Let f(x) = ₀^ x^2 t^2 - 5t + k 2+ t dt , where k is a real parameter. The number of integral values of k for which f(x) has exactly 5 points of local extrema is :

Step-by-step solution

By Leibnitz's rule, the derivative of f(x) is given by: f'(x) = (x^2)^2 - 5(x^2) + k 2+ (x^2) d dx (x^2) - 0 f'(x) = 2x(x^4 - 5x^2 + k) 2+ (x^2) For f(x) to have exactly 5 points of local extrema, f'(x) must change sign at exactly 5 distinct real values of x . Clearly, x = 0 is one such point (since the denominator is always positive and 2x changes sign at x=0 ). The remaining 4 points must come from the roots of the biquadratic equation: x^4 - 5x^2 + k = 0 Let x^2 = y . The quadratic equation y^2 - 5y + k = 0 must