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Test: Calculus- 1 - Computer Science Engineering (CSE) MCQ


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20 Questions MCQ Test - Test: Calculus- 1

Test: Calculus- 1 for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Test: Calculus- 1 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Calculus- 1 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calculus- 1 below.
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Test: Calculus- 1 - Question 1

Let  and  denote the area of region bounded by f(x) and the X-axis, when  varies from -1 to 1. Which of the following statements is/are TRUE?

I.  is continuous in [-1, 1]
II.  is not bounded in [-1, 1]
III.  is nonzero and finite

Detailed Solution for Test: Calculus- 1 - Question 1

I. False.
II. True.
III. True. An area is always positive, while the definite integral might be composed of several regions, some positive and some negative. A definite integral gets you the net area, because any part of the graph that is below the x-axis will give you a negative area. So, a definite integral is not necessarily the area under the curve, but the value of the area above the x-axis less the area under the x-axis. So, A is non-zero and finite.

Test: Calculus- 1 - Question 2

The function y=|2 - 3x|

Detailed Solution for Test: Calculus- 1 - Question 2


as y is polynomial it is continuous and differentiable at all points but don't know at = 2/3

LL = RL = f(a) so y is continuous 

Test: Calculus- 1 - Question 3

Which one of the following functions is continuous at x = 3?

Detailed Solution for Test: Calculus- 1 - Question 3

For continuity, Left hand limit must be equal to right hand limit. For continuity at x = 3, the value of f(x) just above and just below 3 must be the same. 

A. f(3) = 2. f(3+) = x - 1 = 2. f(3-) = (x+3)/3 = 6/3 = 2. Hence continuous.
B. f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. So, not continuous.
C. f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1. So, not continuous.
D. f(3) is not existing. So, not continuous.

Test: Calculus- 1 - Question 4

A function f(x) is continuous in the interval [0,2]  f(0) = f(2) = -1 and f(1) = 1.  Which one of the following statements must be true?

Detailed Solution for Test: Calculus- 1 - Question 4

Let's define a new function g,
g(y) = f(y) -f(y+1)
Since function f is continuous in [0,2], therefore g would be continuous in [0,1] g(0) = -2, g(1) = 2
since g is continuous and goes from negative to positive value in [0,1]. therefore at some point g would be 0 in (0,1).
g=0 ⇒ f(y) = f(y+1) for some y in (0,1)

Therefore, correct answer would be (A).

Test: Calculus- 1 - Question 5

 and , then the constants  and  are

Detailed Solution for Test: Calculus- 1 - Question 5



*Answer can only contain numeric values
Test: Calculus- 1 - Question 6

Let f(x) be a polynomial and   be its derivative. If the degree of is 10, then the degree of 


Detailed Solution for Test: Calculus- 1 - Question 6

F is some function where the largest even degree term is having degree  10. no restiction on odd degree terms.
since f(x)+f(-x)= degree 10
even power gets converted to odd in derivative.
then the the degrre of required expression =9.
the odd powers in F will become even in derivative and G(X)-G(-X) retains only odd powers.

Test: Calculus- 1 - Question 7

Consider the following two statements about the function

  • P. f(x) is continuous for all real values of .
  • Q .f(x) is differentiable for all real values of  .

Which of the following is TRUE?

Detailed Solution for Test: Calculus- 1 - Question 7

f(x)=|x| here for all values of x, f(x) exists. therefore it is continuous for all real values of x. At x=0, f(x) is not differentiable. Because if we take the left hand limit here, it is negative while the right hand limit is positive.  

Test: Calculus- 1 - Question 8

Consider the function  . In this interval, the function is

Detailed Solution for Test: Calculus- 1 - Question 8

It is continuous but not differentiable at x=0 as left hand limit will be negative while the right hand limit will be positive but for differentiation, both must be same.

*Answer can only contain numeric values
Test: Calculus- 1 - Question 9

Let f be a function defined by

Find the values for the constants a, b,c  and d so that f is continuous and differentiable everywhere on the real line.


Detailed Solution for Test: Calculus- 1 - Question 9



Test: Calculus- 1 - Question 10

Let the function

where denote the derivative of f with respect to θ. Which of the following statements is/are TRUE? 

Detailed Solution for Test: Calculus- 1 - Question 10

We need to solve this by rolle's theorem, to apply rolle's theorem following 3 conditions should be satisfied:
1) f(x) should be continuous in interval [a, b],
2) f(x) should be differentiable in interval (a, b), and 3) f(a) = f(b)
If these 3 conditions are satisfied simultaneously then, there exists at least one 'x' such that f '(x) = 0
So, for the above question, it satisfies all the three conditions, so we can apply rolle's theorem, i.e, there exists 'at least one' theta that gives f '(theta) = 0
Also, the given function is also not a constant function, i.e f '(theta) ≠ 0
So, answer is C

*Answer can only contain numeric values
Test: Calculus- 1 - Question 11

The function  satisfies the following equation:

. The value of  is______.


Detailed Solution for Test: Calculus- 1 - Question 11



Test: Calculus- 1 - Question 12

The formula used to compute an approximation for the second derivative of a function f at a point x0 is

Test: Calculus- 1 - Question 13

In the interval  [0,π] the equation x = cos x has 

Detailed Solution for Test: Calculus- 1 - Question 13

f you consider x=0 then cosx=1
now if x= PI/4 = 0.785 then cosx=0.7071
for some x value x=cosx
after this x is increasing and cosx is decreasing. so we can say exactly 1 solution.

EDIT-
It is very easy to show that the equation x = cos x has a unique solution. For example take f(x) = c - cos x and notice that  (equality holding in isolated points) so f(x) is strictly increasing and hence the equation can have at most one solution.

At and function is continious (difference of two continuous functions is continuous). Therefore there is solution in , hence there is a solution in [0,π]

Test: Calculus- 1 - Question 14

then the range of f(x) is 

Detailed Solution for Test: Calculus- 1 - Question 14



Test: Calculus- 1 - Question 15

Detailed Solution for Test: Calculus- 1 - Question 15

as f(xi).f(xi+1) <0
means one of them is positive and one of them in negative . as their multiplication is negative
so when you draw the graph for f(x) where xi<=x<=xi+1. Definitely F(x) will cut the X- axis.
so there will definitely a root of F(x) between xi and xi+1

Test: Calculus- 1 - Question 16

If  f(x) = x2 and g(x) = x sinx +cosx then

Detailed Solution for Test: Calculus- 1 - Question 16


We are asked to make this two function equal and see at how many points they meet.

This F(x) is even and we just need to check at how many points F(x) crosses the X - axis for x > 0.

Here 2-cos x is greater than 0 and  (2 - cosx) will be also greater than  for x > 0.

As we have already seen that F(x) is even, therefore F(x) has only two real roots.

Test: Calculus- 1 - Question 17

Given

 what will be the evaluation of the definite integral

Detailed Solution for Test: Calculus- 1 - Question 17

Test: Calculus- 1 - Question 18

What is the value of 

Detailed Solution for Test: Calculus- 1 - Question 18

Put  x - π = t then limit 0 changes to  - π and upper limit 2π changes to π.

Integration of  t2 for limit  to . One is an odd function and one is even and product of odd and even functions is odd function and integrating an odd function from the same negative value to positive value gives 0.

Test: Calculus- 1 - Question 19

If for non-zero x,  where a

Detailed Solution for Test: Calculus- 1 - Question 19


Integrating both sides,




 (2) and (3) by multiplying (2) by  and (3) by  and subtracting

Test: Calculus- 1 - Question 20

Detailed Solution for Test: Calculus- 1 - Question 20

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