Let f : R → R be a continuous function. Then lim _(x → π/4) (π/4∫₂^(sec ²x) f(x) dx)/(x² - (π ²)/16) is equal to :

Solution

Maths sir · 4 steps

Top and bottom both vanish at π/4, so differentiate each, with the chain rule on the upper limit.

1. Check the form. At x = π/4, sec² x = 2, so the integral runs from 2 to 2 and equals 0. The denominator x² - π²/16 is also 0, so this is a 0/0 form and L'Hopital's rule applies. The variable inside the integral is only a dummy, so read it as ∫₂^(sec² x) f(t) dt.
2. Differentiate the numerator. By the Leibniz rule, the derivative of ∫₂ᵍ⁽ˣ⁾ f(t) dt is f(g(x)) g'(x). Here g(x) = sec² x and g'(x) = 2 sec² x tan x.

   lim(x → π/4) (π/4 f(sec² x)· 2 sec² x tan x)/2x

3. Put in the values. At x = π/4, sec² x = 2 and tan x = 1. The numerator is π/4 × f(2) × 2 × 2 × 1 = π f(2). The denominator is 2 × π/4 = π/2.
4. So. The limit is (π f(2))/(π/2) = 2f(2), which is option (b).

The idea to keep. An integral with a variable limit inside a 0/0 limit calls for L'Hopital plus Leibniz: f at the upper limit, times the derivative of that limit.

Answer from the NTA final answer key. Solution by PrepWiser.