Let g(x) = f(x) + f(1 - x) and f''(x) > 0, x ∈ (0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan-1 (2α) + tan-1 (1/α) + tan-1 (α+1/α) is equal to

The portion of the line $ 4x + 5y = 20 $ in the first quadrant is trisected by the lines $ L_1 $ and $ L_2 $ passing through the origin. The tangent of an angle between the lines $ L_1 $ and $ L_2 $ is:

Let g(x) = f(x) + f(1 - x) and f''(x) > 0, x ∈ (0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan-1 (2α) + tan^-1 (\(\frac{1}{α}\)) + tan^-1\( (\frac{α+1}{α})\) is equal to

The Correct Option is B

Solution and Explanation

Top Questions on types of differential equations

Questions Asked in JEE Main exam

Let g(x) = f(x) + f(1 - x) and f''(x) > 0, x ∈ (0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan-1 (2α) + tan-1 (\(\frac{1}{α}\)) + tan-1\( (\frac{α+1}{α})\) is equal to

The Correct Option is B

Solution and Explanation

Top Questions on types of differential equations

Questions Asked in JEE Main exam

Let g(x) = f(x) + f(1 - x) and f''(x) > 0, x ∈ (0,1). If g is decreasing in the interval (0, α) and increasing in the interval (α, 1), then tan-1 (2α) + tan^-1 (\(\frac{1}{α}\)) + tan^-1\( (\frac{α+1}{α})\) is equal to