JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the Base Parameter of a Logarithmic Function from its Range

Let f: ℝ → ℝ be a function defined by f(x) = log_(√(m)) {√(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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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

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 $\sqrt{2} (\sin x - \cos x) = 2 \sin (x - π / 4)$ , 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 $m - 2 = 1 ⟹ m = 3$ . 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 $\log_{\sqrt{m}} ({arg}_{\min}) = 0$ , giving ${arg}_{\min} = (\sqrt{m})^{0} = 1$ .

Reviewed route

Solution

StepWorking

01Given

The function $f (x) = \log_{\sqrt{m}} (\sqrt{2} (\sin x - \cos x) + m - 2)$ 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 $- \sqrt{1^{2} + (- 1)^{2}} \leq \sin x - \cos x \leq \sqrt{1^{2} + (- 1)^{2}}$ , we have $- \sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$ . Multiplying by $\sqrt{2}$ yields $- 2 \leq \sqrt{2} (\sin x - \cos x) \leq 2$ . Adding $m - 2$ to all parts gives $m - 4 \leq \sqrt{2} (\sin x - \cos x) + m - 2 \leq m$ .

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: $\log_{\sqrt{m}} (m - 4) = 0 ⟹ m - 4 = (\sqrt{m})^{0} = 1 ⟹ m = 5$ .

✓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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 - 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 $(\sqrt{m})^{0} = 1$ . Write $\sqrt{2} (\sin x - \cos x) = 2 \sin (x - π / 4)$ , 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 $m - 2 = 1 ⟹ m = 3$ . 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 $\log_{\sqrt{m}} ({arg}_{\min}) = 0$ , giving ${arg}_{\min} = (\sqrt{m})^{0} = 1$ .

Step-by-step solution

Given: The function $f (x) = \log_{\sqrt{m}} (\sqrt{2} (\sin x - \cos x) + m - 2)$ 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 $- \sqrt{1^{2} + (- 1)^{2}} \leq \sin x - \cos x \leq \sqrt{1^{2} + (- 1)^{2}}$ , we have $- \sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$ . Multiplying by $\sqrt{2}$ yields $- 2 \leq \sqrt{2} (\sin x - \cos x) \leq 2$ . Adding $m - 2$ to all parts gives $m - 4 \leq \sqrt{2} (\sin x - \cos x) + m - 2 \leq m$ .
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: $\log_{\sqrt{m}} (m - 4) = 0 ⟹ m - 4 = (\sqrt{m})^{0} = 1 ⟹ m = 5$ .
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 $f (x) = \log_{\sqrt{m}} (\sqrt{2} (\sin x - \cos x) + m - 2)$ .

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 $m - 4 = 1 ⟹ m = 5$ .

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

Finding the Base Parameter of a Logarithmic Function from its Range

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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Suppose $f : R \to (0, \infty)$ be a differentiable function such that…

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