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The Domain Of Definition Of The Function Is

is 1) [1, ∞) 2) [2, ∞) 3) [3, ∞) 4) None of these Solution: (3) ≥ 0 22x + 22(x-2) - 22(x-1) ≥ 52 22x [1 + 2-4 - 2-2] ≥ 52 22x (13 /... View Article

Let F X 1 X 1 1 X 3 And G X 2 X 1 2 X 2 Then Fog X

1) x + 1 2) − 1 − x 3) x − 1 4) Solution: (4) f (x) = − 1 + | x − 1 | g (x) = 2 − | x + 1| f (g (x)) = -1 + |g (x) - 1, -1 ≤ g (x) ≤ 3... View Article

If F X X 2 X X 2 2x Then Domain Of F Is

Solution: (2) f (x) = [x2 - x] / [x2 + 2x] = x [x - 1] / x [ x + 2] = [x - 1] / [x + 2] f (x) = [x - 1] / [x + 2] ≠ y x - 1 = yx + 2y x (1... View Article

If G F X Sinx And F G X Sin Root X 2 Then

1) f (x) = sin2 x and g (x) = √x 2) f (x) = sinx and g (x) = [x] 3) f (x) = x2 and g (x) = sin √x 4) f and g cannot be determined Solution:... View Article

Let Z1 Z2 Be Two Complex Numbers Such That Z1 Z2 And Z1z2 Both Are Real Then

(1) z1 = -z2 (2) z1 = bar z2 (3) z1 = -bar z2 (4) z1 = z2 Solution: Given z1+z2 and z1z2 both are real. Let z1= a+ib z2 = c+id... View Article

The Number Of Real Values Of A Satisfying The Equation

Solution: Given a2-2a sin x+1 = 0 This is a quadratic equation in a. So a = [2 sin x±√(4 sin2x-4)]/2 = [2 sin x ±2√-(1- sin2x)]/2 = sin x... View Article

If X Iy P Iq X2 Y2 I Then

Solution: (x+iy)(p+iq) = (x2+y2)i xp+pyi+xqi-yq = (x2+y2)i xp-yq+(py+xq)i = (x2+y2)i Comparing real and imaginary part xp-yq = 0 xp = yq... View Article

If Z 1 A B Ic And A2 B2 C2 1 Then 1 Iz 1 Iz Is Equal To

Solution: Given z(1+a) = b+ic z = (b+ic)/(1+a) (1+iz)/(1-iz) = (1+i(b+ic)/(1+a))/(1-i(b+ic)/(1+a)) = (1+a-c+ib)/(1+a+c-ib) =... View Article

If Sigma K 0 To 100 Ik X Iy Then The Values Of X And Y Are

Solution: ∑k=0100 ik = x+iy 1+i+i2+i3+i4+....i100 = x+iy This is a G.P with common ratio i and first term 1. => 1(1-i101)/(1-i) = x+iy... View Article

If 1 I 1 Plus I 100 A Ib Then

Solution: ((1-i)/(1+i)) = ((1-i)(1-i)/(1+i)(1-i)) = (1-2i-1)/(1+1) = -2i/2 = -i ((1-i)/(1+i))100 = -i100 = (i4)25 = 1 Comparing... View Article

The Multiplicative Inverse Of A Number Is The Number Itself Then Its Initial Value Is

Solution: Check the options. If we consider, -1, the multiplicative inverse is 1/-1 = -1 Hence option (2) is the answer.... View Article

If The Imaginary Part Of Is Zero Where A Is A Real Number Then The Value Of A Is Equal To

Solution: (2+i)/(ai-1) = (2+i)(-1-ai)/(ai-1)(-1-ai) = (-2-i-2ai+a)/(1+a2) = (-2+a)+i(-2a-1)/(1+a2) Given imaginary part is zero. So -2a-1... View Article

If Z 1 I Then The Multiplicative Inverse Of Z 2 Is

Solution: Given z = 1+i z2 = (1+i)2 = 1+2i-1 = 2i Multiplicative inverse = 1/2i = -i/2 Hence option (3) is the answer.... View Article

X Iy 1 3 A Ib Then X A Y B Is Equal To

(1) 4(a2+b2) (2) 4(a2-b2) (3) 4(b2-a2) (4) None of these Solution: Given (x+iy)1/3 = a+ib (x+iy) = (a+ib)3 x+iy = a3-ib3+3a2bi-3ab2... View Article

If Z Not Equal To Zero Is A Complex Number Then

(1) Re (z) = 0 => Im (z2) = 0 (2) Re (z2) = 0 => Im (z2) = 0 (3) Re (z) = 0 => Re (z2) = 0 (4) None of these Solution: Let z =... View Article

The Real Values Of X And Y For Which The Equation Is X Iy 2 3i 4i Is Satisfied Are

Solution: Given (x+iy)(2-3i) = 4+i (x+iy)(2-3i) = 2x+2iy-3ix+3y = 2x+3y+i(2y-3x) Comparing real and imaginary parts 2x+3y = 4 -3x+2y = 1... View Article

If A2 B2 1 Then 1 B Ia 1 B Ia Is Equal To

Solution: (1+b+ia)/(1+b-ia) Multiply numerator and denominator with (1+b+ia) = (1+b+ia)(1+b+ia)/(1+b-ia)(1+b+ia) = (1+b+ia)2/((1+b)2+a2) =... View Article

Root 5 12i Root 5 12i Root 5 12i Root 5 12i Is Equal To

Solution: [√(5+12i)+ √(5-12i)]/[√(5+12i)- √(5-12i)] Multiply numerator and denominator with √(5+12i)+√(5-12i)... View Article

A Ib Greater Than C Id Can Be Explained Only When

Solution: Since inequality in complex numbers is not defined, a+ib> c+id is possible if b = 0 and d = 0 Hence option (2) is the answer.... View Article

The Imaginary Part Of 1 I 2 I Is

Solution: (1+i)2/(2-i) = (1+2i-1)/(2-i) = 2i/(2-i) = 2i(2+i)/(2-i)(2+i) = (4i-2)/(4+1) = (4i-2)/5 Imaginary part is 4/5. Hence option... View Article

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