# 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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Attempt Test: Basic Concepts Of Differential And Integral Calculus- 1 | 40 questions in 40 minutes | Mock test for CA Foundation preparation | Free important questions MCQ to study Quantitative Aptitude for CA CPT for CA Foundation Exam | Download free PDF with solutions
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

Solution:

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

QUESTION: 2

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QUESTION: 3

### The derivative of y =

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QUESTION: 4

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QUESTION: 5

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QUESTION: 6

If y = x (x –1) (x – 2) then  is

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QUESTION: 7

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

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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

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QUESTION: 9

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

Solution:

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

QUESTION: 10

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

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QUESTION: 11

The derivative of the function

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QUESTION: 12

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

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QUESTION: 13

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QUESTION: 14

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

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QUESTION: 15

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

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QUESTION: 16

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

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QUESTION: 17

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

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QUESTION: 18

If y =

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QUESTION: 19

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

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QUESTION: 20

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

Solution:

so the point is (√2,√2)

= -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

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QUESTION: 22

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

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QUESTION: 23

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QUESTION: 24

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

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QUESTION: 25

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

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QUESTION: 26

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

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QUESTION: 27

The derivative of x2 log x is

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QUESTION: 28

The derivative of

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QUESTION: 29

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QUESTION: 30

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QUESTION: 31

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QUESTION: 32

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

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QUESTION: 33

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QUESTION: 34

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QUESTION: 35

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QUESTION: 36

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

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QUESTION: 37

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QUESTION: 38

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QUESTION: 39

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QUESTION: 40

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

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