Question

Let $f\left(x\right)=x^{\alpha }+3x^{\beta }$ with $\alpha >1,\beta >1$. If the value of $a$ for which the area of the region bounded by $y=f\left(x\right)$ and straight lines $x=0,x=1\text{and}y=f\left(a\right)$ is minimum is $\lambda$, then value of $3.5\lambda$ is

Answer 1

Using the theorem
If $y=f\left(x\right)$ is monotonic in $[a,b]$, then area bounded by $y=f\left(x\right),x=a,x=b\&y=k$ is minimum if $k=\frac{a+b}{2}f\left(x\right)=x^{\alpha }+3x^{\beta }$
$f^{\prime }\left(x\right)=\alpha x^{\alpha -1}+3\beta x^{\beta -1}\Rightarrow f^{\prime }\left(x\right)>0\text{for all}x\in [0,1]$
So, $f\left(x\right)$ is increasing and concave upwards
So, $\lambda =\frac{0+1}{2}=\frac{1}{2}$
So, $3.5\lambda =1.75$