# Tutor profile: Merin T.

## Questions

### Subject: Pre-Calculus

Find limit of 1/abs(x).

Let us split the function, f(x)=1/x, if x>0 =-1/x, if x<0 limx->infinity f(x)=limx->infinity 1/x= infinity limx->-infinity f(x)=limx->-infinity -1/x= infinity So, limit of 1/abs(x)=infinity.

### Subject: Trigonometry

What is the height of the tower, when the shadow of it increases by 14 meters when the angle of elevation of the sun rays decreases from 85° to 55° , what is the height of the building?

let the base of the first right triangle be x, and its base angle is 85° let the base of the second right triangle be x+14 , and label its base angle as 55° For both let the height be taken as h. We get tan85 = h/x ---> h = xtan85 from the second: tan55 = h/(x+14) ---> h = (x+14)tan55 then xtan85 = (x+14)tan55 xtan85 = xtan55 + 14tan55 xtan85 - xtan55 = 14tan55 x(tan85-tan55) = 14tan55 x = 14tan55/(tan85-tan55) back in h = xtan85 h = 14tan55tan85/(tan85-tan55) .

### Subject: Calculus

If the product of two positive numbers is constant, find them when its sum is minimum.

Let us assume both the numbers as x and y. As per the question, ab=k, where k is a constant. So y=k/x. We need to find, for which values does x+y become minimum. So f(x)=x+k/x Differentiate both sides, f'(x)=1-k/x^2 Set it equal to zero and solve for k. 1-k/x^2=0 => 1=k/x^2 =>x^2=k => x =sqrt(k), as we only need positive number. Find the function's second derivative, f"(x)=2k/x^3 Plug x=sqrt(k), so f"(sqrt(k))=2k/sqrt(k)^3>0 So, f(x) will have a minimum when x=sqrt(k) and y=sqrt(k).