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Function f[(1)/(2)pi, (3)/(2)pi] rarr [-...

Function `f[(1)/(2)pi, (3)/(2)pi] rarr [-1, 1], f(x) = cos x` is

A

Many-one onto

B

Onto

C

One-one onto

D

Many-one into

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The correct Answer is:
To determine the nature of the function \( f(x) = \cos x \) with the domain \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \) and the co-domain \( [-1, 1] \), we will analyze whether it is one-one, onto, many-one, or into. ### Step-by-Step Solution: 1. **Identify the Function and Its Domain:** - We have the function \( f(x) = \cos x \). - The domain is \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \). 2. **Graph the Function:** - The cosine function oscillates between -1 and 1. - For the specified domain, the graph of \( \cos x \) will start at \( \cos\left(\frac{\pi}{2}\right) = 0 \) and end at \( \cos\left(\frac{3\pi}{2}\right) = 0 \). - The minimum value occurs at \( x = \pi \), where \( \cos(\pi) = -1 \). 3. **Determine the Range of the Function:** - From the graph, we can see that as \( x \) moves from \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \), \( f(x) \) decreases from 0 to -1 and then returns to 0. - Therefore, the range of \( f(x) \) over the given domain is \( [-1, 0] \). 4. **Check if the Function is One-One:** - A function is one-one if no two different inputs produce the same output. - In this case, since the graph of \( f(x) \) is decreasing from \( 0 \) to \( -1 \) and then increasing back to \( 0 \), it intersects the horizontal line \( y = c \) (for \( c \) in the range) at two points. - Therefore, the function is **many-one**. 5. **Check if the Function is Onto:** - A function is onto if every element in the co-domain is mapped by at least one element in the domain. - The co-domain is \( [-1, 1] \) while the range is \( [-1, 0] \). - Since the range does not cover the entire co-domain (it does not reach values greater than 0), the function is **not onto**. 6. **Determine if the Function is Into:** - A function is into if it does not cover the entire co-domain. - Since the range \( [-1, 0] \) does not cover the co-domain \( [-1, 1] \), the function is **into**. ### Conclusion: Based on the analysis, we conclude that the function \( f(x) = \cos x \) defined on the domain \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \) and co-domain \( [-1, 1] \) is **many-one and into**. ### Final Answer: The function is many-one into.
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AAKASH INSTITUTE ENGLISH-RELATIONS AND FUNCTIONS -Assignment (Section - A) Objective Type Questions (one option is correct)
  1. Function f : R rarr R, f(x) = x + |x|, is

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  2. Function f : [(pi)/(2), (3pi)/(2)] rarr [-1, 1], f(x) = sin x is

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  3. Function f[(1)/(2)pi, (3)/(2)pi] rarr [-1, 1], f(x) = cos x is

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  4. If f : R rarr R, f(x) = sin^(2) x + cos^(2) x, then f is

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  5. If function f(x) = (1+2x) has the domain (-(pi)/(2), (pi)/(2)) and co-...

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  6. The function f : (0, oo) rarr [0, oo), f(x) = (x)/(1+x) is

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  7. If f(x) = x/(x-1)=1/y then the value of f(y) is

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  8. gof exists, when :

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  9. If f : R rarr R, f(x) = x^(2) + 2x - 3 and g : R rarr R, g(x) = 3x - 4...

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  10. If f : R rarr R, f(x) = x^(2) - 5x + 4 and g : R^(+) rarr R, g(x) = lo...

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  11. If f : R - {1} rarr R, f(x) = (x-3)/(x+1), then f^(-1) (x) equals

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  12. If function f : R rarr R^(+), f(x) = 2^(x), then f^(-1) (x) will be eq...

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  13. If f(x) = 2 sinx, g(x) = cos^(2) x, then the value of (f+g)((pi)/(3))

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  14. The graph of the function y = log(a) (x + sqrt(x^(2) + 1)) is not sym...

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  15. If the function f:[1,\ oo)->[1,\ oo) defined by f(x)=2^(x(x-1)) is inv...

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  16. If f(x)=(1)/((1-x)),g(x)=f{f(x)}andh(x)=f[f{f(x)}]. Then the value of ...

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  17. If f(x)=sin^2x+sin^2(x+pi/3)+cosxcos(x+pi/3)a n dg(5/4=1, then (gof)(x...

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  18. If g(f(x))=|sinx|a n df(g(x))=(sinsqrt(x))^2 , then f(x)=sin^2x ,g(x)...

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  19. Let g(x)=1+x-[x] "and " f(x)={{:(-1","x l...

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  20. If F :[1,oo)->[2,oo) is given by f(x)=x+1/x ,t h e n \ f^(-1)(x) equal...

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