What is the value of log (1+x) if |x|<1
Answers
Answer:
ANSWER.
Option [ b ] is correct answer.
EXPLANATION.
\begin{gathered} \sf : \implies \: { \underline{conditions \: of \: one - one \: and \: onto}} \\ \\ \sf : \implies \: \: one - one \: we \: must \: check \: f {}^{' } (x) > 0 \: \: or \: f {}^{' } (x) < 0 \: in \: given \: domain \\ \\ \sf : \implies \: in \: onto \: function \: range = codomain\end{gathered}
:⟹
conditionsofone−oneandonto
:⟹one−onewemustcheckf
′
(x)>0orf
′
(x)<0ingivendomain
:⟹inontofunctionrange=codomain
\begin{gathered} \sf : \implies \: f \ratio \: (0 , 3) \to \: (1 , 29) \\ \\ \sf : \implies \: f(x) = 2 {x}^{3} - 15 {x}^{2} + 36x + 1 \\ \\ \sf : \implies \: we \: can \: differentiate \: the \: equation \\ \\ \sf : \implies \: f {}^{ ' } (x) = 6 {x}^{2} - 30x + 36 \\ \\ \sf : \implies f {}^{ ' } (x) = 6( {x}^{2} - 5x + 6) \\ \\ \sf : \implies \: 6( {x}^{2} - 3x - 2x + 6) \end{gathered}
:⟹f:(0,3)→(1,29)
:⟹f(x)=2x
3
−15x
2
+36x+1
:⟹wecandifferentiatetheequation
:⟹f
′
(x)=6x
2
−30x+36
:⟹f
′
(x)=6(x
2
−5x+6)
:⟹6(x
2
−3x−2x+6)
\begin{gathered} \sf : \implies \: f {}^{ ' } (x) = 6(x - 2)( x - 3) \\ \\ \sf : \implies \: we \: can \: put \: the \: value \: in \:wavy \: curve \: method \\ \\ \sf : \implies \: zeroes \: of \: equation \: are =( 0,2,3) \\ \\ \sf: \implies \: the \: value \: given \: is \\ \\ \sf : \implies \:for \: given \: domain \: (0,3) \: \\ \\ \sf : \implies \: f(x) \: is \: increasing \: as \: well \: as \: decreasing \: \implies \: many \: one \: \end{gathered}
:⟹f
′
(x)=6(x−2)(x−3)
:⟹wecanputthevalueinwavycurvemethod
:⟹zeroesofequationare=(0,2,3)
:⟹thevaluegivenis
:⟹forgivendomain(0,3)
:⟹f(x)isincreasingaswellasdecreasing⟹manyone
\begin{gathered} \sf : \implies \: now \: put \: the \: value \: f {}^{'}(x) = 0 \\ \\ \sf : \implies \: 2(0 {}^{3} ) - 15( {0}^{2} ) + 36(0) + 1 \\ \\ \sf : \implies \: f(x) =1 \\ \\ \sf : \implies \: put \: f(2) \\ \\ \sf : \implies \: 2(2 {}^{3}) - 15(2 {}^{2} ) + 36(2) + 1 \\ \\ \sf : \implies \: 16 - 60 + 72 + 1 = 29 \\ \\ \sf : \implies \: f(2) = 29 \end{gathered}
:⟹nowputthevaluef
′
(x)=0
:⟹2(0
3
)−15(0
2
)+36(0)+1
:⟹f(x)=1
:⟹putf(2)
:⟹2(2
3
)−15(2
2
)+36(2)+1
:⟹16−60+72+1=29
:⟹f(2)=29
\begin{gathered} \sf : \implies \: f(3) \\ \\ \sf : \implies \: 2(3 {}^{3}) - 15(3 {}^{2} ) + 36(3) + 1 \\ \\ \sf : \implies \: 54 - 135 + 108 + 1 \\ \\ \sf : \implies \: f(3) = 28\end{gathered}
:⟹f(3)
:⟹2(3
3
)−15(3
2
)+36(3)+1
:⟹54−135+108+1
:⟹f(3)=28
\begin{gathered} \sf : \implies \: range \: = (1 , 29) \\ \\ \sf : \implies \: \green{{ \underline{function \: is \: onto \: but \: not \: one - one}}}\end{gathered}
:⟹range=(1,29)
:⟹
functionisontobutnotone−one