Solve this following
Question:

Mark $(\sqrt{ })$ against the correct answer in each of the following:

$\int \frac{d x}{\sqrt{x^{2}+6 x+5}}=?$

A. $\log \left|x+\sqrt{x^{2}+6 x+5}\right|+C$

B. $\log \left|x-\sqrt{x^{2}+6 x+5}\right|+C$

C. $\log \left|(x+3)+\sqrt{x^{2}+6 x+5}\right|+C$

D. none of these

Solution:

Solve this following
Question:

Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c)$,

$(2, b)$ and $(a, b)$ be $\left(\frac{10}{3}, \frac{7}{3}\right)$. If $\alpha, \beta$ are the roots

of the equation $a x^{2}+b x+1=0$, then the value of $\alpha^{2}+\beta^{2}-\alpha \beta$ is :

1. $\frac{71}{256}$

2. $\frac{69}{256}$

3. $-\frac{69}{256}$

4. $-\frac{71}{256}$

Correct Option: , 4

Solution:

$\frac{a+2+a}{3}=\frac{10}{3}$

$a=4$

and $\frac{\mathrm{c}+\mathrm{b}+\mathrm{b}}{3}=\frac{7}{3}$

$c+2 b=7$

also $2 b=a+c$

$2 b-a+2 b=7$

$\mathrm{b}=\frac{11}{4}$

now $4 x^{2}+\frac{11}{4} x+1=0<\beta$

$\alpha^{2}+\beta^{2}-\alpha \beta=(\alpha+\beta)^{2}-3 \alpha \beta$

$=\left(\frac{-11}{16}\right)^{2}-3\left(\frac{1}{4}\right)$

$=\frac{121}{256}-\frac{3}{4}=\frac{-71}{256}$