A=∫01xdx∫12[x]dx∫23{x}dx∫01sinπxdx∫12[sinπx]dx∫23{sinπx}dx∫0π/2(sin2x)dx∫0π(sin2x)dx∫0π/2a12dx(where [p] denotes greatest integer less than or equal to p and {p} denotes the fraction part of p).If value of A=0 then value of a is

|∫01x dx∫121dx∫01x dx∫01sinπxdx∫12(−1)dx∫23sin(πx)dx∫0π/2sin2x dx∫0πsin2x dx∫0π/2a12dx|=0{p}=p if p∈[0,1)Also ∫23sin(πx)dx=∫01sin(πx)dx [period=2]|121122π−12ππ4π2a24π|=0|1211−π21612a|=0⇒|0210−π216−a12a|=(6−a)(2+π2)=0a=6

A=∫01xdx∫12[x]dx∫23{x}dx∫01sinπxdx∫12[sinπx]dx∫23{sinπx}dx∫0π/2(sin2x)dx∫0π(sin2x)dx∫0π/2a12dx(where [p] denotes greatest integer less than or equal to p and {p} denotes the fraction part of p).If value of A=0 then value of a is

|∫01x dx∫121dx∫01x dx∫01sinπxdx∫12(−1)dx∫23sin(πx)dx∫0π/2sin2x dx∫0πsin2x dx∫0π/2a12dx|=0{p}=p if p∈[0,1)Also ∫23sin(πx)dx=∫01sin(πx)dx [period=2]|121122π−12ππ4π2a24π|=0|1211−π21612a|=0⇒|0210−π216−a12a|=(6−a)(2+π2)=0a=6

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$A = |\begin{matrix} \int_{0}^{1} x d x & \int_{1}^{2} [x] d x & \int_{2}^{3} {x} d x \\ \int_{0}^{1} s i n π x d x & \int_{1}^{2} [\sin π x] d x & \int_{2}^{3} {\sin π x} d x \\ \int_{0}^{π / 2} (s i n^{2} x) d x & \int_{0}^{π} (s i n^{2} x) d x & \int_{0}^{π / 2} \frac{a}{12} d x \end{matrix}|$
(where [p] denotes greatest integer less than or equal to p and {p} denotes the fraction part of p).If value of A=0 then value of a is

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Detailed Solution

$\begin{array}{l} | \begin{matrix} \int_{0}^{1} x d x & \int_{1}^{2} 1 d x & \int_{0}^{1} x d x \\ \int_{0}^{1} \sin π x d x & \int_{1}^{2} (- 1) d x & \int_{2}^{3} \sin (π x) d x \\ \int_{0}^{π / 2} \sin^{2} x d x & \int_{0}^{π} \sin^{2} x d x & \int_{0}^{π / 2} \frac{a}{12} d x \end{matrix} | = 0 \\ {p} = p i f p \in [0, 1) \\ A l s o \int_{2}^{3} \sin (π x) d x = \int_{0}^{1} \sin (π x) d x [p e r i o d = 2] \\ | \begin{matrix} \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{2}{π} & - 1 & \frac{2}{π} \\ \frac{π}{4} & \frac{π}{2} & \frac{a}{24} π \end{matrix} | = 0 \\ | \begin{matrix} 1 & 2 & 1 \\ 1 & - \frac{π}{2} & 1 \\ 6 & 12 & a \end{matrix} | = 0 \Rightarrow | \begin{matrix} 0 & 2 & 1 \\ 0 & - \frac{π}{2} & 1 \\ 6 - a & 12 & a \end{matrix} | \\ = (6 - a) (2 + \frac{π}{2}) = 0 \\ a = 6 \end{array}$