Functions Question 5
Question 5 - 25 January - Shift 2
Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x)=$ $\log _{\sqrt{m}}{\sqrt{2}(\sin x-\cos x)+m-2}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is
(1) 5
(2) 3
(3) 2
(4) 4
Show Answer
Answer: (1)
Solution:
Formula: Logarithmic function, Domain and Range of Trigonometric Functions
Since,
$-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2}$
$\therefore-2 \leq \sqrt{2}(\sin x-\cos x) \leq 2$
(Assume $\sqrt{2}(\sin x-\cos x)=k$ )
$-2 \leq k \leq 2$
$f(x)=\log _{\sqrt{m}}(k+m-2)$
Given,
$ \begin{aligned} & 0 \leq f(x) \leq 2 \\ & 0 \leq \log _{\sqrt{m}}(k+m-2) \leq 2 \\ & 1 \leq k+m-2 \leq m \\ & -m+3 \leq k \leq 2 \end{aligned} $
From eq. (i) and (ii), we get $-m+3=-2$
$ \Rightarrow m=5 $
