If the area of the region {(x,y):a/x²≤ y≤ 1/x,1≤ x≤ 2,0<a<1} is (logₑ 2)-1/7 then the value of 7a-3 is equal to :

Solution

Maths sir · 4 steps

Integrate the upper curve minus the lower curve from 1 to 2, then match the result to fix a.

1. Which curve on top. For 1 ≤ x ≤ 2 and 0 < a < 1, we have a < x. Dividing by x² gives a/x² < 1/x, so y = 1/x is the upper curve throughout.

   Area = ∫₁²(1/x - a/x²)dx = [ln x]₁² + a[1/x]₁²

2. Evaluate. The first part is ln 2 - ln 1 = ln 2. The second is a(1/2 - 1) = -a/2. So the area is ln 2 - a/2.
3. Match the given area. Setting ln 2 - a/2 = ln 2 - 1/7 gives a/2 = 1/7, so a = 2/7. This lies in (0, 1), as required.
4. So. 7a - 3 = 2 - 3 = -1, which is option (d).

The idea to keep. Area between two curves is the integral of (upper minus lower). Always check which curve is on top over the whole interval before you integrate.

Answer from the NTA final answer key. Solution by PrepWiser.