MCQ
$\int\limits_{0}^{\frac{\pi }{2}}{\frac{{{\sin }^{2}}nx}{\sin x}\,\,dx=........\forall n\in N}$
- A$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
- ✓$1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}$
- C$1$
- D$0$
$n = 1$ લેતાં, $\int_{0}^{\frac{\pi}{2}} \frac{sin^2nx}{sin x} = \int_{0}^{\frac{\pi}{2}} sin x dx = {[-cosx]_0^{\frac{\pi}{2}}} = 1$
$n=2$ લેતા $\int_{0}^{\frac{\pi}{2}} \frac{sin^2x}{sin x} = \int_{0}^{\frac{\pi}{2}} \frac{4 sin^2 x cos^2 x}{sin x} dx $
$= - 4 \int_{0}^{\frac{\pi}{2}} cos^2x(-sin x)dx = - 4 \left[\frac{cos^3x}{3}\right]_0^{\frac{\pi}{2}} = -4 \left[0 - \frac{1}{3}\right] = \frac{4}{3} = 1 + \frac{1}{3}$
$n =1$ અને $2$ માટે $(B)$ સત્ય છે.
$\therefore \ \forall \ n \ \in \ N$ ધારો કે $\int_{0}^{\frac{\pi}{2}} \frac{sin^2 nx}{sin x} dx = 1 + \frac{1}{3} + \frac{1}{5} + ..... + \frac{1}{2n - 1}$
ધારો કે $p(k)$ સત્ય છે. $K \ \in N$
$\therefore p(k) : \int_{0}^{\frac{\pi}{2}} \frac{sin^2kx}{sin x} dx = 1 + \frac{1}{3} + \frac{1}{5} + .... + \frac{1}{2k - 1}$
અહી, $\int_{0}^{\frac{\pi}{2}} \frac{sin^2 (k + 1)x}{sin x} dx - \int_{0}^{\frac{\pi}{2}} \frac{sin^2k x}{sin x} dx = \int_{0}^{\frac{\pi}{2}} \frac{sin^2(k + 1)x - sin^2 kx}{sin x} dx $
$= \int_{0}^{\frac{\pi}{2}} \frac{sin (2k + 1) x sinx}{sin x} dx$
$= \int_{0}^{\frac{\pi}{2}} sin (2k + 1)x dx = \left[\frac{-cos (2k + 1)x}{(2k + 1)}\right]_0^{\frac{\pi}{2}} = \frac{-1}{2k + 1} \left[cos (2k + 1)\frac{\pi}{2} - cos 0\right] = \frac{1}{2k + 1} $
$\therefore \int_{0}^{\frac{\pi}{2}} \frac{sin^2 (k + 1)x}{sin x} dx = \int_{0}^{\frac{\pi}{2}} \frac{sin^2 k x}{sin x} dx + \frac{1}{2k + 1}$
$= 1 + \frac{1}{3} + \frac{1}{5} + .... + \frac{1}{2k - 1} + \frac{1}{2k + 1}$
$\therefore \ p(k + 1)$ સત્ય છે.
$\therefore \ p(n)$ દરેક $n \ \in N$ માટે સત્ય છે.
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વિધાન $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
વિધાન $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$