Question

1 point
14. Which of the following has its area and
perimeter numerically equal?
an equilateral triangle of side 1 cm
O
a circle of radius 2 cm
O
a square of side 1 cm​

Answer 1

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