First slide

Functions (XII)

Question

Consider the functions
$f (x) = \{\begin{cases} x + 1, x \leq 1 \\ 2 x + 1, 1 < x \leq 2 \end{cases}$ and $g (x) = \{\begin{cases} x^{2}, - 1 \leq x < 1 \\ x + 2, 2 \leq x \leq 2 \end{cases}$

Moderate

Question

The domain of the function $f (g (x))$ is

A

$[0, \sqrt{2}]$
B

$[- 1, 2]$
C

$[- 1, \sqrt{2}]$
D

none of these

Solution

$\begin{array}{l} f (x) = \{\begin{cases} x + 1, x \leq 1 \\ 2 x + 1, 1 < x \leq 2 \end{cases} \\ g (x) = \{\begin{cases} x^{2}, - 1 \leq x < 2 \\ x + 2, 2 \leq x \leq 3 \end{cases} \\ ∴ f (x) = \{\begin{cases} g (x) + 1, g (x) \leq 1 \\ 2 g (x) + 1, 1 < g (x) \leq 2 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, x^{2} \leq 1, - 1 \leq x < 2 \\ x + 2 + 1, x + 2 \leq 1, 2 \leq x \leq 3 \\ 2 x^{2} + 1, 1 < x^{2} \leq 2, - 1 \leq x < 2 \\ 2 (x + 2) + 1, 1 < x + 2 \leq 2, 2 \leq x \leq 3 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, - 1 \leq x \leq 1 \\ 2 x^{2} + 1, 1 < x \leq \sqrt{2} \end{cases} \end{array}$
Hence, the domain of $f (x) is [- 1, \sqrt{2}]$

Question

The range of the function f(g(x)) is

A

[1, 5]
B

[2, 3]
C

$[1, 2] \cup [3, 5]$
D

none of these

Solution

$\begin{array}{l} f (x) = \{\begin{cases} x + 1, x \leq \\ 2 x + 1, 1 < x \leq 2 \end{cases} \\ g (x) = \{\begin{cases} x^{2}, - 1 \leq x < 2 \\ x + 2, 2 \leq x \leq 3 \end{cases} \\ ∴ f (x) = \{\begin{cases} g (x) + 1, g (x) \leq 1 \\ 2 g (x) + 1, 1 < g (x) \leq 2 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, x^{2} \leq 1, - 1 \leq x < 2 \\ x + 2 + 1, x + 2 \leq 1, 2 \leq x \leq 3 \\ 2 x^{2} + 1, 1 < x^{2} \leq 2, - 1 \leq x < 2 \\ 2 (x + 2) + 1, 1 < x + 2 \leq 2, 2 \leq x \leq 3 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, - 1 \leq x \leq 1 \\ 2 x^{2} + 1, 1 < x \leq \sqrt{2} \end{cases} \end{array}$
$\begin{array}{l} For - x \leq x \leq 1, we have x^{2} \in [0, 1] or x^{2} + 1 \in [1, 2] . \\ For 1 < x \leq \sqrt{2}, we have x^{2} \in (1, 2] or 2 x^{2} + 1 \in (3, 5] . \\ Hence, the range is [1, 2] \cup (3, 5] . \end{array}$

Question

The number of roots of the equation f(g(x)) = 2 is

A

1
B

2
C

4
D

none of these

Solution

$\begin{array}{l} f (x) = \{\begin{cases} x + 1, x \leq \\ 2 x + 1, 1 < x \leq 2 \end{cases} \\ g (x) = \{\begin{cases} x^{2}, - 1 \leq x < 2 \\ x + 2, 2 \leq x \leq 3 \end{cases} \\ ∴ f (x) = \{\begin{cases} g (x) + 1, g (x) \leq 1 \\ 2 g (x) + 1, 1 < g (x) \leq 2 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, x^{2} \leq 1, - 1 \leq x < 2 \\ x + 2 + 1, x + 2 \leq 1, 2 \leq x \leq 3 \\ 2 x^{2} + 1, 1 < x^{2} \leq 2, - 1 \leq x < 2 \\ 2 (x + 2) + 1, 1 < x + 2 \leq 2, 2 \leq x \leq 3 \end{cases} \\ or f (g (x)) = \{\begin{cases} x^{2} + 1, - 1 \leq x \leq 1 \\ 2 x^{2} + 1, 1 < x \leq \sqrt{2} \end{cases} \end{array}$
$\begin{array}{l} For f (g (x)) = 2, x^{2} + 1 = 2 and 2 x^{2} + 1 = 2, i . e ., \\ x = \pm 1 or x = \pm \frac{1}{\sqrt{2}} . \\ Therefore, x = \pm 1 only . hence, two roots . \end{array}$

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