Mathematics - Continuity - Differentiability Question with Solution | TestHub

For $f(x)$ to be differentiable at non-integer points, $g(x)$ must be differentiable, which is true for all given options. The critical points are integers.

For $f(x)$ to be continuous at an integer $k$, we need ${\lim }_{x\to k^{-}}f(x)={\lim }_{x\to k^{+}}f(x)=f(k)$. This implies $(k-1)g(k)=kg(k)=kg(k)$, so $(k-1)g(k)=kg(k)$, which means $g(k)=0$ for all integers $k$. For $f(x)$ to be differentiable at an integer $k$, we need LHD = RHD. LHD at $x=k$: ${\lim }_{h\to 0^{-}}\frac{[k+h]g(k+h)-[k]g(k)}{h}={\lim }_{h\to 0^{-}}\frac{(k-1)g(k+h)-kg(k)}{h}$. Since $g(k)=0$, this becomes ${\lim }_{h\to 0^{-}}\frac{(k-1)g(k+h)}{h}=(k-1)g^{\prime }(k)$. RHD at $x=k$: ${\lim }_{h\to 0^{+}}\frac{[k+h]g(k+h)-[k]g(k)}{h}={\lim }_{h\to 0^{+}}\frac{kg(k+h)-kg(k)}{h}$. Since $g(k)=0$, this becomes ${\lim }_{h\to 0^{+}}\frac{kg(k+h)}{h}=k{g}^{\prime }(k)$. For differentiability, $(k-1)g^{\prime }(k)=k{g}^{\prime }(k)$, which implies $g^{\prime }(k)=0$ for all integers $k$. Let's check the options for $g(k)=0$ and $g^{\prime }(k)=0$ for all integers $k$: A) $g(x)=\sin (\pi x)$. $g(k)=\sin (\pi k)=0$. $g^{\prime }(x)=\pi \cos (\pi x)$, so $g^{\prime }(k)=\pi \cos (\pi k)=\pi (-1{)}^k\ne 0$. (Fails) B) $g(x)=\cos (\pi x)$. $g(k)=\cos (\pi k)=(-1{)}^k\ne 0$. (Fails continuity) C) $g(x)=(\sin (\pi x){)}^2$. $g(k)=(\sin (\pi k){)}^2=0$. $g^{\prime }(x)=2\sin (\pi x)\cos (\pi x)\pi =\pi \sin (2\pi x)$, so $g^{\prime }(k)=\pi \sin (2\pi k)=0$. (Works) D) $g(x)=x^2-[x{]}^2$. $g(k)=k^2-k^2=0$. $g^{\prime }(x)=2x$ for $x\notin \mathbb{Z}$. So $g^{\prime }(k)=2k$. This implies $2k=0$, so $k=0$. This holds only for $k=0$, not for all integers. (Fails) Thus, only $g(x)=(\sin (\pi x){)}^2$ satisfies the conditions.