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Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 21 Maths Textbook Solution.

Answers (1)

Answer:

                f is not invertible

Given:

                f\left ( x \right )= \cos \left ( x+2 \right )
Hint:
If the function is invertible, that should fulfil the one-one, onto condition.
Solution:
Let x,y be two elements in the domain \left ( R \right )

                f\left ( x \right )= f\left ( y \right )

                \cos \left ( x+2 \right )=\cos \left ( y+2 \right )

                \left ( x+2 \right )=2\pi - \left ( y+2 \right )

                x=2\pi -y-4

So, we can’t say x=y

For example,

                \cos \frac{\pi }{2}= \cos \frac{3\pi }{2}=0

So, \frac{\pi }{2} and \frac{3\pi }{2}=0

So, \frac{\pi }{2} and \frac{3\pi }{2} have the same image 0

f is not one-one, bijection, surjection.

Hence, f is not invertible.

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