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If k = lim x → 0 + ( Min. ( t 2 − 6 ( t + x ) + 12 ) + [ ln ( 1 + 2 x ) x ] ) k=\lim_{x\rightarrow 0^+}\left(\text{Min.}(t^2-6(t+x)+12)+\left[\frac{\ln(1+2x)}{x}\right]\right) k = lim x → 0 + ( Min. ( t 2 − 6 ( t + x ) + 12 ) + [ x l n ( 1 + 2 x ) ] ) …
Mathematics
Problem If k = lim x → 0 + ( Min. ( t 2 − 6 ( t + x ) + 12 ) + [ ln ( 1 + 2 x ) x ] ) k=\lim_{x\rightarrow 0^+}\left(\text{Min.}(t^2-6(t+x)+12)+\left[\frac{\ln(1+2x)}{x}\right]\right) k = lim x → 0 + ( Min. ( t 2 − 6 ( t + x ) + 12 ) + [ x l n ( 1 + 2 x ) ] ) where t ∈ R t\in\mathbb{R} t ∈ R and [ y ] [y] [ y ] denotes greatest integer less than or equal to y y y , then the value of 3 2 k \frac{3}{2}k 2 3 k is equal to
(a) 4
(b) 3
(c) 6
(d) 9
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