Question

Match the given statement in Column-I with one or more statement(s) in Column-II.

Answer 1

(A) $IA=rcosec\frac{A}{2}$,...

so, $\frac{S^2}{abc}.\frac{\left(r^3\right)}{sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}}=\frac{S^2{r}^{3}4R}{r.abc}$

$=\frac{S^2.r^2.4R}{abc}$

$=(Sr{)}^2.\frac{4R}{abc}$

$={\Delta }^3=1$

(B) $b=ar,c=a{r}^2,d=a{r}^3$ & $a-2+a{r}^2-9=2ar-14$

$\Rightarrow a+a{r}^2=2ar-3$ ...(i)

$ar-7+a{r}^3-5=2a{r}^2-18$

$\Rightarrow ar+a{r}^3=2a{r}^2-6$ ...(ii)

$\Rightarrow r(2ar-3)=2a{r}^2-6$

$\Rightarrow$ r = 2 & a = -3

(C) $\underset{x\to 0}{lim}\ell n\left[\frac{e^{ef\left(x\right)}[1-\frac{1}{e^{ef\left(x\right)}}]}{f\left(x\right)}\right]=1$

$\Rightarrow \underset{x\to 0}{lim}\ell n\left[\frac{e^{ef\left(x\right)}-1}{e^{f\left(x\right)}}\right]=1$

$\Rightarrow \underset{x\to 0}{lim}\left[\frac{e^{ef\left(x\right)}-1}{f\left(x\right)}\right]=e$ so, f(0) = 0

(D) Total page = 30, torn page 1 & 2

(E) $\underset{n\to 0}{lim}\left[\frac{(1+n{)}^{\frac{1}{n}}-e}{n}\right]$

$=\underset{n\to 0}{lim}\frac{e^{\frac{1}{n}\ell n(1+n)}-e}{n}=\underset{n\to 0}{lim}\frac{e[e^{\frac{\ell n(1+n)}{n}-1}-1]}{n}$

$=\underset{n\to 0}{lim}e\left[\frac{\ell n(1+n)-n}{n^2}\right]=e\left[\frac{n-\frac{n^2}{2}+....}{n^2}\right]$

$=-\frac{e}{2}$

so, L = 2