Mathematics - Continuity - Differentiability Question with Solution | TestHub

For $f(x)$ to be differentiable everywhere, it must be continuous and differentiable at $x=0$ and $x=1$. At $x=0$: Continuity: ${\lim }_{x\to 0^{+}}\frac{e^{x^2}-\cos (2x)}{x^2}$. Using Taylor series expansions for $e^{x^2}=1+x^2+\frac{x^4}{2!}+\dots$ and $\cos (2x)=1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\cdots =1-2x^2+\frac{2x^4}{3}-\dots$.${\lim }_{x\to 0^{+}}\frac{(1+x^2+x^4/2)-(1-2x^2+2x^4/3)}{x^2}={\lim }_{x\to 0^{+}}\frac{3x^2-x^4/6}{x^2}=3$. Thus, $f(0)=B=3$. Differentiability: $f^{\prime }(x)=A$ for $x<0$. The left derivative at $x=0$ is $A$. For $0<x\le 1$, let $g(x)=\frac{e^{x^2}-\cos (2x)}{x^2}=3-\frac{1}{6}{x}^2+O(x^4)$. Then $g^{\prime }(x)=-\frac{1}{3}x+O(x^3)$.${\lim }_{x\to 0^{+}}{g}^{\prime }(x)=0$. Thus, $A=0$. At $x=1$: Continuity: ${\lim }_{x\to 1^{-}}\frac{e^{x^2}-\cos (2x)}{x^2}=e-\cos (2)$. Also, $f(1)=C\ln (1)+D(1)=D$. So, $D=e-\cos (2)$. Differentiability: For $0<x\le 1$, $g^{\prime }(x)=\frac{x^2(2x{e}^{x^2}+2\sin (2x))-2x(e^{x^2}-\cos (2x))}{x^4}=\frac{x(2x{e}^{x^2}+2\sin (2x))-2(e^{x^2}-\cos (2x))}{x^3}$. The left derivative at $x=1$ is $g^{\prime }(1)=\frac{1(2e+2\sin (2))-2(e-\cos (2))}{1}=2e+2\sin (2)-2e+2\cos (2)=2\sin (2)+2\cos (2)$. For $x>1$, $f^{\prime }(x)=C/x+D$. The right derivative at $x=1$ is $C+D$. So, $C+D=2\sin (2)+2\cos (2)$. Substitute $D=e-\cos (2)$: $C+(e-\cos (2))=2\sin (2)+2\cos (2)⟹C=2\sin (2)+3\cos (2)-e$. Finally, $A+B+C+D=0+3+(2\sin (2)+3\cos (2)-e)+(e-\cos (2))=3+2\sin (2)+2\cos (2)$. Using $\sin (2)\approx 0.909297$ and $\cos (2)\approx -0.416147$: $A+B+C+D\approx 3+2(0.909297)+2(-0.416147)\approx 3+1.818594-0.832294\approx 3+0.9863\approx 3.9863$. Rounding to two decimal places, the value is $3.99$.