1.Let f(x)={x+1,0≤x≤12x2−6x+6,1<x≤2f(x)=\begin{cases} x+1, & 0 \le x \le 1 \\ 2x^2-6x+6, & 1 < x \le 2 \end{cases}f(x)={x+1,2x2−6x+6,0≤x≤11<x≤2 and g(t)=∫t−1tf(x) dxg(t)=\int_{t-1}^{t} f(x)\,dxg(t)=∫t−1tf(x)dx for t∈[1,2]t\in[1,2]t∈[1,2]. Which of the following hold(s) good?a.f(x)f(x)f(x) is continuous and differentiable in [0,2][0,2][0,2]b.g′(t)g'(t)g′(t) vanishes for t=32t=\frac{3}{2}t=23 and 222c.g(t)g(t)g(t) is maximum at t=32t=\frac{3}{2}t=23d.g(t)g(t)g(t) is minimum at t=1t=1t=1Login to continueOnly logged in users canattempt or see the solution.
