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The correct OPTION is (d) 9

The best explanation: COMPARING the coefficients of the above equation we get,

1/1 = b/3 = c/3

This means b = 3 and c = 3

Therefore, b * c = 9.

The correct answer is (b) 4x^2 – 49x + 118 =0

The best I can explain: Let, y = x^2 + 2

Then, 2x^2 – 3X – 6 = 0

So, (3x)^2 = (2x^2 – 6)^2

[2(y-2) – 6]^2 = 9(y-2)

= 4x^2 – 49x + 118 = 0.

Right CHOICE is (c) [10/3, ∞)

To explain: We have, (2x^2 – 2x – 3) / (x^2 + x + 1) ≤ a ᵾ x € R

=> 2x^2 – 2x – 3 ≤ a(x^2 + x + 1) ᵾ x € R

=> (2 – a)x^2 – (2 – a)x – (3 – a) ᵾ x € R

2 – a < 0 and (2 – a)x^2 – 4(2 – a)(3 – a) ≤ 0 ᵾ x € R

So, a > 2 and a ≤ 2 or a ≥ 10/3

=> a ≥ 10/3

Therefore, a € [10/3, ∞)

Right choice is (C) 6

Explanation: COMPARING the coefficients of the above equation we get,

1/1 = b/3 = c/3

This means b = 3 and c = 3

Therefore, b + c = 6.

The correct option is (d) Not possible

Explanation: For, equal ROOTS, D = 0

Where, D = b^2 – 4ac

In the equation, (a + 1)x^2 + 2(a+1)x + (a – 2) = 0

D = [2(a+1)]^2 – 4 (a + 1)(a – 2)

= 4a^2 + 4 + 8a – 4{a^2 – 2a + a – 2}

= 4a^2 + 4 + 8a – 4a^2 + 4 a + 8 > 0

=> 12A + 12 = 0

=> 12a = -12

=> a = -1

So, from here it is clear that a = -1 is not possible because the equation is BECOMING linear.

The correct answer is (d) (1, 0) ∪ (1, ∞)

The EXPLANATION: If, X^2 > 1, then x + 1 > 0

So, x > 0

 x € (1, ∞)

 If, 0 < x < 1, the 0 < x + 1 < 1

 x € (-1, 0)

THUS, x € (1, ∞) ∪ (1, ∞)

The correct answer is (c) 7

To explain I would say: Here, f(x) = x^4 + 4x^3 + 6ax^2 + 6bx + c so, let its roots be, α, β, γ, δ and

g(x) = x^3 + 3x^2 + 9x + 3 so, let its roots be, α, β, γ

So, from here we can conclude,

α + β + γ + δ = -4 and α + β + γ = -3

Thus, δ = -1

This means (x + 1)(x^3 + 3x^2 + 9x + 3)

On SOLVING this equation in simpler FORM we get,

x^4 + 4x^3 + 12X^2 + 12x + 3

=> 6a = 12 => a = 2

=> 6b = 12 => b = 2

=> c = 3

=> a + b + c = 7

The correct option is (a) (-∞, -1)

For explanation I would say: For, imaginary roots, D > 0

Where, D = b^2 – 4ac

In the equation, (a + 1)x^2 + 2(a+1)x + (a – 2) = 0

D = [2(a+1)]^2 – 4 (a + 1)(a – 2)

= 4a^2 + 4 + 8a – 4{a^2 – 2A + a – 2}

= 4a^2 + 4 + 8a – 4a^2 + 4 a + 8 < 0

=> 12a + 12 < 0

=> 12a < -12

=> a < -1

Therefore, a € (-∞, -1)

Right CHOICE is (B) 1/4

Easiest explanation: If the COEFFICIENTS of (x^2 + x + c) = 0, then

x will always be = 1

Therefore, here, (a + 2b – 3c) + (b + 2c – 3a) + (c + 2a – 3b) = 0

So, x = 1.

As, one of its ROOT is 1 so, we will calculate the other one.

As, a, b, c are in A.P so,

b = (a + c)/2

Thus, PRODUCT of the roots αβ = (c + 2a – 3b)/(a + 2b – 3c)

As, a root say α = 1, then,

β = (c + 2a – 3(a + c)/2) / (a + 2(a + c)/2 – 3c)

We get the value of β = 1/4

The CORRECT CHOICE is (d) Does not exist

The explanation is: Since, x^2, 5|x| and 6 are POSITIVE,

So, x^2 + 5|x| + 6 = 0 does not have any REAL root

Therefore, sum does not exist.

