Question

Let $f:R\rightarrow R$ be a function defined by $f(x)=log_{\sqrt{m}}(\sqrt{2}(sinx-cosx)+m-2)$, for some m, such that the range of f is [0, 2]. Then the value of m is:

• A. 5
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

5

Answer 2

The argument of the logarithm is $A(x) = \sqrt{2}(\sin x - \cos x) + m - 2$. The range of $\sqrt{2}(\sin x - \cos x)$ is $[-2, 2]$. Thus, the range of $A(x)$ is $[m-4, m]$. For $f(x)$ to be defined, the base $\sqrt{m} > 0, \neq 1$ and the argument $A(x) > 0$. The condition $m-4 > 0 \implies m > 4$ ensures these. Since $m > 4$, $\sqrt{m} > 1$, so $\log_{\sqrt{m}}$ is increasing. The range of $f(x)$ is $[\log_{\sqrt{m}}(m-4), \log_{\sqrt{m}}(m)]$. Given range is $[0, 2]$. Equating minimums: $\log_{\sqrt{m}}(m-4) = 0 \implies m-4 = 1 \implies m = 5$. Verifying with maximums: $\log_{\sqrt{5}}(5) = 2$, which is correct.