Differential Calculus NAT Level - 1 - IIT JAM MCQ

# Differential Calculus NAT Level - 1 - IIT JAM MCQ

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## 10 Questions MCQ Test - Differential Calculus NAT Level - 1

Differential Calculus NAT Level - 1 for IIT JAM 2024 is part of IIT JAM preparation. The Differential Calculus NAT Level - 1 questions and answers have been prepared according to the IIT JAM exam syllabus.The Differential Calculus NAT Level - 1 MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Differential Calculus NAT Level - 1 below.
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Differential Calculus NAT Level - 1 - Question 1

### If f"(x) > 0 and f'(1) = 0 such that g(x) = f(cot2 x + 2cot x + 2) where 0 < x < π, then g(x) decreasing in (a, b) where  is

Detailed Solution for Differential Calculus NAT Level - 1 - Question 1

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Differential Calculus NAT Level - 1 - Question 2

### If f(x) has a maximum or a minimum at a point x0 inside the interval, then f '(x0) equals :

Detailed Solution for Differential Calculus NAT Level - 1 - Question 2

It is necessary that at the point of maxima or minima of a function, say ff ' will become zero.

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Differential Calculus NAT Level - 1 - Question 3

### If 1" = α radians, then the approximate value of cos 60°1' is given as  Find the value of  λ.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 3

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Differential Calculus NAT Level - 1 - Question 4

The maximum value of u is, where u = axy2z2 - x2y2z3 - xy3z3 - xy2z4 is  Find the value of α.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 4

Solving, we get

Thus, max value is

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Differential Calculus NAT Level - 1 - Question 5

The sum of one number and three times a second number is 60. Find the pair, where product is maximum.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 5

Let the two numbers be x & y

⇒ x + 3y = 60 ....(1)

Now, let z = xy = product of two numbers.

Putting x = 30 in (1), we get y = 10 and

Hence, z is maximum when x = 30 and y = 10.

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Differential Calculus NAT Level - 1 - Question 6

Let  be increasing for all real values of x, then range of a is  (α, ∞). Find value of α.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 6

Since f(x) is a function so,

Value of α = 5

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Differential Calculus NAT Level - 1 - Question 7

The greatest and the least value of the function f(x) = x3 – 18x2 + 96x in the interval [0, 9] are :

Detailed Solution for Differential Calculus NAT Level - 1 - Question 7

f(x) = x3 – 18x2 + 96x

= 3(x – 4)(x – 8)
Put f'(x) = 0, we get x = 4, 8
f''(x) = 6x – 36 = 6(x – 6)
which is positive at x = 8 and negative at x = 4

Now, f(0) = 0,  f(8) = 128
f(4) = 160,      f(9) = 135
∴  least value is 0 and greatest value is 160.

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Differential Calculus NAT Level - 1 - Question 8

Let f(x, y) = x4 + y4 - 2x2 + 4xy - 2y2 has a minimum at (-√α, √α) and (√α, - √α) Find the value of α.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 8

⇒

Solving, we get

r = 12x2 - 4
s = 4
t = 12y2 – 4
at (-√2, √2)
r = 20, s = 4, t = 20
rt - s2 > 0 and r > 0  ∴ minimum

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Differential Calculus NAT Level - 1 - Question 9

If the function  is downward concave is (α, β) the [β - α] is

Detailed Solution for Differential Calculus NAT Level - 1 - Question 9

∵  [.] greatest integer function

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Differential Calculus NAT Level - 1 - Question 10

The minimum value of  Find the value of α.

Detailed Solution for Differential Calculus NAT Level - 1 - Question 10

For max. or min value of u,

Solving these equations, we get
x = y = a

at x = ay = a
r = 2, s = 1, t = 2
rt – s2 = 4 – 1 = 3 > 0 and r > 0

Therefore, u is min at x = ay = a and min. value is 3a2.