JEE MainMathematics

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Sets, Relations and Functions: Mathematics | JEE Main

Let

f : \mathbb{R} \rightarrow \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

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What feels right?

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A5B4C3D2

Step-by-step solutionView

Correct answer

The value of $m$ is 5, found by bounding the argument of the logarithm between $m - 4$ and $m$ , and matching it to the range $[0, 2]$ .

Option analysis

Why each option works or fails

A · 5

None. This is the correct value. Setting the argument's maximum value $m$ equal to $(\sqrt{m})^2 = m$ and minimum value $m - 4$ equal to $(\sqrt{m})^0 = 1$ consistently yields $m = 5$ .

B · 4

The student forgets to multiply the amplitude $\sqrt{2}$ of $(\sin x - \cos x)$ by the outer $\sqrt{2}$ , taking the amplitude of the whole term as $\sqrt{2}$ instead of $2$ . Write $\sqrt{2}(\sin x - \cos x) = 2\sin(x - \pi/4)$ , so its range is $[-2, 2]$ , not $[-\sqrt{2}, \sqrt{2}]$ .

C · 3

The student equates the lower bound of the argument $m - 2$ instead of $m - 4$ to $(\sqrt{m})^0 = 1$ , arriving at $m - 2 = 1 \implies m = 3$ . Remember that the minimum value of $\sqrt{2}(\sin x - \cos x)$ is $-2$ , making the minimum argument $(m - 2) - 2 = m - 4$ , not $m - 2$ .

D · 2

The student misinterprets the base $\sqrt{m}$ as having to be strictly greater than 1 without properly solving the lower bound condition, or conflates $m - 4 = 1$ with $m - 4 = -2$ . Ensure the argument at the range minimum satisfies $\log_{\sqrt{m}}(\text{arg}_{\min}) = 0$ , giving $\text{arg}_{\min} = (\sqrt{m})^0 = 1$ .

Reviewed route

Solution

StepWorking

01given

The function $f(x) = \log_{\sqrt{m}} \left(\sqrt{2}(\sin x - \cos x) + m - 2\right)$ has a range of $[0, 2]$ .

02approach

Find the range of the argument $g(x) = \sqrt{2}(\sin x - \cos x) + m - 2$ using the bounds of $a\sin x + b\cos x$ , then apply the monotonicity of the logarithm base $\sqrt{m}$ (where $m > 1$ ) to equate the lower and upper bounds to $0$ and $2$ .

03execute

Since $-\sqrt{1^2 + (-1)^2} \le \sin x - \cos x \le \sqrt{1^2 + (-1)^2}$ , we have $-\sqrt{2} \le \sin x - \cos x \le \sqrt{2}$ . Multiplying by $\sqrt{2}$ yields $-2 \le \sqrt{2}(\sin x - \cos x) \le 2$ . Adding $m - 2$ to all parts gives $m - 4 \le \sqrt{2}(\sin x - \cos x) + m - 2 \le m$ .

04execute

Assuming base $\sqrt{m} > 1$ , the function $\log_{\sqrt{m}}(u)$ is strictly increasing. Thus, the range of $f(x)$ is $[\log_{\sqrt{m}}(m - 4), \log_{\sqrt{m}}(m)]$ . We are given that the range is $[0, 2]$ . For the upper limit: $\log_{\sqrt{m}}(m) = \log_{m^{1/2}}(m) = 2$ , which holds identically for any $m > 1, m \neq 1$ . For the lower limit: $\log_{\sqrt{m}}(m - 4) = 0 \implies m - 4 = (\sqrt{m})^0 = 1 \implies m = 5$ .

✓verify

Check for $m = 5$ : base is $\sqrt{5} > 1$ . The argument is $\sqrt{2}(\sin x - \cos x) + 3$ , which ranges from $-2 + 3 = 1$ to $2 + 3 = 5$ . Then $\log_{\sqrt{5}}(1) = 0$ and $\log_{\sqrt{5}}(5) = 2$ . Range is precisely $[0, 2]$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why must m - 4 = 1 and not 0?

For any valid logarithmic base b > 0 (b ≠ 1), log_b(y) = 0 if and only if y = 1. Therefore, the minimum value of the argument must equal 1, leading to m - 4 = 1.

Answer

The value of $m$ is 5, found by bounding the argument of the logarithm between $m - 4$ and $m$ , and matching it to the range $[0, 2]$ .

