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
#absolute value
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The values of x which statisfied |3x| ≥ |6 − 3x| A. (0, 1] B. [1, 4] C. (4, ∞) D. (−1, 0) E. (−∞, 0) Choose the correct
The logic is straightforward: complete solution: write formula, plug data, do arithmetic. Final value matches C and E only. Simple and accurate.
Ref: NCERT Mathematics Class XI and XII
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
Evaluate: _-/2^/2| x| dx
The trick is after simplifying, 2 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
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) = x|x| and g(x) = sin x then gof(x) then at x = 0
Here's the idea in simple terms: solving fully: apply relevant formula, put numbers, compute. Result comes as gof(x) is differentiable. Easy verification shows it satisfies condition.
Ref: NCERT Mathematics Class XII — Applications of Derivatives
What is the value of _-/2^/2| x|:dx?
The trick is we verify 2 by full calculation: use standard identity, substitute, simplify step by step to reach 2.
Ref: NCERT Mathematics Class XII — Integrals
What is the minimum value of |x|?
Let's break it down: |x|≥0 always. Minimum occurs at x=0, value 0. Simple absolute value property, complete.
Ref: NCERT Mathematics Class XII — Applications of Derivatives
