ক) AI-তৈরি নমুনা উত্তর
ধরি, I = ∫ sin x sin x − cos x d x I = \int \frac{\sin x}{\sin x - \cos x} dx I = ∫ s i n x − c o s x s i n x d x
আমরা লবকে নিম্নরূপে সাজাতে পারি:sin x = 1 2 [ ( sin x − cos x ) + ( sin x + cos x ) ] \sin x = \frac{1}{2} [(\sin x - \cos x) + (\sin x + \cos x)] sin x = 2 1 [( sin x − cos x ) + ( sin x + cos x )]
অতএব,I = 1 2 ∫ ( sin x − cos x ) + ( sin x + cos x ) sin x − cos x d x I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx I = 2 1 ∫ s i n x − c o s x ( s i n x − c o s x ) + ( s i n x + c o s x ) d x = 1 2 ∫ 1 d x + 1 2 ∫ sin x + cos x sin x − cos x d x = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\sin x + \cos x}{\sin x - \cos x} dx = 2 1 ∫ 1 d x + 2 1 ∫ s i n x − c o s x s i n x + c o s x d x = 1 2 x + 1 2 ln ∣ sin x − cos x ∣ + c = \frac{1}{2} x + \frac{1}{2} \ln |\sin x - \cos x| + c = 2 1 x + 2 1 ln ∣ sin x − cos x ∣ + c
যেখানে c c c
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( x ) = sin 4 x f(x) = \sin^4 x f ( x ) = sin 4 x I = ∫ 0 π x sin 4 x d x I = \int_0^\pi x \sin^4 x \, dx I = ∫ 0 π x sin 4 x d x
আমরা জানি, ∫ 0 a g ( x ) d x = ∫ 0 a g ( a − x ) d x \int_0^a g(x) dx = \int_0^a g(a-x) dx ∫ 0 a g ( x ) d x = ∫ 0 a g ( a − x ) d x ∴ I = ∫ 0 π ( π − x ) sin 4 ( π − x ) d x = ∫ 0 π ( π − x ) sin 4 x d x \therefore I = \int_0^\pi (\pi - x) \sin^4(\pi - x) \, dx = \int_0^\pi (\pi - x) \sin^4 x \, dx ∴ I = ∫ 0 π ( π − x ) sin 4 ( π − x ) d x = ∫ 0 π ( π − x ) sin 4 x d x ⇒ I = π ∫ 0 π sin 4 x d x − ∫ 0 π x sin 4 x d x \Rightarrow I = \pi \int_0^\pi \sin^4 x \, dx - \int_0^\pi x \sin^4 x \, dx ⇒ I = π ∫ 0 π sin 4 x d x − ∫ 0 π x sin 4 x d x ⇒ I = π ∫ 0 π sin 4 x d x − I \Rightarrow I = \pi \int_0^\pi \sin^4 x \, dx - I ⇒ I = π ∫ 0 π sin 4 x d x − I ⇒ 2 I = π ∫ 0 π sin 4 x d x = 2 π ∫ 0 π / 2 sin 4 x d x \Rightarrow 2I = \pi \int_0^\pi \sin^4 x \, dx = 2\pi \int_0^{\pi/2} \sin^4 x \, dx ⇒ 2 I = π ∫ 0 π sin 4 x d x = 2 π ∫ 0 π /2 sin 4 x d x ⇒ I = π ∫ 0 π / 2 sin 4 x d x \Rightarrow I = \pi \int_0^{\pi/2} \sin^4 x \, dx ⇒ I = π ∫ 0 π /2 sin 4 x d x
এখন,sin 4 x = ( sin 2 x ) 2 = ( 1 − cos 2 x 2 ) 2 = 1 4 ( 1 − 2 cos 2 x + cos 2 2 x ) \sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x) sin 4 x = ( sin 2 x ) 2 = ( 2 1 − c o s 2 x ) 2 = 4 1 ( 1 − 2 cos 2 x + cos 2 2 x ) = 1 4 ( 1 − 2 cos 2 x + 1 + cos 4 x 2 ) = 1 8 ( 3 − 4 cos 2 x + cos 4 x ) = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{1}{8}(3 - 4\cos 2x + \cos 4x) = 4 1 ( 1 − 2 cos 2 x + 2 1 + c o s 4 x ) = 8 1 ( 3 − 4 cos 2 x + cos 4 x )
অতএব,I = π 8 ∫ 0 π / 2 ( 3 − 4 cos 2 x + cos 4 x ) d x I = \frac{\pi}{8} \int_0^{\pi/2} (3 - 4\cos 2x + \cos 4x) dx I = 8 π ∫ 0 π /2 ( 3 − 4 cos 2 x + cos 4 x ) d x = π 8 [ 3 x − 2 sin 2 x + sin 4 x 4 ] 0 π / 2 = \frac{\pi}{8} \left[ 3x - 2\sin 2x + \frac{\sin 4x}{4} \right]_0^{\pi/2} = 8 π [ 3 x − 2 sin 2 x + 4 s i n 4 x ] 0 π /2 = π 8 ( 3 π 2 − 0 + 0 ) = 3 π 2 16 = \frac{\pi}{8} \left( \frac{3\pi}{2} - 0 + 0 \right) = \frac{3\pi^2}{16} = 8 π ( 2 3 π − 0 + 0 ) = 16 3 π 2
