Test: Basic Concepts Of Differential And Integral Calculus- 1

# Test: Basic Concepts Of Differential And Integral Calculus- 1

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## 40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Basic Concepts Of Differential And Integral Calculus- 1

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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 1

### Choose the most appropriate option (a) (b) (c) or (d) The gradient of the curve y = 2x3 –3x2 – 12x +8 at x = 0 is

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 1

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

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 2

### The gradient of the curve y = 2x3 –5x2 – 3x at x = 0 is

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 3

### The derivative of y =

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 4

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 5

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 6

If y = x (x –1) (x – 2) then  is

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 7

The gradient of the curve y – xy + 2px + 3qy = 0 at the point (3, 2 ) is and q are

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 7

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 8

The curve y2 = ux3 + v passes through the point P(2, 3) and  = 4 at P. The values of u and v are

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 8

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 9

The gradient of the curve y + px +qy = 0 at (1, 1) is 1/2. The values of p and q are

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 9

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

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 10

If xy = 1 then y2 + dy/dx is equal to

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 11

The derivative of the function

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 12

Given e-xy –4xy = 0,  can be proved to be

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 13

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 13

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 14

If log (x / y) = x + y,   may be found to be

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 15

If f(x, y) = x3 + y3 – 3axy = 0,  can be found out as

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 16

Given x = at2, y = 2at;  is calculated as

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 17

Given x = 2t + 5, y = t2 – 2;  is calculated as

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 18

If y =

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 19

If x = 3t2 –1, y = t3 –t, then   is equal to

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 20

The slope of the tangent to the curve y = at the point, where the ordinate and the abscissa are equal, is

Detailed Solution for Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 20

so the point is (√2,√2)

= -1

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 21

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

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 22

For the curve x2 + y2 + 2gx + 2hy = 0, the value of  at (0, 0) is

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 23

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 24

If xy.yx = M, M is constant then is equal to

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 25

Given x = t + t–1 and y = t – t–1 the value of  at t = 2 is

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 26

If x3 –2x2 y2 + 5x +y –5 =0 then  at x = 1, y = 1 is equal to

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 27

The derivative of x2 log x is

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 28

The derivative of

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 29

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 30

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 31

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 32

If f(x) = 3x2, then F(x) =

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 33

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 34

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 35

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 36

If f(x) = x2 – 6x+8 then f ’(5) – f ’(8) is equal to

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 37

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 38

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 39

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 40

If f(x) = xk and f’(1) = 10 the value of k is

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