Why each option works or fails

A · correct: 5 - None. This is the correct value. Setting the argument's maximum value $m$ equal to $(\sqrt{m})^2 = m$ and minimum value $m - 4$ equal to $(\sqrt{m})^0 = 1$ consistently yields $m = 5$ .
B: 4 - The student forgets to multiply the amplitude $\sqrt{2}$ of $(\sin x - \cos x)$ by the outer $\sqrt{2}$ , taking the amplitude of the whole term as $\sqrt{2}$ instead of $2$ . Write $\sqrt{2}(\sin x - \cos x) = 2\sin(x - \pi/4)$ , so its range is $[-2, 2]$ , not $[-\sqrt{2}, \sqrt{2}]$ .
C: 3 - The student equates the lower bound of the argument $m - 2$ instead of $m - 4$ to $(\sqrt{m})^0 = 1$ , arriving at $m - 2 = 1 \implies m = 3$ . Remember that the minimum value of $\sqrt{2}(\sin x - \cos x)$ is $-2$ , making the minimum argument $(m - 2) - 2 = m - 4$ , not $m - 2$ .
D: 2 - The student misinterprets the base $\sqrt{m}$ as having to be strictly greater than 1 without properly solving the lower bound condition, or conflates $m - 4 = 1$ with $m - 4 = -2$ . Ensure the argument at the range minimum satisfies $\log_{\sqrt{m}}(\text{arg}_{\min}) = 0$ , giving $\text{arg}_{\min} = (\sqrt{m})^0 = 1$ .

Step-by-step solution

given: The function $f(x) = \log_{\sqrt{m}} \left(\sqrt{2}(\sin x - \cos x) + m - 2\right)$ has a range of $[0, 2]$ .
approach: Find the range of the argument $g(x) = \sqrt{2}(\sin x - \cos x) + m - 2$ using the bounds of $a\sin x + b\cos x$ , then apply the monotonicity of the logarithm base $\sqrt{m}$ (where $m > 1$ ) to equate the lower and upper bounds to $0$ and $2$ .
execute: Since $-\sqrt{1^2 + (-1)^2} \le \sin x - \cos x \le \sqrt{1^2 + (-1)^2}$ , we have $-\sqrt{2} \le \sin x - \cos x \le \sqrt{2}$ . Multiplying by $\sqrt{2}$ yields $-2 \le \sqrt{2}(\sin x - \cos x) \le 2$ . Adding $m - 2$ to all parts gives $m - 4 \le \sqrt{2}(\sin x - \cos x) + m - 2 \le m$ .
execute: Assuming base $\sqrt{m} > 1$ , the function $\log_{\sqrt{m}}(u)$ is strictly increasing. Thus, the range of $f(x)$ is $[\log_{\sqrt{m}}(m - 4), \log_{\sqrt{m}}(m)]$ . We are given that the range is $[0, 2]$ . For the upper limit: $\log_{\sqrt{m}}(m) = \log_{m^{1/2}}(m) = 2$ , which holds identically for any $m > 1, m \neq 1$ . For the lower limit: $\log_{\sqrt{m}}(m - 4) = 0 \implies m - 4 = (\sqrt{m})^0 = 1 \implies m = 5$ .
verify: Check for $m = 5$ : base is $\sqrt{5} > 1$ . The argument is $\sqrt{2}(\sin x - \cos x) + 3$ , which ranges from $-2 + 3 = 1$ to $2 + 3 = 5$ . Then $\log_{\sqrt{5}}(1) = 0$ and $\log_{\sqrt{5}}(5) = 2$ . Range is precisely $[0, 2]$ .

Shortcut:

When to use it: Fast option elimination using the lower limit property

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

approach: For the minimum value of $f(x)$ to be 0, the minimum value of the argument inside the logarithm must be equal to 1.

execute: Minimum value of $\sqrt{2}(\sin x - \cos x)$ is $\sqrt{2}(-\sqrt{2}) = -2$ . Thus, the minimum value of the argument is $-2 + m - 2 = m - 4$ . Setting this to 1 gives $\mathbb{N} \times \mathbb{N}$ .

verify: With $m = 5$ , base $\sqrt{5} > 1$ , function is well-defined and positive over the whole domain.

Sets, Relations and Functions: Mathematics | JEE Main

What feels right?

The value of $m$ is 5, found by bounding the argument of the logarithm between $m - 4$ and $m$ , and matching it to the range $[0, 2]$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

✓verify

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