For x > 0 , 0 ≤ t ≤ 2 π , K > 2 3 + 2 , K being a fixed real number the minimum value of x 2 + x 2 K 2 − 2 { ( 1 + cos t ) x + x K ( 1 + s i n t ) } + 3 + 2 cos t + 2 sin t is

Given expansion = { x − ( 1 + cos t ) } 2 + { x K − ( 1 + sin t ) } 2

Start Free Mock LoginAlready registered? Login here

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaHard2 minQB

MathematicsHardsingle choice

For $x > 0, 0 \leq t \leq 2 π, K > \frac{3}{2} + 2$ , K being a fixed real number the minimum value of $x^{2} + \frac{K ^{2}}{x ^{2}} - 2 {(1 + cos t) x + \frac{K ( 1 + s i n t )}{x}} + 3 + 2 cos t + 2 sin t$ is

Options:

Answer:

Solution:

Given expansion $= {x - (1 + cos t)}^{2} + {\frac{K}{x} - (1 + sin t)}^{2}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...