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


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

Test: Basic Concepts Of Differential And Integral Calculus- 1 for GRE 2024 is part of GRE preparation. The Test: Basic Concepts Of Differential And Integral Calculus- 1 questions and answers have been prepared according to the GRE exam syllabus.The Test: Basic Concepts Of Differential And Integral Calculus- 1 MCQs are made for GRE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts Of Differential And Integral Calculus- 1 below.
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 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

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

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

Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 2
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The gradient of the curve y = 2x3 –5x2 – 3x at x = 0 is

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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 3
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The derivative of y = 

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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 5
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 6
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If y = x (x –1) (x – 2) then  is
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 7
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The gradient of the curve y – xy + 2px + 3qy = 0 at the point (3, 2 ) is and q are
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 8
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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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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 9
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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
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If xy = 1 then y2 + dy/dx is equal to
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 11
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The derivative of the function 
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 12
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Given e-xy –4xy = 0,  can be proved to be
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 14
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If log (x / y) = x + y,   may be found to be
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 15
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If f(x, y) = x3 + y3 – 3axy = 0,  can be found out as
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 16
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Given x = at2, y = 2at;  is calculated as
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 17
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Given x = 2t + 5, y = t2 – 2;  is calculated as 
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 18
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If y = 
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 19
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If x = 3t2 –1, y = t3 –t, then   is equal to
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 20
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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
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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
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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
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 24
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If xy.yx = M, M is constant then is equal to
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 25
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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
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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
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The derivative of x2 log x is
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 28
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The derivative of 
Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 29
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Test: Basic Concepts Of Differential And Integral Calculus- 1 - Question 30
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