JEE Mathematics 2023 Question with Solution

Given: $f(x) = \log_{\sqrt{m}}\left(\sqrt{2}(\sin x - \cos x) + m - 2\right)$ and the range of $f$ is $[0,2]$.

Find: the value of $m$.

Let

$$k = \sqrt{2}(\sin x - \cos x)$$

Since $\sin x - \cos x \in [-\sqrt{2}, \sqrt{2}]$, we get

$$-2 \le k \le 2$$

Now

$$f(x) = \log_{\sqrt{m}}(k + m - 2)$$

Given that

$$0 \le f(x) \le 2$$

So,

$$0 \le \log_{\sqrt{m}}(k + m - 2) \le 2$$

Using the conversion shown in the solution,

$$1 \le k + m - 2 \le m$$

Hence,

$$-m + 3 \le k \le 2$$

But already,

$$-2 \le k \le 2$$

For the range of $f$ to be exactly $[0,2]$, the left endpoints must match as shown:

$$-m + 3 = -2$$

Therefore,

$$m = 5$$

So the correct option is A.

The solution labels the option as B, but it also states the numerical value is $5$. Since $5$ is option A in the given options, the correct answer here is A.

Given: $f(x) = \log_{\sqrt{m}}\left(\sqrt{2}(\sin x - \cos x) + m - 2\right)$ and range of $f$ is $[0,2]$.

Find: $m$.

First use the standard range

$$-\sqrt{2} \le \sin x - \cos x \le \sqrt{2}$$

Multiplying by $\sqrt{2}$ gives

$$-2 \le \sqrt{2}(\sin x - \cos x) \le 2$$

If we denote

$$k = \sqrt{2}(\sin x - \cos x)$$

then

$$-2 \le k \le 2$$

and

$$f(x) = \log_{\sqrt{m}}(k+m-2)$$

Since the range of $f$ is from $0$ to $2$, the argument of the logarithm must vary from

$$(\sqrt{m})^0 = 1$$

to

$$(\sqrt{m})^2 = m$$

So the solution uses

$$1 \le k+m-2 \le m$$

which implies

$$3-m \le k \le 2$$

Now compare this with the actual range of $k$:

$$-2 \le k \le 2$$

Therefore,

$$3-m = -2$$

so

$$m = 5$$

Hence the required value is $5$, which corresponds to option A.