TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the function
f(x)=x3+2x2−x
The function is continuous and differentiable at [-1,2] and(−1,2).
Therefore, by Lagrange's mean volume theorem, there exist c∈(−1,2) such that,
f′(c)=

f(2)−f(1)

2−(−1)

3c2+4c−1=

14−2

3

3c2+4c−1=4
3c2+4c−5=0
Solve further,
c=

−4±√16+60

6

c=

−4±√76

6

since,
8<√76<9
Therefore,
c=

−4+√76

6

=

−2+√19

3