- fx- cases x-1- -1 - x-0 krx - 2- - 0 - x - 1 andgx-sin x - Considerthef unctionh1 x-f-gx- and h- 2x-fgx- -Which of the following is not true about h 1 x -A- It is periodic function with period -B- Domain is RC- Range is -0-1-D- None of these

The correct option is D None of these|g(x)|=|sin x|, ∀ x ∈ R f(|g(x)|)= |sin x| 1, amp; 1 ≤ |sin x| lt;0 (|sin x|)^2, amp; 0 ≤ |sin x|≤ 1 ⇒ f(|g(x)| ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

$ fx= cases x...

Question

$f (x) = {\begin{matrix} x - 1, & - 1 \leq x < 0 x^{2}, & 0 \leq x \leq 1 \end{matrix}$ and $g (x) = sin x$ . Consider the function $h_{1} (x) = f (| g (x) |)$ and $h_{2} (x) = | f (g (x)) |$
Which of the following is not true about $h_{1} (x)$ ?

It is periodic function with period

π

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Range is

[0, 1]

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Domain is

R

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None of these

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Solution

The correct option is D None of these
$| g (x) | = | sin x |, \forall x \in R$
$f (| g (x) |) = {\begin{matrix} | sin x | - 1, & - 1 \leq | sin x | < 0 (| sin x |)^{2}, & 0 \leq | sin x | \leq 1 \end{matrix} \Rightarrow f (| g (x) |) = {sin}^{2} x$

Clearly, $h_{1} (x) = f (| g (x) |) = {sin}^{2} x$ has a period $π$ , range $[0, 1]$ , and domain $R$

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Similar questions

f (x) = {\begin{matrix} x - 1, & - 1 \leq x < 0 x^{2}, & 0 \leq x \leq 1 \end{matrix}

and

g (x) = sin x

. Consider the function

h_{1} (x) = f (| g (x) |)

and

h_{2} (x) = | f (g (x)) |

Which of the following is not true about

h_{2} (x)

f (x) = {\begin{matrix} x - 1, & - 1 \leq x < 0 x^{2}, & 0 \leq x \leq 1 \end{matrix}

and

g (x) = sin x

. Consider the function

h_{1} (x) = f (| g (x) |)

and

h_{2} (x) = | f (g (x)) |

For when

h_{1} (x)

and

h_{2} (x)

are identical functions, then which of the following is not true?

Identify the function from the descriptions given below.

1. Its domain is R and Range is [0, 1)

2. f(x) = x, if x $ϵ$ [0, 1)

3. f(x) = 0 if x $ϵ$ I (set of integers)

4. f(x) = 1 + x if x $ϵ$ [-1, 0)

5. f(x) is periodic with period 1

Q. Assertion :Consider the function

f (x) = x^{2} - 2 x

and

g (x) = - | x |

The composite function

F (x) = f (g (x))

is not derivable at

x = 0

Reason:

f^{'} (0^{+}) = 2

and

f^{'} (0^{-}) = - 2

Q. Let

f (x) = f_{1} (x) - 2 f_{2} (x)

,

where,

f_{1} (x) = {\begin{matrix} min {x^{2}, | x |}, & | x | \leq 1 max {x^{2}, | x |}, & | x | > 1 \end{matrix}

and,

f_{2} (x) = {\begin{matrix} min {x^{2}, | x |}, & | x | > 1 max {x^{2}, | x |}, & | x | \leq 1 \end{matrix}

and,

g (x) = {\begin{matrix} min {f (t) : - 3 \leq t \leq x, - 3 \leq x < 0} max {f (t) : 0 \leq t \leq x, 0 \leq x \leq 3} \end{matrix}

For

- 3 \leq x \leq - 1

, the range of

g (x)

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f(x)={x−1, −1≤x<0 x2, 0≤x≤1 and g(x)=sinx. Consider the function h1(x)=f(|g(x)|) and h2(x)=|f(g(x))| Which of the following is not true about h1(x)?

The correct option is D None of these|g(x)|=|sinx|, ∀ x∈R f(|g(x)|)={|sinx|−1, −1≤|sinx|<0 (|sinx|)2, 0≤|sinx|≤1 ⇒f(|g(x)|)=sin2x Clearly, h1(x)=f(|g(x)|)=sin2x has a period π, range [0,1], and domain R

$f (x) = {\begin{matrix} x - 1, & - 1 \leq x < 0 x^{2}, & 0 \leq x \leq 1 \end{matrix}$ and $g (x) = sin x$ . Consider the function $h_{1} (x) = f (| g (x) |)$ and $h_{2} (x) = | f (g (x)) |$
Which of the following is not true about $h_{1} (x)$ ?