Question

Let $f\left(x\right)=\frac{\left(\log \right(1+x)-\log 2)({3.4}^{x-1}-3x)}{\left\{\right(7+x{)}^{1/3}-(1+3x{)}^{1/2}\}\sin \pi x},x\ne 1$
The value of f(1) so that f is continuous at $x=1$ is

• A. an algebraic number
• B. a rational number
• C. a transcendental number
• D. $-\frac{9}{\pi }\log \frac{e}{4}$

Answer 1

Answer: (C), (D)

The correct options are
C a transcendental number
D $-\frac{9}{\pi }\log \frac{e}{4}$

$\underset{x\to 1}{lim}\frac{\left(log\right(1+x)-log2)({3.4}^{x-1}-3x)}{\left\{\right(7+x{)}^{1/3}-(1+3x{)}^{1/2}\}\sin \pi x}$
$=\underset{y\to 0}{lim}\frac{\left(\log \right(2+y)-log2)({3.4}^y-3y-3)}{\left\{\right(8+y{)}^{1/3}-(1+3y+3{)}^{1/2}\}\sin \pi (y+1)}$ ...Using $[y=x-1]$
$=-\underset{y\to 0}{lim}\frac{\log (1+\frac{y}{2}).\left(3\right(4^y-1)-3y)}{2\{{(1+\frac{y}{8})}^{1/3}-{(1+\frac{3}{4}y)}^{1/2}\}\sin \pi y}$ ...$\because sin(\pi +\theta )=-sin\theta$
$=-\frac{1}{\pi }\underset{y\to 0}{lim}\frac{\log (1+y/2)}{y}.\{\frac{3(4^y-1)}{y}-3\}\times \frac{\pi y}{\sin \pi y}\frac{y}{[1+\frac{y}{24}-1-\frac{3}{8}y+O(y^2\left)\right]}$
$=-\frac{1}{\pi }[3\log 4-3](-3)=\frac{9}{\pi }\log \frac{4}{e}$
which is transcendental number