Question
If f, $\text{f, g : R}\rightarrow \text{R}$ be two functions defined as $\text{f}(x) = |x| + x \text{ and } \text{g} (x) = |x| - x, \forall \text{ }x \in \text{R}.$Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).
✓
Answer
$\text{Given f}(x) = |\text{x}| + \text{x}$
$\text{and g} (x) = |{\text{x}}| - x , \forall x \in \text{R}$
$\text{fog} = \text{f(g(x))} = |\text{x}| + \text{g(x)}$
$ = \text{||x| - x| + (|x| - x)}$
Therefore,
$\text{f(g(x))} = \begin{cases} 0 & \text{x} \geq 0\\ 4x & \text{x} < 0\\ \end{cases}$
$\text{fog} = \begin{cases} 4\text{x} & \text{x} < 0\\ 0 & \text{x} \geq 0\\ \end{cases}$
$\text{gof = g(f(x)) = |f(x)| - f(x)}$
$= ||\text{x| + x| - (|x| + x)}$
Therefore, $\text{g(f(x)) = gof = 0}$
$\text{and g} (x) = |{\text{x}}| - x , \forall x \in \text{R}$
$\text{fog} = \text{f(g(x))} = |\text{x}| + \text{g(x)}$
$ = \text{||x| - x| + (|x| - x)}$
Therefore,
$\text{f(g(x))} = \begin{cases} 0 & \text{x} \geq 0\\ 4x & \text{x} < 0\\ \end{cases}$
$\text{fog} = \begin{cases} 4\text{x} & \text{x} < 0\\ 0 & \text{x} \geq 0\\ \end{cases}$
$\text{gof = g(f(x)) = |f(x)| - f(x)}$
$= ||\text{x| + x| - (|x| + x)}$
Therefore, $\text{g(f(x)) = gof = 0}$
| Now, fog(−3) =(4)(−3) = −12 | (since, fog = 4x for x < 0) |
| fog(5) = 0 | (since, fog = 0 for x ≥ 0) |
| gof(−2) = 0 | (since, gof = 0 for x < 0) |
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $\text{A} = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix},$ find adj.A and verify that $\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
→Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
→$\sin^{-1}\Big\{\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{2}\Big\},0<\text{x}<1$
Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big),\text{a}>1, -\infty<\text{x}<0$
→$\tan^{-1}\Big(\frac{2\text{a}^{\text{x}}}{1-\text{a}^{2\text{x}}}\Big),\text{a}>1, -\infty<\text{x}<0$
Find the values of a and b such that the function f defined by $\text{f(x)}=\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if x}<4\\\text{a+}\text{b},&\text{if x}=4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if x}>4\end{cases}$ is a continuous function at x = 4.
→Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=1,\text{y}'(0)=1$Function $\text{y}=\sin\text{x}+\cos\text{x}$
→The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
If $\text{y}=\text{e}^{2\text{x}}(\text{ax}+\text{b}),$ show that $\text{y}_2-\text{4}\text{y}_1+4\text{y}=0$
→Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from,
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from,
- Box A?
- Box B?
- Box C?