Differential Calculus NAT Level - 1
10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Differential Calculus NAT Level - 1
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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
is
The correct answer is: 3.14
If f(x) has a maximum or a minimum at a point x0 inside the interval, then f '(x0) equals :
It is necessary that at the point of maxima or minima of a function, say f, f ' will become zero.
The correct answer is: 0
If 1" = α radians, then the approximate value of cos 60°1' is given as 
Find the value of λ.
Find the value of λ.
The correct answer is: 0.008
The maximum value of u is, where u = axy2z2 - x2y2z3 - xy3z3 - xy2z4 is 
Find the value of α.
Solving, we get
Thus, max value is 
The correct answer is: 1
The sum of one number and three times a second number is 60. Find the pair, where product is maximum.
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.
The correct answer is: 300
Let 
be increasing for all real values of x, then range of a is (α, ∞). Find value of α.
Since f(x) is a function so, 
Value of α = 5
The correct answer is: 5
The greatest and the least value of the function f(x) = x3 – 18x2 + 96x in the interval [0, 9] are :
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.
The correct answer is: 160
Let f(x, y) = x4 + y4 - 2x2 + 4xy - 2y2 has a minimum at (-√α, √α) and (√α, - √α) Find the value of α.
⇒ 
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
The correct answer is: 2
If the function 
is downward concave is (α, β) the [β - α] is
∵ [.] greatest integer function
The correct answer is: 1
Find the value of α.For max. or min value of u,
Solving these equations, we get
x = y = a
at x = a, y = a
r = 2, s = 1, t = 2
rt – s2 = 4 – 1 = 3 > 0 and r > 0
Therefore, u is min at x = a, y = a and min. value is 3a2.
The correct answer is: 3
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