Test: Basic Concepts Of Differential And Integral Calculus- 1
40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Basic Concepts Of Differential And Integral Calculus- 1
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Choose the most appropriate option (a) (b) (c) or (d)
The gradient of the curve y = 2x3 –3x2 – 12x +8 at x = 0 is
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
The gradient of the curve y = 2x3 –5x2 – 3x at x = 0 is
The derivative of y = 
If y = x (x –1) (x – 2) then 
is
The gradient of the curve y – xy + 2px + 3qy = 0 at the point (3, 2 ) is 
and q are
The curve y2 = ux3 + v passes through the point P(2, 3) and = 4 at P. The values of u and v are
The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values of p and q are
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
If xy = 1 then y2 + dy/dx is equal to
The derivative of the function 
Given e-xy –4xy = 0, can be proved to be
If log (x / y) = x + y, 
may be found to be
If f(x, y) = x3 + y3 – 3axy = 0, can be found out as
Given x = at2, y = 2at; is calculated as
Given x = 2t + 5, y = t2 – 2; is calculated as
If x = 3t2 –1, y = t3 –t, then is equal to
The slope of the tangent to the curve y = 
at the point, where the ordinate and the abscissa are equal, is
so the point is (√2,√2)
= -1
The slope of the tangent to the curve y = x2 –x at the point, where the line y = 2 cuts the curve in the Ist quadrant, is
For the curve x2 + y2 + 2gx + 2hy = 0, the value of at (0, 0) is
If xy.yx = M, M is constant then is equal to
Given x = t + t–1 and y = t – t–1 the value of at t = 2 is
If x3 –2x2 y2 + 5x +y –5 =0 then at x = 1, y = 1 is equal to
The derivative of x2 log x is
If f(x) = 3x2, then F(x) =
If f(x) = x2 – 6x+8 then f ’(5) – f ’(8) is equal to
If f(x) = xk and f’(1) = 10 the value of k is
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