The correct choice is (B) (-∞, -√7] ∪ [√7, ∞)

Best explanation: We have, log1/2 x^2 ≥ log1/2 (x + 2)

=> x^2 ≤ x + 2

=> -1 ≤ x ≤ 2

And, 49x^2 – 4a^4 ≤ 0 i.e. x^2 ≤ 4a^4 / 49

=> -2a^2/7 ≤ a ≤ 2a^2/7

From the above EQUATIONS,

-2a^2/7 ≤ -1 and 2 ≤ 2a^2/7

i.e. a^2 €7/2 and a^2 ≥ 7

=> a € (-∞, -√7] ∪ [√7, ∞)

So, S = (-∞, -√7] ∪ [√7, ∞)

Correct answer is (d) -2

Explanation: By WAVY curve METHOD

(x^2 – 1)/(x – 2)(x – 3) > 0

So, x = -1, 1, 2, 3

Thus, x €(-∞, -1) ∪ (1, 2) ∪ (3, ∞)

THEREFORE, the largest negative integer is -2.

Correct option is (C) r^1/n [cos(2kπ + θ)/n + i sin(2kπ + θ)/n]

The best I can explain: If n is any integer, then (cosθ + isinθ)^n = cos(nθ) + i sin(nθ).

WRITING the binomial EXPANSION of (cosθ + isinθ)^n and equating real parts of cos(nθ) and the imaginary part to sin(nθ), we get,

cos(nθ) = cos^nθ – ^nC2 cos^n-2θ sin^2θ + ^nC4 cos^n-4θ sin^4θ + ……….

sin(nθ) = ^nC1 cos^n-1θ sinθ – ^nC3 cos^n-3θ sin^3θ + ……….

If, n is a rational number, then one of the value of (cosθ + isinθ)^n = cos(nθ) + i sin(nθ).

If, n = p/q, where, p and q are integers (q>θ) and p, q have no common factor, then (cosθ + isinθ)^n has q distinct values one of which is cos(nθ) + i sin(nθ)

If, z^1/n = r^1/n [cos(2kπ + θ)/n + i sin(2kπ + θ)/n], where K = 0, 1, 2, ……….., n – 1.

Correct option is (d) Less than 12

For EXPLANATION: As, we know | z1 + z2 + …….. +ZN| ≤ |z1| + |z2| + ………. + |zn|

So, |z1 + z2 + 3 + 4i| ≤ |z1| + |z2| + |3 + 4i|

Now, putting the given values in the equation, we GET,

=> |z1 + z2 + 3 + 4i| ≤ 4 + 3 + √(9 + 16)

=> |z1 + z2 + 3 + 4i| ≤ 4 + 3 + 5

=> |z1 + z2 + 3 + 4i| ≤ 12.

The correct answer is (b) b, a, C are in A.P

To explain I WOULD say: Given, (a – b)x^2 + (b – c)x + (c – a) = 0 and x^2 + px + 1 =0

So, 1 / (a – b) = p / (b – c) = 1 / (c – a)

EQUATING the above equation, we GET,

(b – c) = p(a – b)and

(b – c) = p(c – a)

So, p(a – b) = p(c – a)

=> a – b = c – a

So, 2a = b + c which means that b, a, c are in A.P.

Correct choice is (a) 4

Explanation: In z=a+bi, a is REAL PART and b is imaginary part.

So, in 4+i, real part is 4.

The CORRECT option is (d) 5-8I

For explanation I would SAY: (-i) (8+5i) = -8i – 5 i^2

= -8i -5(-1) = 5-8i.

Right option is (a) x-AXIS

Easiest EXPLANATION: To convert polar representation (R, θ) into argand plane (x, y), substitute x = r cos θ and y = r sin θ.

x = r cos 2π = r and y = r sin 2π = 0.

Argand plane representation is (r, 0). SINCE imaginary part is zero, so it LIES on real axis i.e. x-axis.

Right answer is (a) x-AXIS

Easy explanation: CONJUGATE complex NUMBERS MEANS their real part is same and imaginary part is inverted i.e. same x part and opposite imaginary part. So, they are MIRROR image on real axis i.e. x-axis.

Correct answer is (c) -1

Easy explanation: SUBTRACTING the equation x^2 + ax + b = 0 to x^2 + bx + a = 0 by solving the equation simultaneously, we get,

(a – b)x + (b – a) = 0

So, (a – b)x = (a – b)

Therefore, x = 1

Now, putting the value of x = 3 in any one of the equation, we get,

1 + a + b = 0

Therefore, a + b = -1.

Right option is (a) \(\FRAC{-1±i\SQRT{11}}{6\sqrt{3}}\)

The explanation: \(\sqrt{3}\)X^2 + x + \(\sqrt{3}\) = 0

=>3x^2 + √3x + 3 = 0

=>D = (√3)^2 – 4.3.3 = 3-36 = -33.

