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Question: If \(\frac{x + 5}{3}\) then \(f:R \rightarrow R\)is...

If x+53\frac{x + 5}{3} then f:R→Rf:R \rightarrow Ris

A

f(x)=3x−4,f(x) = 3x - 4,

B

e3

C

f−1(x)f^{- 1}(x)

D

3x+43x + 4

Answer

f(x)=3x−4,f(x) = 3x - 4,

Explanation

Solution

lim⁡x→∞f(x)=lim⁡x→∞(x2+5x+3x2+x+2)x=lim⁡x→∞(1+4x+1x2+x+2)x\lim _ { x \rightarrow \infty } f ( x ) = \lim _ { x \rightarrow \infty } \left( \frac { x ^ { 2 } + 5 x + 3 } { x ^ { 2 } + x + 2 } \right) ^ { x } = \lim _ { x \rightarrow \infty } \left( 1 + \frac { 4 x + 1 } { x ^ { 2 } + x + 2 } \right) ^ { x }

= lim⁡x→∞[(1+4x+1x2+x+2)x2+x+24x+1]α\lim _ { x \rightarrow \infty } \left[ \left( 1 + \frac { 4 x + 1 } { x ^ { 2 } + x + 2 } \right) ^ { \frac { x ^ { 2 } + x + 2 } { 4 x + 1 } } \right] ^ { \alpha } ……………..(i)

where α=x(4x+1)x2+x+2=4+1x1+1x+2x2→4\alpha = \frac { x ( 4 x + 1 ) } { x ^ { 2 } + x + 2 } = \frac { 4 + \frac { 1 } { x } } { 1 + \frac { 1 } { x } + \frac { 2 } { x ^ { 2 } } } \rightarrow 4as x → α

∴ (1) ⇒ lim⁡x→∞f(x)=e4\lim _ { x \rightarrow \infty } f ( x ) = e ^ { 4 }