Kindly tell que.2.....least integral value...........integer type questions

hello tanisha this is a bit tricky ques, but thankfully easy to understand.

let x= tantheta which implies that the range of x is (0, infinity).

so expression at hand becomes

3(x^2+1/x^2) – 8(x+1/x) + 10

now let y= x+1/x, and since x is positive, range of y is [2, infinity). [note: y has to be greater than or equal to 2 because of AM- GM inequality we can write (x+1/x)/2 greater than equal to root(x.1/x)= root(1)=1 hence (x+1/x) greater than equal to 2].

so, y^2= x^2+1/x^2 + 2 or y^2 – 2= x^2+1/x^2

so the expression becomes

3(y^2 – 2) – 8y + 10 or 3y^2 – 8y + 4= f(y) (say) which is a quadratic in y :)

now f’(y)= 6y – 8= 6(y – 4/3) which is obviously greater than zero since y is greater than 2.

so f is an increasing fn in [2, infinity).

hence f(2) is the least value in this interval.

but f(2)= 3*4 – 8*2+4= 0. surprisingly, this value also happens to be an integer.

so, zero is the least integral value.