x – c – β ) = c

⇒ (x–c)² – (α + β)(x – c) + α β – c = 0

(x – c)²– (a + b)(x – c) + ab = 0

(x – c)²– a(x – c) – b(x – c) + ab = 0

⇒ (x – c – a)(x – c – b) = 0

x = c + a and b + c

Discussion

2.	If a ∈ Z and the equation (x – a)(x – 10) +1 = 0 has integral roots, then the values of a are(a) 10, 8(b) 12, 10(c) 12, 8(d) None of these
Answer» Correct option (c) 12, 8 Explanation: ⇒ (x – a)(x – 10) = – 1 x – a = 1 & x – 10 = –1 OR x – a = –1 & x – 10 =1 9 – a = 1 x = 10 – 1 11 – a = – 1 x = 11 a = 8 x = 9 a =12

2.

If a ∈ Z and the equation (x – a)(x – 10) +1 = 0 has integral roots, then the values of a are(a) 10, 8(b) 12, 10(c) 12, 8(d) None of these

Answer»

Correct option (c) 12, 8

Explanation:

⇒ (x – a)(x – 10) = – 1

x – a = 1 & x – 10 = –1

OR x – a = –1 & x – 10 =1

9 – a = 1 x = 10 – 1 11 – a = – 1 x = 11

a = 8 x = 9 a =12

Discussion

3.	The polynomial equation (ax2+bx+c)(ax2–dx–c) = 0 , ac ≠ 0 has (a) four real roots(b) atleast two real roots(c) atmost two real roots(d) No real roots
Answer» Correct option (b) atleast two real roots Explanation: ac ≠ 0 ac > 0 or ac < 0 Now D₁ = b²– 4ac & D₂ = d²+ 4ac When ac > 0 D₂> 0 but D₁ may positive or negative When ac < 0 D₁ > 0 but D₂ may be positive or negative In either case the polynomial has atleast two real roots

3.

The polynomial equation (ax2+bx+c)(ax2–dx–c) = 0 , ac ≠ 0 has (a) four real roots(b) atleast two real roots(c) atmost two real roots(d) No real roots

Answer»

Correct option (b) atleast two real roots

Explanation:

ac ≠ 0

ac > 0 or ac < 0

Now D₁ = b²– 4ac & D₂ = d²+ 4ac

When ac > 0

D₂> 0 but D₁ may positive or negative

When ac < 0

D₁ > 0 but D₂ may be positive or negative

In either case the polynomial has atleast two real roots

Discussion

4.	If a, b, c ∈ R , and the equation ax2 + bx + c = 0 has no real roots, then (a) (a + b + c) > 0(b) a(a + b + c) > 0(c) b (a + b + c) > 0(d) c(a + b + c) < 0
Answer» Correct option (b) a(a + b + c) > 0 Explanation: a > 0 ƒ(0) ⇒ c > 0 ƒ(2) > 0 ⇒ a + b + c > 0 aƒ(1) > 0 & cƒ(1) > 0 a(a + b + c) > 0 and c (a + b + c) > 0 a < 0 ƒ(0) < 0 ⇒ c < 0 ƒ(1) < 0 ⇒ a + b + c < 0 aƒ(1) > 0 & c.ƒ(1) > 0 a(a + b + c) > 0 and c (a + b + c) > 0

4.

If a, b, c ∈ R , and the equation ax2 + bx + c = 0 has no real roots, then (a) (a + b + c) > 0(b) a(a + b + c) > 0(c) b (a + b + c) > 0(d) c(a + b + c) < 0

Answer»

Correct option (b) a(a + b + c) > 0

Explanation:

a > 0

ƒ(0) ⇒ c > 0

ƒ(2) > 0 ⇒ a + b + c > 0

aƒ(1) > 0 & cƒ(1) > 0

a(a + b + c) > 0 and c (a + b + c) > 0

a < 0

ƒ(0) < 0 ⇒ c < 0

ƒ(1) < 0 ⇒ a + b + c < 0

aƒ(1) > 0 & c.ƒ(1) > 0

a(a + b + c) > 0 and c (a + b + c) > 0

Discussion

5.	The values of p for which 6 lies between the roots of the equation x2 + 2(p – 3)x + 9 = 0 is(a) (-∞, -3/4)(b) (-∞, -3/4)(c) (-∞, 1)(d) none of these
Answer» Correct option (a) (-∞, -3/4) Explanation: D > 0 (– 2(p – 3))2 – 4.1.9 > 0 p²– 6p > 0 p (p – 6) > 0 p < 0, p > 6 .....(1) a.ƒ(6) < 0 1.(36 + 12(p – 3) + 9)) < 0 12p + 9 < 0 P < -3/4 ...(2) From (1) and (2) P ∈ (-∞, -3/4)

5.

