SOLUTION

Solution

f(x)=−x^3?/ 3 + x²sin1.5 a−xsina⋅sin2a−5arcsin(a²−8a+17)

(x)=−x^3?/ 3 + x²sin6 −xsin4sin8−5sin⁻¹((a−4)²+1) 

f′(x)=−x²+2xsin6−sin4.sin8 .......( a=4 considering the domain of inverse sine function) 
f′(sin8)=−sin²8+2sin6sin8−sin4sin8
=sin8(−sin8+2sin6−sin4)
=−sin8(sin8+sin4−2sin6)
=−sin8(2sin6cos2−2sin6)
=2sin8sin6(1−cos2) 

Now, 
sin8>0           (Since,2π<8<3π )
sin6<0          (Since, π<6<2π)
(1−cos2)>0      
Hence, f′(sin8)<0

SOLUTION

SOLUTION

Clearly, f(x) must be of the from

f(x)=a₀?[xⁿ+(k−x)ⁿ]+a₁?[xⁿ⁻¹+(k−x)ⁿ⁻¹]+...+a_n−1?[x+(k−x)]+a_n.?

It may be noted that n must be even for otherwise f(x) will become a polynomial of degree n−1.

Clearly, f′(x) is a polynomial of degree n−1.

SOLUTION

SOLUTION

∴  m₁ = |1 + 4i|;m₂ = |3 + i|;m₃ = |1 — i|;
m₄ = |2 — 3i|

∴ m₁ = √1+6 = √17 ; m₂ =√9+1 = √10

m₃ =√1+1 =√2;            m₄=√9+4 = √13
∴ m₃ < m₂ < m₄ < m₁
Here m₃ will be the smallest

Solution

Let α and β be the roots of the equation.
Hence
α+β=4
and α³+β³=28
Now α³+β³ can be written as

=(α+β)³−3αβ(α+β)
Hence
28=64−12αβ
12αβ=36
αβ=3
Therefore,
x²−(α+β)x+αβ=0
x²−(4x)+3=0

SOLUTION
 

The equation is $$10\left( z\bar{z} \right)-3i\left\{ {{z}^{2}}-{{\left( {\bar{z}} \right)}^{2}} \right\}-6=0$$

Or 10 ( x² + y² ) - 3i ( 2x ) ( 2iy ) - 6 = 0

⇒ 5 ( x² + y² ) + 6xy - 8 = 0 ...........(1)

Let ( r cos θ, r sin θ ) be a point on (1) , then

5r² + 6r² sin θ cos θ - 8 = 0 ⇒ r² = $$ cfrac{8}{5+3\sin 2\theta }$$ 

Clearly, 1 ≤ r² ≤ 4 ⇒ 1 ≤ | r | ≤ 2

∴ r₁ = | r |_max = 2 and r₂ = | r |_min = 1 ⇒ r₁ + r₂ = 3

SOLUTION

A and B are non-singular  
⇒ AB is non-singular  
AB adj  (AB) = |AB| I ...(1)
AB (adj  B adj A) = A (B adj  B) adj  A  
= A (|B| I) adj  A  
= |B| (A adj  A)) = |B| |A| I ...(2)  

adj  (AB) = (adj  B) (adj  A).  

SOLUTION