Question

The minimum value of (4tan2θ+9cot2θ) is

Answer 1

Answer:

Arithmetic Mean >= Geometric mean

=>( 4tan^2 theta + 9cot^2 theta )/2 >=

Sq root (4tan^2 theta × 9cot^2 theta ) = 6

{ since tan^2 theta × cot^2 theta =1

=> 4tan^2 theta + 9cot^2 theta >= 12

Hence,minimum value of 4tan^2 theta + 9cot^2 theta = 12 Ans .

hope this helps

Answer 2

Answer:

p(X)=p(-3)

p(X)=p(-3)therefore X=-3

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-2

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-29-3k-2

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-29-3k-27-3k

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-29-3k-27-3k-7=-3k

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-29-3k-27-3k-7=-3k7=3k

p(X)=p(-3)therefore X=-3x+3=0 ( you told me)let put the Value of X=-3X²+kx-2 = -3×-3+k×-3-29-3k-27-3k-7=-3k7=3kk=7/3