Question

If f(x) = sin[x²]x + sin[-π²]x where [] denotes the greatest integer less than or equal to x then?

[A] none of these
[B] f(π/2) = 1
[C] f(π) = 2
[D] f(π/4) = -1

Answer 1

Appropriate Question :-

If f(x) = sin[π²]x + sin[-π²]x where [] denotes the greatest integer less than or equal to x then

[A] none of these

[B] f(π/2) = 1

[C] f(π) = 2

[D] f(π/4) = -1

$\large\underline{\sf{Solution-}}$

Given that

$\rm \: f(x) = sin[ {\pi}^{2}]x + sin[ - {\pi}^{2}]x$

We know

Greatest Integer function reduces any real number to its nearest lowest Integer.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = [x]\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf [0,1) & \sf 0 \\ \\ \sf [1,2) & \sf 1 \\ \\ \sf [2,3) & \sf 2\\ \\ \sf [ - 1,0) & \sf - 1\\ \\ \sf [ - 2, - 1) & \sf - 2 \end{array}} \\ \end{gathered}$

Now, We know

$\rm \: \pi = 3.14$

$\rm\implies \: {\pi}^{2} = 9.8596$

So,

$\rm\implies \:[{\pi}^{2}] = [9.8596] = 9$

and

$\rm\implies \:[ - {\pi}^{2}] = [ - 9.8596] = - 10$

Thus, given function

$\rm \: f(x) = sin[ {\pi}^{2}]x + sin[ - {\pi}^{2}]x$

can be rewritten as

$\rm \: f(x) = sin9x + sin( - 10x)$

$\rm \: f(x) = sin9x - sin10x$

So,

$\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \dfrac{9\pi}{2} \bigg) - sin\bigg( \dfrac{10\pi}{2} \bigg)$

$\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \pi + \dfrac{\pi}{2} \bigg) - sin5\pi$

$\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = sin\bigg( \dfrac{\pi}{2} \bigg) - 0$

$\rm\implies \:\rm \: f\bigg( \dfrac{\pi}{2} \bigg) = 1$

So, option [B] is correct.

Now, Consider

$\rm \: f(\pi) = sin9\pi - sin10\pi$

$\rm \: f(\pi) = 0 - 0$

$\rm\implies \:\rm \: f(\pi) = 0$

It implies, option [C] is not correct.

Now, Consider

$\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg( \dfrac{9\pi}{4} \bigg) - sin\bigg( \dfrac{10\pi}{4} \bigg)$

$\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg(2\pi + \dfrac{\pi}{4} \bigg) - sin\bigg( \dfrac{5\pi}{2} \bigg)$

$\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = sin\bigg(\dfrac{\pi}{4} \bigg) - sin\bigg(2\pi + \dfrac{\pi}{2} \bigg)$

$\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } - sin\bigg( \dfrac{\pi}{2} \bigg)$

$\rm \: f\bigg( \dfrac{\pi}{4} \bigg) = \dfrac{1}{ \sqrt{2} } - 1$

It implies, option [D] is not correct.

Hence, from above we concluded that

$\rm\implies \: \:\boxed{\tt{ \: \rm \: f\bigg( \dfrac{\pi}{2} \bigg) = 1 \: \: }}$

It implies, option [B] is correct.