The values of p for which 6 lies between the roots of the equation x2 + 2(p – 3)x + 9 = 0 is(a) (-∞, -3/4)(b) (-∞, -3/4)(c) (-∞, 1)(d) none of these

Answer»

Correct option (a) (-∞, -3/4)

Explanation:

D > 0

(– 2(p – 3))2 – 4.1.9 > 0

p²– 6p > 0

p (p – 6) > 0

p < 0, p > 6 .....(1)

a.ƒ(6) < 0

1.(36 + 12(p – 3) + 9)) < 0

12p + 9 < 0

P < -3/4 ...(2)

From (1) and (2) P ∈ (-∞, -3/4)

Discussion

6.	The values of m(m ∈ R), for which both roots of the equation x2–6mx + 9m2 – 2m + 2 = 0 exceed 3 is(a) (-∞, 1)(b) (-∞, 1)(c) (1, ∞)(d) none of these
Answer» Correct option (d) none of these Explanation: D ≥ 0 (–6m)² –4.1.(9m² –2m + 2) ≥ 0 8m – 8 ≥ 0 m ≥ 1 ........(1) a.f(3) > 0 1.(32 – 18m + 9m²– 2m + 2) > 0 9m² – 20m + 11 > 0 9m² – 9m – 11m + 11 > 0 9m – 11) (m–1) > 0 m < 1 & m > 11/9 ...(2) α + β > 6 6m > 6 From (1), (2) and (3) m ∈(11/9, ∞)

6.

The values of m(m ∈ R), for which both roots of the equation x2–6mx + 9m2 – 2m + 2 = 0 exceed 3 is(a) (-∞, 1)(b) (-∞, 1)(c) (1, ∞)(d) none of these

Answer»

Correct option (d) none of these

Explanation:

D ≥ 0

(–6m)² –4.1.(9m² –2m + 2) ≥ 0

8m – 8 ≥ 0

m ≥ 1 ........(1)

a.f(3) > 0

1.(32 – 18m + 9m²– 2m + 2) > 0

9m² – 20m + 11 > 0

9m² – 9m – 11m + 11 > 0

9m – 11) (m–1) > 0

m < 1 & m > 11/9 ...(2)

α + β > 6

6m > 6

From (1), (2) and (3)

m ∈(11/9, ∞)

Discussion

7.	The values of m for which both roots of the equation x2 – mx + 1 = 0 are less than unity is(a) (-∞, -2)(b) (-∞, -2)(c) (-2, -∞)(d) none of these
Answer» Correct option (b) (–∞,–2] Explanation: D ≥ 0 (–m)² – 4.1.1 ≥ 0 (m – 2) (m + 2) ≥ 0 ⇒ m ∈ (–∞,–2] ∪ [2,∞) .... (1) a. f(1) > 0 1.(1 – m + 1) > 0 m < 2 m ∈(–∞,–2] ...(2) α + β < 2 m < 2 m ∈(–∞,–2] ...(3) Form (1), (2) and (3) we have m ∈(–∞,–2]

7.

The values of m for which both roots of the equation x2 – mx + 1 = 0 are less than unity is(a) (-∞, -2)(b) (-∞, -2)(c) (-2, -∞)(d) none of these

Answer»

Correct option (b) (–∞,–2]

Explanation:

D ≥ 0

(–m)² – 4.1.1 ≥ 0

(m – 2) (m + 2) ≥ 0

⇒ m ∈ (–∞,–2] ∪ [2,∞) .... (1)

a. f(1) > 0

1.(1 – m + 1) > 0

m < 2

m ∈(–∞,–2] ...(2)

α + β < 2

m < 2

m ∈(–∞,–2] ...(3)

Form (1), (2) and (3) we have

m ∈(–∞,–2]

Discussion

8.	If α ,β are roots of the equation x2 – 2x + 3 = 0 Then the equation whose roots are α3–3 α2 + 5α – 2 and β3 – β2 + β + 5 is(a) x2 + 3x + 2 = 0(b) x2– 3x – 2 = 0(c) x2 – 3x + 2 = 0(d) None
Answer» Correct option (c) x²– 3x + 2 = 0 Explanation: α²– 2α + 3 = 0 and β²– 2β + 3 = 0 α³– 2α² + 3α and β³– 2β² + 3β P = α³– 3α² + 5α - 2 = 2α²– 3α - 3α² + 5α - 2 = α²– 2α - 2 = 3 - 2 = 1 Similarly, we can show that Q = β³– β² + β + 5 = 2 Sum = 1 + 2 = 3 and product = 1 × 2 = 2 Hence x²– 3x + 2 = 0

8.

