Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tanx(cosx – y). If the curve passes through the point (π/4, 0) then the value of
\(\int_{0}^{\frac{\pi}{2}} y \,dx\)
is equal to :
Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tanx(cosx – y). If the curve passes through the point (π/4, 0) then the value of
\(\int_{0}^{\frac{\pi}{2}} y \,dx\)
is equal to :
\((2-\sqrt2)+\frac{π}{\sqrt2}\)
\(2-\frac{π}{\sqrt2}\)
\((2+\sqrt2)+\frac{π}{\sqrt2}\)
\(2+\frac{π}{\sqrt2}\)
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(2-\frac{π}{\sqrt2}\)
\(\frac{dy}{dx}=2tanx(cosx−y)\)
\(⇒\frac{dy}{dx}+2tanxy=2sinx\)
\(I.F=e^{∫2tanxdx}=sec^2x\)
∴ Solution of D.E. will be
\(y(x)sec^2x=∫2sinxsec^2xdx\)
\(ysec^2x=2secx+c\)
∵ Curve passes through
\((\frac{π}{4},0)\)
\(∴c=−2\sqrt2\)
\(∴y=2cosx−2\sqrt2cos^2x\)
\(\therefore \int_{0}^{\frac{\pi}{2}} y \,dx = \int_{0}^{\frac{\pi}{2}} (2\cos x - 2\sqrt{2}\cos^2 x) \,dx\)
\(=2−2\sqrt2⋅\frac{π}{4}\)
\(=2−\frac{π}{\sqrt2}\)
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Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
- Given a function y = f(x), the equation of the tangent for this curve at x = x0
- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
