Let for $x \in R$ $f(x)=\frac{x+|x|}{2} \text { and } g(x)=\begin{cases}x, & x<0 \\x^2, & x \geq 0\end{cases} $
Then area bounded by the curve $y=(f \circ g)(x)$ and the lines $y=0,2 y-x=15$ is equal to
Let for $x \in R$ $f(x)=\frac{x+|x|}{2} \text { and } g(x)=\begin{cases}x, & x<0 \\x^2, & x \geq 0\end{cases} $
Then area bounded by the curve $y=(f \circ g)(x)$ and the lines $y=0,2 y-x=15$ is equal to
Show Hint
Correct Answer: 72
Approach Solution - 1
The correct answer is 72.
2y−x=15
A=0∫3(2x+15−x2)dx+21×215×15
4x2+215x−3x3∣∣03+4225
=49+245−9+4225=499−36+225
=4288=72
Approach Solution -2
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Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
