The greatest and the least value of the function, f (x) = 1-2 x + x ^ 2 - 1+2 x + x ^ 2 …
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The greatest and the least value of the function,
f (x) = \(\sqrt{1-2 \mathrm{x}+\mathrm{x}^{2}}-\sqrt{1+2 \mathrm{x}+\mathrm{x}^{2}}\), x ∈ (−∞, ∞) are
f (x) = \(\sqrt{1-2 \mathrm{x}+\mathrm{x}^{2}}-\sqrt{1+2 \mathrm{x}+\mathrm{x}^{2}}\), x ∈ (−∞, ∞) are
- 2, –2
- 2, –1
- 2, 0
- none
Answer
(A) 2, –2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The interval in which the function x 3 increases less rapidly than 6x 2 + 15x + 5 is :
- For what values of x is the rate of increase of x 3 – 5x 2 + 5x + 8 is twice the rate of increase of x ?
- The maximum value of the function y = x (x –1) 2 , 0 ≤ x ≤ 2 is
- The difference between the greatest and least values of the function, f (x) = cos x + 1 2 cos 2x – 1 3 cos 3x…
- Let f: R R be such that f(1)=3 and f^ (1)=6 . Then _ x 0 ( f(1+x) f(1) )^ 1 / x equals
- If px 2 + qx + r = 0, p, q, r ∈ R has no real zero and the line y + 2 = 0 is tangent to f (x) = px 2 + qx + r…
- Consider a function f( x )= ( - 1 - x ) (4 – 3x 2 ) where ‘α’ is a positive parameter Least possible value of…
- The set of points where the function f (x) = x |x| is differentiable is
More Limit, Continuity and Differentiability questions · Browse all practice questions