Tutor profile: Saumya G.
If cos A + cos^2 A = 1 and a sin^12 A + b sin^10 A + c sin^8 A + d sin^6 A - 1 = 0. Find the value of a+b/c+d
We are given that cos A = 1 - cos^2 A Using the trigonometric identities, cos A = sin^2 A cos^2 A = sin^4 A 1 - sin^2 A = sin^4 A 1 = sin^4 A + sin^2 A Cubing both sides, 1 = sin^12 A + 3 * sin^10 A + 3 sin^8 A + sin^6 A Comparing the equations, a=1, b=3, c=3, d=1 Thus (a+b)/(c+d) = 1
Find the derivative of y = (x^x) for x>0.
The above function (x^x) is neither a power function like (x^k) nor an exponential function like (k^x) so the normal derivative formulas cannot be used. Taking log on both sides, ln y = ln ((x^x)) which is ln y = x ln x using the property (ln (a^b))= b ln a. Carrying out the differentiation, (1/y) * (dy/dx) = ln x + 1 Thus, dy/dx = (ln x + 1) * y Thus dy/dx = (ln x + 1)* (x^x) is the final answer.
What is the expansion of (a+b)^7?
For a binomial expansion, the first term ('a' in the above problem) starts with the highest power and goes on decreasing each time whereas the second term ('b' in the above problem) increases each time. Using Pascal's triangle phenomenon (used to derive the coefficients of expansion), we find the coefficients to be 1,7,21,35,35,21,7,1. Thus the solution to the above problem is: (a^7)+7*(a^6)*b+21*(a^5)*(b^2)+35*(a^4)*(b^3)+35*(a^3)*(b^4)+21*(a^2)*(b^5)+7*a*(b^6)+(b^7)