11Q1)
When x-->0, then {x} also -->0
So we can take the limit as
\lim_{\left\{x \right\}\rightarrow 0}\frac{1}{\frac{tan\left\{x \right\}}{\left\{x \right\}}}
which turns out to be 1...
so i think (a) is right
10 Answers
11
106@ani. as x--> 0- {x} --> 1 isnt it? so limit shudnt exist naa?.. btw ans IS given 1 :(
11i think it was supposed to be
\lim_{n\rightarrow \infty}tan\alpha + \frac{tan \frac{\alpha }{2}}{2}+\frac{tan \frac{\alpha }{2^2}}{2^2}+.......+ \frac{tan \frac{\alpha }{2^n}}{2^n}
11Q4
Observe that it is an even function so it will be symmetric about x=0 so
\frac{\partial y}{\partial x}=\frac{2x(1-ln(x^{2}+1)}{ln(x^{2}+1)^{2}}
At x=0 the function is not defined
i suspect that the min value will be when
(x^{2}+1)^{2}=e
we get
x=\sqrt{\sqrt{e}-1}
12.
Use
-2-n Cot[2-n a] + 2-n Tan[2-n a]=-2(1 - n) Cot[2(1 - n) a]
So the sum is -2 Cot[2 a] + 2^{-n} Cot[2^{-n} a]
ans=1/a - 2 Cot[2 a]