(i) Length of f(x)
Given,
f ( x ) = x − 2 , f o r 2 ≤ x ≤ 4 f(x)=x-2, \quad for 2 \leq x \leq 4 f ( x ) = x − 2 , f or 2 ≤ x ≤ 4
The length would be the length of hypotenuse of right angled isosceles triangle of side length =2 units
⇒ h 2 = b 2 + h 2 ⇒ h 2 = 2 2 + 2 2 ⇒ h 2 = 8 ⇒ h = 2 2 \begin{aligned}
&\Rightarrow h^{2}=b^{2}+h^{2} \\
&\Rightarrow h^{2}=2^{2}+2^{2} \\
&\Rightarrow h^{2}=8 \\
&\Rightarrow h=2 \sqrt{2}
\end{aligned} ⇒ h 2 = b 2 + h 2 ⇒ h 2 = 2 2 + 2 2 ⇒ h 2 = 8 ⇒ h = 2 2
Length of graph contributed by f ( x ) = 2 2 f(x)=2 \sqrt{2} f ( x ) = 2 2
(ii) Length of g(x)
Given,
g ( x ) = 4 − ( x − 6 ) 2 + 2 , for 4 ≤ x ≤ 8 ⇒ g ( x ) − 2 = 4 − ( x − 6 ) 2 ⇒ ( g ( x ) − 2 ) 2 = 4 − ( x − 6 ) 2 ⇒ ( x − 6 ) 2 + ( g ( x ) − 2 ) 2 = 2 2 \begin{aligned}
&g(x)=\sqrt{4-(x-6)^{2}}+2, \quad \text { for } 4 \leq x \leq 8 \\
&\Rightarrow g(x)-2=\sqrt{4-(x-6)^{2}} \\
&\Rightarrow(g(x)-2)^{2}=4-(x-6)^{2}\\
&\Rightarrow (x-6)^{2}+(g(x)-2)^{2}=2^2
\end{aligned} g ( x ) = 4 − ( x − 6 ) 2 + 2 , for 4 ≤ x ≤ 8 ⇒ g ( x ) − 2 = 4 − ( x − 6 ) 2 ⇒ ( g ( x ) − 2 ) 2 = 4 − ( x − 6 ) 2 ⇒ ( x − 6 ) 2 + ( g ( x ) − 2 ) 2 = 2 2
So, g(x) is a circle with centre at (6,2) and radius = 2 units
Length contributed by g(x) would be circumference of semicircle
Length contributed by g(x) = π r = 2 π \pi r=2\pi π r = 2 π
(iii) Length of h ( x ) \mathrm{h}(\mathrm{x}) h ( x )
Given,
f ( x ) = x − 6 , f o r 8 ≤ x ≤ 10 f(x)=x-6, \quad for \ 8 \leq x \leq 10 f ( x ) = x − 6 , f or 8 ≤ x ≤ 10
The length would be the length of hypotenuse of right angled triangle of base = height =2 units
⇒ h 2 = b 2 + h 2 ⇒ h 2 = 2 2 + 2 2 ⇒ h 2 = 8 ⇒ h = 2 2 \begin{aligned}
&\Rightarrow h^{2}=b^{2}+h^{2} \\
&\Rightarrow h^{2}=2^{2}+2^{2} \\
&\Rightarrow h^{2}=8 \\
&\Rightarrow h=2 \sqrt{2}
\end{aligned} ⇒ h 2 = b 2 + h 2 ⇒ h 2 = 2 2 + 2 2 ⇒ h 2 = 8 ⇒ h = 2 2
Length of graph contributed by h ( x ) = 2 2 h(x)=2 \sqrt{2} h ( x ) = 2 2
Total length of graph
L = length contributed by f ( x ) + length contributed by g ( x ) + length contributed by h ( x ) ⇒ L = 2 2 + 2 π + 2 2 ⇒ L = 4 2 + 2 π \begin{aligned}
&L=\text { length contributed by } f(x)+\text { length contributed by } g(x)+\text { length contributed by } h(x) \\
&\Rightarrow L=2 \sqrt{2}+2 \pi+2 \sqrt{2} \\
&\Rightarrow L=4 \sqrt{2}+2 \pi
\end{aligned} L = length contributed by f ( x ) + length contributed by g ( x ) + length contributed by h ( x ) ⇒ L = 2 2 + 2 π + 2 2 ⇒ L = 4 2 + 2 π
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