সৃজনশীলঃ ‘f(x)=sin⁡xg(x,y)=9×2+25y2−225h(x)=x−3…’ দিনাজপুর বোর্ড ২০২৩ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\sin x$

$\therefore f\left(3x\right)=\sin 3x$

$\therefore {\int }_0^{\frac{\pi }{2}}\left\{f\right(x\left){\}}^{2}f\right(3x)dx={\int }_0^{\frac{\pi }{2}}{\sin }^{2}x\sin 3xdx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2{\sin }^{2}x\cdot \sin 3xdx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(1-\cos 2x)\sin 3xdx$

$=\frac{1}{2}[{\int }_0^{\frac{\pi }{2}}\sin 3xdx-{\int }_0^{\frac{\pi }{2}}\sin 3x\cos 2xdx]$

$=\frac{1}{2}[{\int }_0^{\frac{\pi }{2}}\sin 3xdx-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2\sin 3x\cos 2xdx]$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sin 3xdx-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\{\sin (3x+2x)$

$+\sin (3x-2x)\}dx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sin 3xdx-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin 5xdx-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin xdx$

$=\frac{-1}{2}{\left[\frac{\cos 3x}{3}\right]}_0^{\frac{\pi }{2}}+\frac{1}{4}{\left[\frac{\cos 5x}{5}\right]}_0^{\frac{\pi }{2}}+\frac{1}{4}[\cos x{]}_0^{\frac{\pi }{2}}$

$=-\frac{1}{6}(0-1)+\frac{1}{20}(0-1)+\frac{1}{4}(0-1)$

$=\frac{1}{6}-\frac{1}{20}-\frac{1}{4}$

$=\frac{10-3-15}{60}=\frac{-8}{60}=-\frac{2}{15}$

নির্ণেয় মান : $-\frac{2}{15}$.

দেওয়া আছে, $g(x,y)=9x^2+25y^2-225$

এবং $h\left(x\right)=x-3$

আবার, $g(x,y)=0$

বা, $9x^2+25y^2-225=0$ ........ (1)

বা, $9x^2+25y^2=225$

বা, $\frac{9x^2}{225}+\frac{25y^2}{225}=1$

বা, $\frac{9x^2}{25}+\frac{y^2}{9}=1$

বা, $\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

এবং $h\left(x\right)=0$

বা, $x-3=0$

বা, $x=3$

(1) নং হতত পাই,

$25y^2=225-9x^2$

বা, $25y^2=9(25-x^2)$

বা, $y^2=\frac{9}{25}(25-x^2)$

বা, $y=\pm \frac{3}{5}\sqrt{25-x^2}$

প্রথম চতুর্ভাগে $y$ ধনাত্মক অর্থাৎ $y=\frac{3}{5}\sqrt{25-x^2}$

চিত্র হতে, $B(3,0)$ এবং $C(5,0)$

(1) নং ও (2) নং দ্বারা আবদ্ধ ক্ষুদ্রতর অংশের ক্ষেত্রফল

$=2\times \mathrm{ABC}$ অংশের ক্ষেত্রফল

$=2{\int }_3^5\frac{3}{5}\sqrt{25-x^2}\mathrm{dx}$

$=\frac{6}{5}{\int }_3^5\sqrt{5^2-x^2}dx$

ধরি, $x=5\sin \theta$

বা $\mathrm{dx}=5\cos \theta d\theta$

$=\frac{6}{5}{\int }^{\frac{\pi }{2}}\sqrt{25-25{\sin }^2\theta }\cdot 5\cos \theta d\theta$${\sin }^{-1}\frac{3}{5}$

$=\frac{6}{5}{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}\sqrt{25(1-{\sin }^2\theta )}\cdot 5\cos \theta$

$=\frac{6}{5}{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}\sqrt{25{\cos }^2\theta }\cdot 5\cos \theta d\theta$

$=\frac{6}{5}{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}25{\cos }^2\theta d\theta =30{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta$

$=15{\int }^{\frac{\pi }{2}}2{\cos }^2\theta d\theta$${\sin }^{-1}\frac{3}{5}$

$=15{\int }^{\frac{\pi }{2}}(1+\cos 2\theta )d\theta$

$=15\left\{\right[\theta {]}_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}+{\left[\frac{\sin 2\theta }{2}\right]}_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}\}$

$=15(\frac{\pi }{2}-{\sin }^{-1}\frac{3}{5})+15{\left[\frac{2\sin \theta \cos \theta }{2}\right]}_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}$

$=15(\frac{\pi }{2}-{\sin }^{-1}\frac{3}{5})+15[0-\sin ({\sin }^{-1}\frac{3}{5})\cos ({\sin }^{-1}\frac{3}{5})]$

$=\frac{15\pi }{2}-15{\sin }^{-1}\frac{3}{5}-15\times \frac{3}{5}\cos ({\cos }^{-1}\frac{4}{5})$

$=\frac{15\pi }{2}-15{\sin }^{-1}\frac{3}{5}-15\times \frac{3}{5}\times \frac{4}{5}$

$=\frac{15\pi }{2}-15{\sin }^{-1}\frac{3}{5}-\frac{36}{5}$

নির্ণেয় ক্ষুদ্রতর অংশের ক্ষেত্রফল : $(\frac{15\pi }{2}-15{\sin }^{-1}\frac{3}{5}-\frac{36}{5})$ বর্গ একক।