Question

1. (a) Let A = 1, 3, 5, 7 and B = X : X is the first three multiple of 7 . Find (A−B).

Answer 1

Answer:

a) A-B = { 1 , 3, 5 }

b) The remainder is 50

c) f + g)(x) = x⁴+ 5x³

d) $\frac{dy}{dx} = \frac{2x- y^3(3x^2)) }{ 1- x^3 3y^2}$

Step-by-step explanation:

Step(I):-

1) a)

Let A = { 1, 3, 5, 7 } and B = { 7 , 14 , 21 }

A - B = {1,3,5,7} - { 7,14,21}

A - B  = { 1 , 3, 5}

b)

Given f (x)=3x³−11x−48

    x-1  ) 3 x³ - 11x -48 (- 3x²-3x -2

            -3x³ +3x²

                 3x²-11x-48

                -3x²+9x

                     2x -48

                    -2x +2  

                                       

                         50

The remainder is 50

   

c)  Given f(x) = x³ and g(x) = x + 5

(f. g)(x) = f(x) .g(x) = x³ . (x+5) = x³(x) + 5 x³ = x⁴+ 5x³

d)

 Given function   y = 3 + x²+x³y³  ...(I)

By using derivative formulas

            y = xⁿ

            $\frac{dy}{dx} = n x^{n-1}$

Apply d/dx ( UV) = U V¹ + V U¹



Differentiating equation (I) with respective to 'x' , we get

                      $\frac{dy}{dx} = 0 + 2x + ( x^3 3y^2(\frac{dy}{dx} )+ y^3(3x^2))$

                     $\frac{dy}{dx} - ( x^3 3y^2(\frac{dy}{dx} )= 2x- y^3(3x^2))$

                   $\frac{dy}{dx} = \frac{2x- y^3(3x^2)) }{ 1- x^3 3y^2}$