Differential Calculus NAT Level - 2 - IIT JAM MCQ

Let rectangle has width b and height h.
Area = h·b
Also,  b² + h² = (2R)² = 4R²

Area is maximum when A² is maxima
A² = h²(4R² – h²)]
f(h) = h²(4R² – h²)
For maxima,  
⇒   = h²(–2h) + (4R² – h²)2h = 0

–h² + 4R² – h² = 0
h² = 2R²
h = √2

From physical nature of problem, it is clear that this should be maximum area since minimum area will tend towards zero.

Hence, value of α = 2
The correct answer is: 2

h = αr + c
α = 3
h = 3r + c
h = 6, r = 1
c = 3


= 33 cm³/second
The correct answer is: 33

⇒ 
For maxima or minima of y.

⇒ 
and 

Hence, y attains maximum value at 
The value of λ = 1.

The correct answer is: 1

f(x) is decreasing in  and increasing in 

The correct answer is: 7

Absolute Maxima = 3
and The Absolute Minima = 1/3

The correct answer is: 9

⇒ 
⇒ 
⇒ 
Hence,    value of α = 1
The correct answer is: 1

Area of 
S = r²sin θ
⇒ 
Area of circle = 
Least area = πS

So,               Value of α = 1

The correct answer is: 1

Given   


D < 0

m = 1, n = – 6
m + n + 7 = 1 – 6 + 7
= 8 – 6 = 2

The correct answer is: 2

Let  

π/4  is point of minima

The correct answer is: 0.844

⇒  6a + 4b + 3c + 3d = 0
Let  

f(0) = e

Since f(x) is continuous and differentiable in (0,2) and f(0) = f(2) = e

Hence, according to Rolle's Theorem, the equation

 has at least one root in (0,2). Thus, value of α = 2.

The correct answer is: 2