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If f(x)={{:(-1, x lt 0),(0, x=0 and g(x)...

If `f(x)={{:(-1, x lt 0),(0, x=0 and g(x)=x(1-x^(2))", then"),(1, x gt 0):}`

A

`fog(x)={{:(-1, -1 lt x lt 0 or x gt 1),(0, x =0 , 1,-1),(1, 0 lt x lt 1):}`

B

`fog(x)={{:(-1, -1 lt x lt 0),(0, x=0,1,-1),(1, 0 lt x lt 1):}`

C

`fog(x)={{:(-1, -1 lt x lt 0 or x gt 1),(0, x=0, 1,-1),(1, 0 lt x lt 1 or x lt -1):}`

D

`fog(x)={{:(-1 , x lt 0 or x gt 1),( 0, x =0, 1,-1),(1, 0 lt x lt 1 or x lt -1):}`

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AI Generated Solution

The correct Answer is:
To solve the problem, we need to evaluate \( f(g(x)) \) given the piecewise function \( f(x) \) and the function \( g(x) \). ### Step 1: Define the functions The function \( f(x) \) is defined as: - \( f(x) = -1 \) for \( x < 0 \) - \( f(x) = 0 \) for \( x = 0 \) - \( f(x) = 1 \) for \( x > 0 \) The function \( g(x) \) is defined as: \[ g(x) = x(1 - x^2) \] ### Step 2: Analyze \( g(x) \) We can rewrite \( g(x) \): \[ g(x) = x(1 - x^2) = x - x^3 \] ### Step 3: Find the roots of \( g(x) \) To find where \( g(x) = 0 \): \[ g(x) = 0 \Rightarrow x(1 - x^2) = 0 \] This gives us the roots: - \( x = 0 \) - \( 1 - x^2 = 0 \Rightarrow x = \pm 1 \) Thus, the roots of \( g(x) \) are \( x = -1, 0, 1 \). ### Step 4: Determine the intervals for \( g(x) \) Next, we analyze the sign of \( g(x) \) in the intervals defined by its roots: 1. For \( x < -1 \): \( g(x) < 0 \) 2. For \( -1 < x < 0 \): \( g(x) > 0 \) 3. For \( 0 < x < 1 \): \( g(x) > 0 \) 4. For \( x > 1 \): \( g(x) < 0 \) ### Step 5: Evaluate \( f(g(x)) \) Now we can evaluate \( f(g(x)) \) based on the intervals: - **For \( x < -1 \)**: \( g(x) < 0 \) so \( f(g(x)) = -1 \) - **For \( -1 < x < 0 \)**: \( g(x) > 0 \) so \( f(g(x)) = 1 \) - **For \( x = 0 \)**: \( g(0) = 0 \) so \( f(g(0)) = 0 \) - **For \( 0 < x < 1 \)**: \( g(x) > 0 \) so \( f(g(x)) = 1 \) - **For \( x > 1 \)**: \( g(x) < 0 \) so \( f(g(x)) = -1 \) ### Step 6: Summarize the results The final result for \( f(g(x)) \) can be summarized as: - \( f(g(x)) = -1 \) for \( x < -1 \) and \( x > 1 \) - \( f(g(x)) = 1 \) for \( -1 < x < 0 \) and \( 0 < x < 1 \) - \( f(g(0)) = 0 \) ### Conclusion Thus, the piecewise function \( f(g(x)) \) can be expressed as: \[ f(g(x)) = \begin{cases} -1 & \text{if } x < -1 \text{ or } x > 1 \\ 1 & \text{if } -1 < x < 0 \text{ or } 0 < x < 1 \\ 0 & \text{if } x = 0 \end{cases} \]
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OBJECTIVE RD SHARMA ENGLISH-REAL FUNCTIONS -Chapter Test
  1. If N denotes the set of all positive integers and if f : N -> N is def...

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  2. The set of value of a for which the function f(x)=sinx+[(x^(2))/(a)] d...

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  3. If f(x)={{:(-1, x lt 0),(0, x=0 and g(x)=x(1-x^(2))", then"),(1, x gt ...

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  4. Find the equivalent definition of f(x)=max.{x^(2),(1-x)^(2),2x(1-x)...

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  5. If f(x) is defined on [0,1], then the domain of f(3x^(2)) , is

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  6. The function f(x) is defined in [0,1] . Find the domain of f(t a nx)do...

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  7. The domain of definition of the real function f(x)=sqrt(log(12)x^(2)) ...

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  8. The values of ba n dc for which the identity of f(x+1)-f(x)=8x+3 is sa...

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  9. The function f(x)=sin""(pix)/(2)+2 cos ""(pix)/(3)-tan""(pix)/(4) is p...

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  10. The period of the function sin""((pix)/(2))+cos((pix)/(2)), is

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  11. If x in R, then f(x)=sin^(-1)((2x)/(1+x^(2))) is equal to

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  12. If x in R , then f(x)=cos^(-1)((1-x^(2))/(1+x^(2))) is equal to

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  13. The equivalent definition of the function f(x)=lim(n to oo)(x^(n)-x^(-...

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  14. Find the domain of definitions of the following function: f(x)=log(10)...

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  15. The domain of definition of f(x)=log(0.5){-log(2)((3x-1)/(3x+2))}, is

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  16. Find the domain of the function : f(x)=sqrt(((log)(0. 2)|x-2|)/(|x|))

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  17. The domain of the function y=sqrt(log10(log10x)-log10(4-log10x)-log10 ...

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  18. The function f(x)=log(2x-5)(x^(2)-3x-10) is defined for all belonging ...

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  19. The domain of definition of f(x)=log(1.7)((2-phi'(x))/(x+1))^(1//2), w...

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  20. f(x)=log(100x)((2log(10)x+1)/(-x)) exists, if …. .

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