At which point does f(x) = |x – 1| has itslocal minimum?
The logic is straightforward: after simplifying, They are equal is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XI and XII
#piecewise function
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Find the area between the lines y = |x - 2| and y = 5 - |x + 1|
Think of it like this: after simplifying, 8 is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XII — Integrals
The minimum value of the function f(x) = |x - 4| exists at
Applying the basic concept, we verify x = 4 by full calculation: use standard identity, substitute, simplify step by step to reach x = 4.
Ref: NCERT Mathematics Class XII — Applications of Derivatives
If f( x ) = array*20c 1 + x&if;x < 0 ( 1 - x )( px + q )&if;x 0 array .
The trick is solving fully: apply relevant formula, put numbers, compute. Result comes as (2, 1). Easy verification shows it satisfies condition.
Ref: NCERT Mathematics Class XI and XII
The function f(x) is defined by f( x ) = array*20c | x |x 0, arrayarray*20c ,x 0 x = 0 array . then, at x = 0 it is
Let's break it down: solving fully: apply relevant formula, put numbers, compute. Result comes as Discontinuous at x = 0 and has discontinuity of first kind. Easy verification shows it satisfies condition.
Ref: NCERT Mathematics Class XI and XII
If f(x) = matrix xx - 1 & if x 1 K & if x = 1 matrix . is continuous at x = 1, then the value of K is
Applying the basic concept, after simplifying, 1 is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XII — Matrices and Determinants
