If f - R - S def ined by f x -sin x -3cos x -1 is onto -then the interval of S isA- -1-3-B- -0-1-C- -0-1-D- -1-1-

The correct option is A [–1, 3] √(1+( √(3))^2) ≤ (sinx √(3)cosx)≤√(1+( √(3))^2) 2≤(sinx √(3)cosx)≤ 2 2+1≤(sinx √(3)cosx+1)≤ 2+1 1≤(sinx √(3)cosx+1)≤ 3 i.e. ...

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Standard XII

Mathematics

Differentiation of Inverse Trigonometric Functions

If f : R → S ...

Question

If $f : R \to S d e f i n e d b y f (x) = s i n x - \sqrt{3} c o s x + 1$ is onto ,then the interval of $S$ is

[–1, 3]

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[1, 1]

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[0, 1]

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[0, –1]

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Solution

The correct option is A
[–1, 3]

$- \sqrt{1 + (- \sqrt{3})^{2}} \leq (s i n x - \sqrt{3} c o s x) \leq \sqrt{1 + (- \sqrt{3})^{2}}$

$- 2 \leq (s i n x - \sqrt{3} c o s x) \leq 2$

$- 2 + 1 \leq (s i n x - \sqrt{3} c o s x + 1) \leq 2 + 1$

$- 1 \leq (s i n x - \sqrt{3} c o s x + 1) \leq 3$ i.e., range= $[- 1, 3]$

$∴$ For f to be onto S= $[- 1, 3]$ .

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If f: R→ S defined by f(x)=sinx−√3cosx+1 is onto ,then the interval of S is

The correct option is A [–1, 3] −√1+(−√3)2 ≤ (sinx−√3cosx)≤√1+(−√3)2 −2≤(sinx−√3cosx)≤ 2 −2+1≤(sinx−√3cosx+1)≤ 2+1 −1≤(sinx−√3cosx+1)≤ 3 i.e., range=[−1,3] ∴ For f to be onto S=[−1,3].

If $f : R \to S d e f i n e d b y f (x) = s i n x - \sqrt{3} c o s x + 1$ is onto ,then the interval of $S$ is