গ) AI-তৈরি নমুনা উত্তর
উদ্দীপকে (ii) নং সমীকরণটি বৃত্তের সমীকরণ: x 2 + y 2 = 8 x x^2 + y^2 = 8x x 2 + y 2 = 8 x ( x − 4 ) 2 + y 2 = 4 2 (x-4)^2 + y^2 = 4^2 ( x − 4 ) 2 + y 2 = 4 2 ( 4 , 0 ) (4, 0) ( 4 , 0 ) r = 4 r = 4 r = 4 y = 2 x y = 2x y = 2 x
ছেদবিন্দু নির্ণয়:x 2 + ( 2 x ) 2 = 8 x ⇒ 5 x 2 = 8 x ⇒ x ( 5 x − 8 ) = 0 x^2 + (2x)^2 = 8x \Rightarrow 5x^2 = 8x \Rightarrow x(5x - 8) = 0 x 2 + ( 2 x ) 2 = 8 x ⇒ 5 x 2 = 8 x ⇒ x ( 5 x − 8 ) = 0 ∴ x = 0 \therefore x = 0 ∴ x = 0 x = 8 5 x = \frac{8}{5} x = 5 8 x = 0 x = 0 x = 0 y = 0 y = 0 y = 0 x = 8 5 x = \frac{8}{5} x = 5 8 y = 16 5 y = \frac{16}{5} y = 5 16 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 8 5 , 16 5 ) \left(\frac{8}{5}, \frac{16}{5}\right) ( 5 8 , 5 16 )
(i) ও (ii) রেখাদ্বয় দ্বারা প্রথম চতুর্ভাগে আবদ্ধ ক্ষুদ্রতম অংশের ক্ষেত্রফল:A = ∫ 0 8 / 5 ( y বৃত্ত − y রেখা ) d x = ∫ 0 8 / 5 ( 8 x − x 2 − 2 x ) d x A = \int_0^{8/5} (y_{বৃত্ত} - y_{রেখা}) \, dx = \int_0^{8/5} (\sqrt{8x - x^2} - 2x) \, dx A = ∫ 0 8/5 ( y বৃত্ত − y রেখা ) d x = ∫ 0 8/5 ( 8 x − x 2 − 2 x ) d x ∫ 0 8 / 5 2 x d x = [ x 2 ] 0 8 / 5 = ( 8 5 ) 2 = 64 25 = 2.56 \int_0^{8/5} 2x \, dx = [x^2]_0^{8/5} = \left(\frac{8}{5}\right)^2 = \frac{64}{25} = 2.56 ∫ 0 8/5 2 x d x = [ x 2 ] 0 8/5 = ( 5 8 ) 2 = 25 64 = 2.56 ∫ 16 − ( x − 4 ) 2 d x = x − 4 2 8 x − x 2 + 16 2 arcsin ( x − 4 4 ) \int \sqrt{16 - (x-4)^2} \, dx = \frac{x-4}{2}\sqrt{8x-x^2} + \frac{16}{2}\arcsin\left(\frac{x-4}{4}\right) ∫ 16 − ( x − 4 ) 2 d x = 2 x − 4 8 x − x 2 + 2 16 arcsin ( 4 x − 4 ) 0 0 0 8 5 \frac{8}{5} 5 8 x = 8 5 x = \frac{8}{5} x = 5 8 − 12 / 5 2 ( 16 5 ) + 8 arcsin ( − 3 5 ) = − 96 25 − 8 arcsin ( 0.6 ) \frac{-12/5}{2} \left(\frac{16}{5}\right) + 8\arcsin\left(-\frac{3}{5}\right) = -\frac{96}{25} - 8\arcsin(0.6) 2 − 12/5 ( 5 16 ) + 8 arcsin ( − 5 3 ) = − 25 96 − 8 arcsin ( 0.6 ) x = 0 x = 0 x = 0 − 4 2 ( 0 ) + 8 arcsin ( − 1 ) = − 8 ( π 2 ) = − 4 π \frac{-4}{2}(0) + 8\arcsin(-1) = -8\left(\frac{\pi}{2}\right) = -4\pi 2 − 4 ( 0 ) + 8 arcsin ( − 1 ) = − 8 ( 2 π ) = − 4 π ∫ 0 8 / 5 8 x − x 2 d x = 4 π − 96 25 − 8 arcsin ( 0.6 ) \int_0^{8/5} \sqrt{8x-x^2} \, dx = 4\pi - \frac{96}{25} - 8\arcsin(0.6) ∫ 0 8/5 8 x − x 2 d x = 4 π − 25 96 − 8 arcsin ( 0.6 ) A = 4 π − 96 25 − 8 arcsin ( 0.6 ) − 64 25 = 4 π − 160 25 − 8 arcsin ( 0.6 ) = 4 π − 32 5 − 8 arcsin ( 0.6 ) ≈ 1.018 A = 4\pi - \frac{96}{25} - 8\arcsin(0.6) - \frac{64}{25} = 4\pi - \frac{160}{25} - 8\arcsin(0.6) = 4\pi - \frac{32}{5} - 8\arcsin(0.6) \approx 1.018 A = 4 π − 25 96 − 8 arcsin ( 0.6 ) − 25 64 = 4 π − 25 160 − 8 arcsin ( 0.6 ) = 4 π − 5 32 − 8 arcsin ( 0.6 ) ≈ 1.018