write essay on 'education is the best policy'
Answers
Answer:
f(x)=sin[π
2
]x+sin[−π
2
]x
We know
Greatest Integer function reduces any real number to its nearest lowest Integer.
\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = [x]\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf [0,1) & \sf 0 \\ \\ \sf [1,2) & \sf 1 \\ \\ \sf [2,3) & \sf 2\\ \\ \sf [ - 1,0) & \sf - 1\\ \\ \sf [ - 2, - 1) & \sf - 2 \end{array}} \\ \end{gathered} \\ \end{gathered}
x
[0,1)
[1,2)
[2,3)
[−1,0)
[−2,−1)
y=[x]
0
1
2
−1
−2
Now, We know
\begin{gathered}\rm \: \pi = 3.14 \\ \end{gathered}
π=3.14
\begin{gathered}\rm\implies \: {\pi}^{2} = 9.8596 \\ \end{gathered}
⟹π
2
=9.8596
So,
\begin{gathered}\rm\implies \:[{\pi}^{2}] = [9.8596] = 9 \\ \end{gathered}
⟹[π
2
]=[9.8596]=9
and
\begin{gathered}\rm\implies \:[ - {\pi}^{2}] = [ - 9.8596] = - 10\\ \end{gathered}
⟹[−π
2
]=[−9.8596]=−10
Thus, given function
\begin{gathered}\rm \: f(x) = sin[ {\pi}^{2}]x + sin[ - {\pi}^{2}]x \\ \end{gathered}
f(x)=sin[π
2
]x+sin[−π
2
]x
can be rewritten as
\begin{gathered}\rm \: f(x) = sin9x + sin( - 10x) \\ \end{gathered}
f(x)=sin9x+sin(−10x)
\begin{gathered}\rm \: f(x) = sin9x - sin10x \\ \end{gathered}
f(x)=sin9x−sin10x
So,
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \dfrac{9\pi}{2} \bigg) - sin\bigg( \dfrac{10\pi}{2} \bigg) \\ \end{gathered}
f(
2
π
)=sin(
2
9π
)−sin(
2
10π
)
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \pi + \dfrac{\pi}{2} \bigg) - sin5\pi \\ \end{gathered}
f(
2
π
)=sin(π+
2
π
)−sin5π
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \dfrac{\pi}{2} \bigg) - 0 \\ \end{gathered}
f(
2
π
)=sin(
2
π
)−0
\begin{gathered}\rm\implies \:\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = 1 \\ \end{gathered}
⟹f(
2
π
)=1
So, option [B] is correct.
Now, Consider
\begin{gathered}\rm \: f(\pi) = sin9\pi - sin10\pi \\ \end{gathered}
f(π)=sin9π−sin10π
\begin{gathered}\rm \: f(\pi) = 0 - 0 \\ \end{gathered}
f(π)=0−0
\begin{gathered}\rm\implies \:\rm \: f(\pi) = 0 \\ \end{gathered}
⟹f(π)=0
It implies, option [C] is not correct.
Now, Consider
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg( \dfrac{9\pi}{4} \bigg) - sin\bigg( \dfrac{10\pi}{4} \bigg) \\ \end{gathered}
f(
4
π
)=sin(
4
9π
)−sin(
4
10π
)
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg(2\pi + \dfrac{\pi}{4} \bigg) - sin\bigg( \dfrac{5\pi}{2} \bigg) \\ \end{gathered}
f(
4
π
)=sin(2π+
4
π
)−sin(
2
5π
)
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg(\dfrac{\pi}{4} \bigg) - sin\bigg(2\pi + \dfrac{\pi}{2} \bigg) \\ \end{gathered}
f(
4
π
)=sin(
4
π
)−sin(2π+
2
π
)
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } - sin\bigg( \dfrac{\pi}{2} \bigg) \\ \end{gathered}
f(
4
π
)=
2
1
−sin(
2
π
)
\begin{gathered}\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } - 1\\ \end{gathered}
f(
4
π
)=
2
1
−1
It implies, option [D] is not correct.
Hence, from above we concluded that
\begin{gathered}\rm\implies \: \:\boxed{\tt{ \: \rm \: f\bigg( \dfrac{\pi}{2} \bigg) = 1 \: \: }} \\ \end{gathered}
⟹
f(
2
π
)=1
It implies, option [B] is correct
Answer:
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