JEE 2023 — Mathematics PYQ

1. Analyze the Trigonometric Term

The expression $\sqrt{2}(\sin x-\cos x)$ can be simplified using the identity $a\sin x+b\cos x\in [-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$.

• Here, $a=\sqrt{2}$ and $b=-\sqrt{2}$.

• Range $=[-\sqrt{(\sqrt{2}{)}^2+(-\sqrt{2}{)}^2},\sqrt{(\sqrt{2}{)}^2+(-\sqrt{2}{)}^2}]$

• Range $=[-\sqrt{4},\sqrt{4}]=[-2,2]$.

2. Determine the Range of the Argument

Let $g(x)=\sqrt{2}(\sin x-\cos x)+m-2$.

• Minimum value of $g(x)=-2+m-2=m-4$

• Maximum value of $g(x)=2+m-2=m$

  So, the argument of the log is in the interval $[m-4,m]$.

3. Set up the Logarithmic Range

The problem states the range of $f(x)={\log }_{\sqrt{m}}\{g(x)\}$ is $[0,2]$.

For the minimum value:

Using the property ${\log }_b(a)=c⟹a=b^c$:

4. Verification

Check the maximum value using $m=5$:

Since $5=(\sqrt{5}{)}^2$:

This matches the upper bound of the given range $[0,2]$.

Correct Option: (1) 5