Since D ≤ 0, imaginary roots are there.

=>x = \(\frac{-\sqrt{3}±i\sqrt{33}}{2.3} = \frac{-1±i\sqrt{11}}{6\sqrt{3}}\).

The CORRECT answer is (a) True

The BEST EXPLANATION: Since COMPLEX number x + y i is represented by (x, y) on argand plane, distance of point (x, y) from origin is \(\sqrt{x^2+y^2}\).

Correct ANSWER is (a) \(\frac{-1±i\sqrt{7}}{4}\)

Easy explanation: 2x^2 + x + 1 = 0

D=1^2-4*2*1 = 1-8 = -7 ≤ 0.

Since D ≤ 0, IMAGINARY ROOTS are there.

=>x = \(\frac{-1±\sqrt{1^2-4 &dot; 2.1}}{2.2} = \frac{-1±i\sqrt{7}}{4}\).

The CORRECT option is (d) (X, -y)

The best explanation: Mirror image of point (x, y) on REAL AXIS is (x, -y).

Since real axis is ACTING as mirror x-coordinate remains same whereas y-coordinate gets inverted.

So, (x, -y) is mirror image of (x, y) on real axis.

Right option is (a) (-4, 4\(\sqrt{3}\))

To EXPLAIN I would SAY: To CONVERT polar representation (r, θ) into argand plane (x, y), substitute x=r cos θ and y=r sin θ.

x=8COS 2π/3 = 8cos(π-π/3) = 8(-1/2) = -4.

y=8sin 2π/3 = 8sin(π-π/3) = 8(\(\sqrt{3}\)/2) = 4\(\sqrt{3}\).

Right option is (d) GREATER than or equal to zero

The best I can explain: For a QUADRATIC equation, ax^2+bx+c = 0, discriminant is b^2-4ac.

ROOTS are \(\FRAC{-b±\sqrt{b^2-4ac}}{2a}\). For real roots, radical must be non-negative i.e. discriminant should be greater than or equal to zero.

Correct choice is (C) (-1)^1/2

Best explanation: Iota is used to DENOTE COMPLEX number.

The value of i (iota) is \(\SQRT{-1}\) i.e. (-1)^1/2.

Correct choice is (a) additive inverse

The BEST I can explain: On ADDING negative of complex NUMBER (-Z) to complex number z, we get additive IDENTITY element zero i.e. z+(-z)=0.

Correct option is (a) x-AXIS

Explanation: SINCE imaginary part of complex NUMBER is zero. So, it is plotted on real axis i.e. x-axis.

2+i0 is POINT on x-axis.

Correct OPTION is (b) y-axis

To explain I would say: The plane having a complex NUMBER assigned to each of its POINT is called the

complex plane or the Argand plane. When (x + y i) is PLOTTED in argand plane, y-axis is imaginary axis.

The correct answer is (a) x-axis

Easy EXPLANATION: The plane having a complex number assigned to each of its point is called the

complex plane or the Argand plane. When (x + y i) is PLOTTED in argand plane, x-axis is real axis.

Right answer is (a) True

For explanation: Let Z=a+ bi

=>\(\BAR{z}\) = a-bi

So, z*\(\bar{z}\) = (a+bi) (a-bi) = a^2-(bi)^2 = a^2-(b^2) (-1) = a^2+b^2

|z|=\(\sqrt{a^2+b^2}\) => |z|^2 = a^2+b^2

Hence, z*\(\bar{z}\) = |z|^2.

Correct option is (d) multiplicative inverse

For EXPLANATION I would say: On multiplying RECIPROCAL of complex NUMBER (1/Z) to complex number z, we get multiplying inverse one i.e. z*1=z.

Right answer is (C) less than zero

For explanation: For a quadratic equation, ax^2+bx+c = 0, discriminant is b^2-4ac.

ROOTS are \(\FRAC{-b±\sqrt{b^2-4ac}}{2a}\). For imaginary roots, radical is negative i.e. discriminant should be less than zero.

Correct choice is (a) 2bc/(a^2 + b^2)

Explanation: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)^2 = (c – bsinθ)^2

a^2 cos^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 (1- sin^2 θ) = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 – a^2 sin^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

Now REARRANGING the ELEMENTS,

(a^2 + b^2) sin^2 θ – 2b c sinθ+( c^2 – a^2) = θ

So, as SUM of the roots are in the form –b/a if there is a quadratic equation ax^2 + bx + c = 0

Now, we can CONCLUDE that

sinα + sinβ = 2bc/(a^2 + b^2).

The CORRECT option is (B) x=i, -i

The best explanation: x^2+1 = 0

=>x^2 = -1 => x = ±\(\sqrt{-1}\) = ±i.