Q913: A function $f:A\rightarrow B$ defined as $f(x)=2x$ is both o...

KNOWLEDGE BASED

APPLY

2 Marks 2023 VSA

A function $f:A\rightarrow B$ defined as $f(x)=2x$ is both one-one and onto. If $A=\{1,2,3,4\}$, then find the set $B$.
OR
Evaluate : $\sin^{-1}(\sin\frac{3\pi}{4})+\cos^{-1}(\cos\frac{3\pi}{4})+\tan^{-1}(1)$

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Explanation

To find the set $B$, we need to determine the range of the function $f$.Reasoning:Since the function $f: A \rightarrow B$ is given to be onto (surjective), every element in the codomain ($B$) must have a corresponding element in the domain ($A$). This implies that the set $B$ is equal to the range of $f$.

Calculation:

Given the domain $A = \{1, 2, 3, 4\}$ and the rule $f(x) = 2x$, we calculate the image for each element in $A$:
For $x = 1$: $f(1) = 2(1) = 2$
For $x = 2$: $f(2) = 2(2) = 4$
For $x = 3$: $f(3) = 2(3) = 6$
For $x = 4$: $f(4) = 2(4) = 8$
Hence Collecting these values gives us the set $B$.$$B = \{2, 4, 6, 8\}$$

AI Tutor Explanation

Step-by-Step Solution

Part 1: Finding set B

Given $f(x) = 2x$ and $A = \{1, 2, 3, 4\}$, since $f$ is onto, $B$ is the range of $f$.

Calculate $f(x)$ for each element in $A$:

$f(1) = 2(1) = 2$
$f(2) = 2(2) = 4$
$f(3) = 2(3) = 6$
$f(4) = 2(4) = 8$

Therefore, $B = \{2, 4, 6, 8\}$.

Part 2: Evaluating the expression

Evaluate $\sin^{-1}(\sin\frac{3\pi}{4})+\cos^{-1}(\cos\frac{3\pi}{4})+\tan^{-1}(1)$

First, simplify $\sin^{-1}(\sin\frac{3\pi}{4})$ and $\cos^{-1}(\cos\frac{3\pi}{4})$:

$\sin\frac{3\pi}{4} = \sin(\pi - \frac{\pi}{4}) = \sin\frac{\pi}{4}$
So, $\sin^{-1}(\sin\frac{3\pi}{4}) = \sin^{-1}(\sin\frac{\pi}{4}) = \frac{\pi}{4}$
$\cos\frac{3\pi}{4} = \cos(\pi - \frac{\pi}{4}) = -\cos\frac{\pi}{4}$
So, $\cos^{-1}(\cos\frac{3\pi}{4}) = \cos^{-1}(-\cos\frac{\pi}{4}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
$\tan^{-1}(1) = \frac{\pi}{4}$

Now, add the results:

$\frac{\pi}{4} + \frac{3\pi}{4} + \frac{\pi}{4} = \frac{5\pi}{4}$

Correct Answer: B = {2, 4, 6, 8} OR 5π/4

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because it requires the student to apply the definition of one-one and onto functions to find the range, and to apply the properties of inverse trigonometric functions to evaluate the given expression.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to follow a specific procedure to find the set B by applying the function to each element of set A, and to use the correct steps to simplify the inverse trigonometric functions.

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 functions and inverse trigonometric functions as covered in the textbook.

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