If α ,β are roots of the equation x2 – 2x + 3 = 0 Then the equation whose roots are α3–3 α2 + 5α – 2 and β3 – β2 + β + 5 is(a) x2 + 3x + 2 = 0(b) x2– 3x – 2 = 0(c) x2 – 3x + 2 = 0(d) None

Answer»

Correct option (c) x²– 3x + 2 = 0

Explanation:

α²– 2α + 3 = 0 and β²– 2β + 3 = 0

α³– 2α² + 3α and β³– 2β² + 3β

P = α³– 3α² + 5α - 2 = 2α²– 3α - 3α² + 5α - 2

= α²– 2α - 2

= 3 - 2 = 1

Similarly, we can show that Q = β³– β² + β + 5 = 2

Sum = 1 + 2 = 3 and product = 1 × 2 = 2

Hence x²– 3x + 2 = 0

Discussion

9.	The all possible values of a for which one root of the equation (a – 5)x2 – 2ax + a – 4 = 0 is smaller than 1 and the other greater than 2 is (a) [5, 24) (b) (5, 24] (c) (5, 24) (d) none of these
Answer» Correct option (c) (5, 24) Explanation: D ≥ 0 (–2a)2– 4(a – 5) (a – 4) ≥ 0 ⇒ 9a –20 ≥ 0 a ≥ 20/9 .......(1) (a –5) ƒ(1) < 0 (a –5) (a–5 –2a + a – 4) < 0 (a –5) (–a) < 0 a > 5 ........(2) (a –5) ƒ(2) < 0 (a –5) (4(a – 5) – 4a + a – 4) < 0 ⇒ (a – 5) (a – 24) < 0 ⇒ 5 < a < 24 .........(3) From (1), (2), and (3) a ∈ (5,24)

9.

The all possible values of a for which one root of the equation (a – 5)x2 – 2ax + a – 4 = 0 is smaller than 1 and the other greater than 2 is (a) [5, 24) (b) (5, 24] (c) (5, 24) (d) none of these

Answer»

Correct option (c) (5, 24)

Explanation:

D ≥ 0

(–2a)2– 4(a – 5) (a – 4) ≥ 0

⇒ 9a –20 ≥ 0

a ≥ 20/9 .......(1)

(a –5) ƒ(1) < 0

(a –5) (a–5 –2a + a – 4) < 0

(a –5) (–a) < 0

a > 5 ........(2)

(a –5) ƒ(2) < 0

(a –5) (4(a – 5) – 4a + a – 4) < 0

⇒ (a – 5) (a – 24) < 0

⇒ 5 < a < 24 .........(3)

From (1), (2), and (3) a ∈ (5,24)

Discussion

Explore topic-wise InterviewSolutions in .

If α,β are roots of the equation (x–a)(x–b)+c = 0 (c ≠ 0), then then roots of the equation (x – c – α)(x – c – β) = c are(a) a and b + c(b) a + c and b(c) a + c and b + c(d) None of these

If a ∈ Z and the equation (x – a)(x – 10) +1 = 0 has integral roots, then the values of a are(a) 10, 8(b) 12, 10(c) 12, 8(d) None of these

The polynomial equation (ax2+bx+c)(ax2–dx–c) = 0 , ac ≠ 0 has (a) four real roots(b) atleast two real roots(c) atmost two real roots(d) No real roots

If a, b, c ∈ R , and the equation ax2 + bx + c = 0 has no real roots, then (a) (a + b + c) &gt; 0(b) a(a + b + c) &gt; 0(c) b (a + b + c) &gt; 0(d) c(a + b + c) &lt; 0

The values of p for which 6 lies between the roots of the equation x2 + 2(p – 3)x + 9 = 0 is(a) (-∞, -3/4)(b) (-∞, -3/4)(c) (-∞, 1)(d) none of these

The values of m(m ∈ R), for which both roots of the equation x2–6mx + 9m2 – 2m + 2 = 0 exceed 3 is(a) (-∞, 1)(b) (-∞, 1)(c) (1, ∞)(d) none of these

The values of m for which both roots of the equation x2 – mx + 1 = 0 are less than unity is(a) (-∞, -2)(b) (-∞, -2)(c) (-2, -∞)(d) none of these

If α ,β are roots of the equation x2 – 2x + 3 = 0 Then the equation whose roots are α3–3 α2 + 5α – 2 and β3 – β2 + β + 5 is(a) x2 + 3x + 2 = 0(b) x2– 3x – 2 = 0(c) x2 – 3x + 2 = 0(d) None

The all possible values of a for which one root of the equation (a – 5)x2 – 2ax + a – 4 = 0 is smaller than 1 and the other greater than 2 is (a) [5, 24) (b) (5, 24] (c) (5, 24) (d) none of these

If a, b, c ∈ R , and the equation ax2 + bx + c = 0 has no real roots, then (a) (a + b + c) > 0(b) a(a + b + c) > 0(c) b (a + b + c) > 0(d) c(a + b + c) < 0