Let F X E X E X 2 And If F G X X Then G X
Solution: (3) Since f (g (x)) = x (fog) (x) = x fog = 1 g (x) = f-1 (x)... View Article
Latest Posts
Let F Theta Sin Theta Sin Theta Sin 3 Theta Then F Theta
Solution: (3) f (θ) = sin θ (sin θ + sin 3θ) = sin θ [sin θ + 3 sin... View Article
The Domain Of Definition Of The Function Y X Given By The Equation 2 X 2 Y 2 Is
1) 0 < x ≤ 1 2) 0 ≤ x ≤ 1 3) − ∞ < x ≤ 0 4)... View Article
The Domain Of Definition Of The Function Is
is 1) [1, ∞) 2) [2, ∞) 3) [3, ∞) 4) None of these Solution: (3) ≥ 0 22x +... 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) = −... 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 [... 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) =... 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... 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... 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 =... 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) = (1+a-c+ib)(1+a+c+ib)/(1+a+c-ib)(1+a+c+ib) = (2a+2a2+2ib+2iab)/(2+2ac+2(a+c)) (since a2+b2+c2 =... 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.... 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 with a+ib,... 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... 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 = 0 -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... 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 =... 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... 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 =... 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) = (1+b2-a2+2iab+2b+2ia)/(1+b)2+a2) = (2b2+2b+2ia(b+1))/(1+2b+b2+a2) = (2b(b+1)+2ia(b+1))/ (2+2b) since... View Article
