Question

Let $a=min[x^2+2x],x\in R$ and $b=\underset{x\to 0}{lim}\frac{sinx.cosx}{e^x+e^{-x}}$. Then value of $a+b$ is

• A. $-1$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$-1$

Since, $a=min[x^2+2x]$. Hence for minima $\frac{da}{dx}=0$.

So,

$\frac{da}{dx}=2x+2=0⟹x=-1$.

At $x=-1$;

$a=(-1{)}^2+2(-1)=-1$

$b=\underset{x\to 0}{lim}\frac{sinx.cosx}{e^x+e^{-x}}=\frac{0\times 1}{1+1}=0$

So, $a+b=-1+0=-1$