Q609: The function $f(x)=kx-\sin~x$ is strictly increasing for ...

KNOWLEDGE BASED

APPLY

1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE

The function $f(x)=kx-\sin~x$ is strictly increasing for

(A) $k\gt1$

(B) $k\lt1$

(D) $k\lt-1$

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Explanation

The solution proceeds as follows:

**Step 1: Find the derivative of the function.**
The derivative of $f(x) = kx - \sin~x$ is $f'(x) = k - \cos~x$.

**Step 2: Determine the condition for a strictly increasing function.**
For a function to be strictly increasing, its derivative must be strictly positive, i.e., $f'(x) > 0$.

**Step 3: Apply the condition to the derivative.**
We need $k - \cos~x > 0$ for all $x$. This means $k > \cos~x$ for all $x$.

**Step 4: Find the maximum value of $\cos~x$.**
The maximum value of $\cos~x$ is 1.

**Step 5: Determine the condition for $k$.**
For $k > \cos~x$ to hold for all $x$, $k$ must be greater than the maximum value of $\cos~x$. Therefore, $k > 1$.

**Final Answer:**
The function $f(x)=kx-\sin~x$ is strictly increasing for $k\gt1$.

The final answer is $\boxed{k\gt1}$.

AI Tutor Explanation

Step-by-Step Solution

Given the function $f(x) = kx - \sin x$, we need to find the condition for which it is strictly increasing.
A function is strictly increasing if its derivative is greater than 0, i.e., $f'(x) > 0$.
Find the derivative of $f(x)$ with respect to $x$: \[f'(x) = \frac{d}{dx}(kx - \sin x) = k - \cos x\]
For $f(x)$ to be strictly increasing, $f'(x) > 0$, so we have: \[k - \cos x > 0\] \[k > \cos x\]
Since the maximum value of $\cos x$ is 1, for $k > \cos x$ to hold true for all $x$, we must have: \[k > 1\]
Therefore, the function $f(x) = kx - \sin x$ is strictly increasing for $k > 1$.

Correct Answer: k > 1

AI Suggestion: Option A

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of derivatives and increasing functions to solve the problem. They must differentiate the given function and then use the condition for strictly increasing functions to find the range of values for k.

Knowledge Dimension: PROCEDURAL

Justification: The question requires a series of steps to solve, including differentiating the function, setting up an inequality based on the condition for strictly increasing functions, and solving the inequality to find the range of k.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of concepts related to derivatives and increasing/decreasing functions, which are core topics covered in the textbook.

Core Concept

Analysis of the Given Problem

The problem asks for the condition on the constant $k$ such that the function $f(x) = kx - \sin x$ is strictly increasing. This falls under the domain of Calculus, specifically the Application of Derivatives.

Sub-topic: Monotonicity (Strictly Increasing Functions)

Step-by-Step Rule

Step 1: Differentiate the Function. Find the first derivative of the given function, $f(x)$, with respect to $x$. This gives us the slope of the tangent at any point $x$.For $f(x) = kx - \sin x$, the derivative is $f'(x) = k - \cos x$.
Step 2: Apply the Condition for Strict Increase. For a function to be strictly increasing on its domain (the set of all real numbers $\mathbb{R}$), its first derivative must be strictly greater than zero for all $x$.Set $f'(x) > 0$.
Step 3: Isolate the Constant. Rearrange the inequality to separate the constant $k$ from the variable term.$k - \cos x > 0 \Rightarrow k > \cos x$.
Step 4: Analyze the Bounds. For the inequality $k > \cos x$ to hold true for all real values of $x$, $k$ must be strictly greater than the maximum possible value of $\cos x$. Since the maximum value of $\cos x$ is $1$, $k$ must be greater than $1$.

Key Mathematical Formula or Condition

$$ \frac{d}{dx}f(x) > 0 \quad \text{for all } x \in \mathbb{R} $$

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