The area bounded by the curve $5 {y} = 3 [x] - 2,$ $1 \leq y \leq 2$ $($ where ${.}$ and $[.]$ represent the fractional part and greatest integer function respectively $)$ is

\frac{5}{2}

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\frac{8}{3}

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3

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2

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Solution

The correct option is C $3$

Given curve
$5 {y} = 4 [x] - 2$
$\Rightarrow 5 y - 5 [y] = 3 [x] - 2$
$\Rightarrow 5 y - 5 = 3 [x] - 2 {1 \leq y < 2}$
$\Rightarrow [x] = \frac{5 y - 3}{3}$
$[x]$ always gives an integer value
at $y = \frac{6}{5}, [x] = 1 \Rightarrow x \in [1, 2)$
at $y = \frac{9}{5}, [x] = 2 \Rightarrow x \in [2, 3)$
Area $= \frac{6}{5} \times 1 + \frac{9}{5} \times 1 = \frac{15}{5} = 3$

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The area bounded by the curve 5{y}=3[x]−2, 1≤y≤2 (where {.} and [.] represent the fractional part and greatest integer function respectively) is

The correct option is C 3 Given curve 5{y}=4[x]−2 ⇒5y−5[y]=3[x]−2 ⇒5y−5=3[x]−2{1≤y<2} ⇒[x]=5y−33 [x] always gives an integer value at y=65,[x]=1⇒x∈[1,2) at y=95,[x]=2⇒x∈[2,3) Area =65×1+95×1=155=3

The area bounded by the curve $5 {y} = 3 [x] - 2,$ $1 \leq y \leq 2$ $($ where ${.}$ and $[.]$ represent the fractional part and greatest integer function respectively $)$ is