The least value of a function refers to its minimum point within a given domain. It can be found using differentiation by setting the first derivative to zero and checking the second derivative. For example in f(x) = x² the least value is 0 at x = 0. It is useful in optimization problems.
Class 12 Maths Chapter 6 Applications of Derivatives is an important topic for the CBSE Exam 2024-25. It covers the rate of change of functions increasing and decreasing functions maxima and minima tangents and normals and approximations. These concepts are useful in solving real-life problems related to physics economics and engineering.
Share
We find the minimum value of the function f(x) = 2 cos x + x over the closed interval [0, π/2] by determining its critical numbers and then we check the value of the function at the interval’s endpoints.
Compute the derivative of the function:
f'(x) = d/dx(2 cos x + x) = -2 sin x + 1
Set the derivative equal to zero to find critical points:
-2 sin x + 1 = 0
sin x = 1/2
The solution to sin x = 1/2 in the interval [0, π/2] is x = π/6.
Evaluate f(x) at the critical point and the endpoints of the interval:
– At x = 0:
f(0) = 2 cos(0) + 0 = 2 × 1 + 0 = 2
– At x = π/6:
f(π/6) = 2 cos(π/6) + π/6 = 2 × (√3/2) + π/6 = √3 + π/6
– At x = π/2:
f(π/2) = 2 cos(π/2) + π/2 = 2 × 0 + π/2 = π/2
Compare the values:
– f(0) = 2
– f(π/6) = √3 + π/6 ≈ 1.732 + 0.523 = 2.255
– f(π/2) = π/2 ≈ 1.570
The minimum value of the function in the interval [0, π/2] is attained at f(0) = 2.
Click here for more:
https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-6