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If the functions of `f` and `g` are defined by `f(x)=3x-4` and `g(x)=2+3x` then `g^(-1)(f^(-1)(5))`

A

1

B

`1//2`

C

`1//3`

D

`1//4`

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The correct Answer is:
To solve the problem, we need to find \( g^{-1}(f^{-1}(5)) \) where the functions are defined as \( f(x) = 3x - 4 \) and \( g(x) = 2 + 3x \). ### Step 1: Find \( f^{-1}(x) \) We start by setting \( y = f(x) \): \[ y = 3x - 4 \] Now, we solve for \( x \): \[ 3x = y + 4 \] \[ x = \frac{y + 4}{3} \] Thus, we can express the inverse function as: \[ f^{-1}(x) = \frac{x + 4}{3} \] ### Step 2: Find \( f^{-1}(5) \) Now, we need to evaluate \( f^{-1}(5) \): \[ f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3 \] ### Step 3: Find \( g^{-1}(x) \) Next, we find \( g^{-1}(x) \). We start by setting \( y = g(x) \): \[ y = 2 + 3x \] Now, we solve for \( x \): \[ 3x = y - 2 \] \[ x = \frac{y - 2}{3} \] Thus, we can express the inverse function as: \[ g^{-1}(x) = \frac{x - 2}{3} \] ### Step 4: Find \( g^{-1}(f^{-1}(5)) \) Now we need to evaluate \( g^{-1}(f^{-1}(5)) \): \[ g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3} \] ### Final Answer Thus, the final answer is: \[ g^{-1}(f^{-1}(5)) = \frac{1}{3} \] ---
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OBJECTIVE RD SHARMA ENGLISH-FUNCTIONS-Chapter Test
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  4. Let f:R to R be a function defined by f(x)=(x^(2)-8)/(x^(2)+2). Then f...

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  5. If f:(-oo,2]to (-oo,4] where f(x), then f ^(-1) (x) is given by :

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  6. Find the inverse of the function, (assuming onto). " " ...

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  7. f: R->R is defined by f(x)=(e^(x^2)-e^(-x^2))/(e^(x^2)+e^(-x^2)) is :

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  8. If f(x)=log((1+x)/(1-x))a n dt h e nf((2x)/(1+x^2)) is equal to {f(x)...

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  9. If f(x)=(2^x+2^(-x))/2 , then f(x+y)f(x-y) is equals to 1/2{f(2x)+f(2y...

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  10. The function f:R to R given by f(x)=x^(2)+x is

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  11. Let f:R to R and g:R to R be given by f(x)=3x^(2)+2 and g(x)=3x-1 for ...

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  12. The function of f:R to R, defined by f(x)=[x], where [x] denotes the g...

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  13. Let f(x)=x,g(x)=1/x and h(x)=f(x)g(x) . Then h(x)=1 for a.x in R b. x...

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  14. If the functions of f and g are defined by f(x)=3x-4 and g(x)=2+3x the...

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  15. If f(x)=(sin^(4)x+cos^(2)x)/(sin^(2)x+cos^(4)x)"for "x in R, then f(20...

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  17. A = { x // x in R, x != 0, -4 <= x <= 4 and f: A -> R is defined by f...

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  18. If f: RvecR and g: RvecR are defined by f(x)=2x+3a n dg(x)=x^2+7, then...

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  19. Let f(x) be defined on [-2,2] and be given by f(x)={(-1",",-2 le x l...

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