We have to find the value of λ given in the real function and we are given with the differentiability of the function f(x) = λx² + 7x - 4 at x = 5 which is f ‘(5) = 97, so we will adopt the same process but with a little variation.

So by using the formula, f ‘(c) = , we get,

f ‘(5) =

f ‘(5) =

f ‘(5) =

f ‘(5) =

as the limit has some finite value, then there must be the formation of some indeterminate form like , so if we put the limit value, then the numerator will also be zero as the denominator, but there must be a factor (x - 5) in the numerator, so that this form disappears.

f ‘(5) =

f ‘(5) = lim_x→5 λx + 5λ + 7 = 97

f ‘(5) = 10λ + 7 = 97

10λ = 90

λ = 9