Test: Basic Concepts Of Differential And Integral Calculus- 1 - GRE MCQ

ANSWER :- a

Solution :- 2x^3 - 3x^2 -12x + 8 = 0 

dy/dx = 6x^2 - 6x -12 = 0

(At x=0) = 6(0)^2 - 6(0) - 12 

        = -12

we know that y = √(x+1) = (x+1)^1/2

dy/dx = 1/2 . (x+1)^-1/2 = 1 / 2√(x+1)
hence option c is correct

df/dx = * d/dx(ax² + bx + c)
= 

• 
• 
• 
• 
• Hence, option A is correct.

• firstly multiply all three equations 
• we get , y = x (x² - 3x + 2)
• y = x³ - 3x² + 2x 
• dy/dx = 3x² - 6x + 2

y+px+qy=0
y+qy=-px
(1+q)y=-px
y=-[p/(1+q)]x
This is a linear function, so has constant gradient at all points on the curve. Hence
-p/(1+q)=1/2
2p=-(1+q)
But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1) - but not (-1,1).
For your curve to pass through (-1,1), we would need to add a constant term, like so:
y+px+qy=1/2
 

• xy = 1
• y = 1/x
• y² = 1/x²
• dy/dx = -1/ x² 
• now , y² + dy/dx
• 1/x² - -1/ x² = 0

y = 1 / 2√(x+√x). (1+1/2√x)
Hence, option c is correct
 

Therefore, Option A 

 if we take minus common , dy/dx = (ay - x² )/ (y² - ax)

x = 2t + 5
dx/dt = 2
dy/dx = dy/dt / dx/dt
= 2t / 2 = t

• as we know, 
• y = x^-1/2 
• dy/dx = -1/2. x^-1/2-1
• therefore, the correct and is -1/2(x^-3/2) which is equal to option c as 
• -1/2. 1/(x√x)

• dx / dt = 6t
• dy/dx = dy/dt / dx/dt = (3t² - 1)/6t
• option A is correct.

so the point is (√2,√2)

= -1

hence , the ans will be none of these