Let [t] denote the greatest integer less than or equal to t

Question

Let $[t]$ denote the greatest integer less than or equal to $t$

Accepted Answer

It is given that $f(x)$ is continuous at $x=0$

This mean at $x=0$

L.H.L = R.H.L = f(0) = a

Calculating L.H.L using the definition of $f(x)$ for $x$\underset{x\to {0}\^{-}}{lim}f(x)=\underset{x\to 0}{lim}{b}\^{2}\sin (\frac{\pi }{2}[\frac{\pi }{2}(\cos x+\sin x)\cos x])=a$

$\Rightarrow$ $\underset{x\to 0}{lim}{b}\^{2}\sin (\frac{\pi }{2}[\frac{\pi }{2}({\cos }\^{2}x+\frac{\sin 2x}{2})])=a$

$$$\begin{gathered} {\cos }^{2}({0}^{-})={1}^{-}1 \ \sin 2({0}^{-})={0}^{-}0 \ \Rightarrow a={b}^{2}\sin (\frac{\pi }{2}[\frac{3.14}{2}({1}^{-}+{0}^{-})]) \end{gathered}$$$

$\Rightarrow$ $a={b}\^{2}\sin (\frac{\pi }{2}[1.57({1}\^{-})])$

$\Rightarrow$ $a={b}\^{2}\sin (\frac{\pi }{2}[{1.57}\^{-}]),[{1.57}\^{-}]=1{[t]\to$ G.I.F $}$

$\Rightarrow$ $a={b}\^{2}\sin (\frac{\pi }{2}\times 1)={b}\^{2}$

Calculating R.H.L using the function definition for $x>0$

$\underset{x\to {0}\^{+}}{lim}f(x)=\underset{x\to {0}\^{+}}{lim}\frac{\sin x-\frac{1}{2}\sin 2x}{{x}\^{3}}=a$

$\Rightarrow$ $\underset{x\to {0}\^{+}}{lim}\frac{\sin x-\frac{1}{2}\times 2\sin x\cos x}{{x}\^{3}}=a$

$\Rightarrow$ $\underset{x\to {0}\^{+}}{lim}\frac{\sin x(1-\cos x)}{x\cdot {x}\^{2}}=a$

Using limits result $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$ and $\underset{x\to 0}{lim}\frac{1-\cos x}{{x}\^{2}}=\frac{1}{2}$

Putting these result

Using limits result $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$ and $\underset{x\to 0}{lim}\frac{1-\cos x}{{x}\^{2}}=\frac{1}{2}$

Putting these result

$a=1\cdot \frac{1}{2}=\frac{1}{2}$

So ${a}\^{2}+{b}\^{2}={a}\^{2}+a{{b}\^{2}=a$ from $L.H.L}$

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Solutions · 1