ক) AI-তৈরি নমুনা উত্তর
প্রদত্ত রাশি,sec 2 ( cot − 1 1 ) + sin 2 ( cos − 1 1 2 ) \sec^2(\cot^{-1} 1) + \sin^2\left(\cos^{-1} \frac{1}{2}\right) sec 2 ( cot − 1 1 ) + sin 2 ( cos − 1 2 1 ) = 1 + tan 2 ( cot − 1 1 ) + 1 − cos 2 ( cos − 1 1 2 ) = 1 + \tan^2(\cot^{-1} 1) + 1 - \cos^2\left(\cos^{-1} \frac{1}{2}\right) = 1 + tan 2 ( cot − 1 1 ) + 1 − cos 2 ( cos − 1 2 1 ) = 1 + [ tan ( tan − 1 1 ) ] 2 + 1 − [ cos ( cos − 1 1 2 ) ] 2 = 1 + \left[\tan(\tan^{-1} 1)\right]^2 + 1 - \left[\cos\left(\cos^{-1} \frac{1}{2}\right)\right]^2 = 1 + [ tan ( tan − 1 1 ) ] 2 + 1 − [ cos ( cos − 1 2 1 ) ] 2 = 1 + ( 1 ) 2 + 1 − ( 1 2 ) 2 = 1 + (1)^2 + 1 - \left(\frac{1}{2}\right)^2 = 1 + ( 1 ) 2 + 1 − ( 2 1 ) 2 = 1 + 1 + 1 − 1 4 = 1 + 1 + 1 - \frac{1}{4} = 1 + 1 + 1 − 4 1 = 3 − 1 4 = 3 - \frac{1}{4} = 3 − 4 1 = 11 4 = \frac{11}{4} = 4 11
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপক-১ থেকে পাই,sec α = p x ⟹ cos α = x p \sec \alpha = \frac{p}{x} \implies \cos \alpha = \frac{x}{p} sec α = x p ⟹ cos α = p x sec β = q y ⟹ cos β = y q \sec \beta = \frac{q}{y} \implies \cos \beta = \frac{y}{q} sec β = y q ⟹ cos β = q y
অতএব,sin α = 1 − cos 2 α = 1 − x 2 p 2 \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{x^2}{p^2}} sin α = 1 − cos 2 α = 1 − p 2 x 2 sin β = 1 − cos 2 β = 1 − y 2 q 2 \sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{y^2}{q^2}} sin β = 1 − cos 2 β = 1 − q 2 y 2
দেওয়া আছে,α + β = γ \alpha + \beta = \gamma α + β = γ ⟹ α = γ − β \implies \alpha = \gamma - \beta ⟹ α = γ − β ⟹ cos α = cos ( γ − β ) \implies \cos \alpha = \cos(\gamma - \beta) ⟹ cos α = cos ( γ − β ) ⟹ cos α = cos γ cos β + sin γ sin β \implies \cos \alpha = \cos \gamma \cos \beta + \sin \gamma \sin \beta ⟹ cos α = cos γ cos β + sin γ sin β ⟹ cos α − cos β cos γ = sin β sin γ \implies \cos \alpha - \cos \beta \cos \gamma = \sin \beta \sin \gamma ⟹ cos α − cos β cos γ = sin β sin γ
উভয়পক্ষকে বর্গ করে পাই,( cos α − cos β cos γ ) 2 = sin 2 β sin 2 γ (\cos \alpha - \cos \beta \cos \gamma)^2 = \sin^2 \beta \sin^2 \gamma ( cos α − cos β cos γ ) 2 = sin 2 β sin 2 γ ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β cos 2 γ = ( 1 − cos 2 β ) sin 2 γ \implies \cos^2 \alpha - 2\cos \alpha \cos \beta \cos \gamma + \cos^2 \beta \cos^2 \gamma = (1 - \cos^2 \beta)\sin^2 \gamma ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β cos 2 γ = ( 1 − cos 2 β ) sin 2 γ ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β cos 2 γ = sin 2 γ − cos 2 β sin 2 γ \implies \cos^2 \alpha - 2\cos \alpha \cos \beta \cos \gamma + \cos^2 \beta \cos^2 \gamma = \sin^2 \gamma - \cos^2 \beta \sin^2 \gamma ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β cos 2 γ = sin 2 γ − cos 2 β sin 2 γ ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β ( cos 2 γ + sin 2 γ ) = sin 2 γ \implies \cos^2 \alpha - 2\cos \alpha \cos \beta \cos \gamma + \cos^2 \beta (\cos^2 \gamma + \sin^2 \gamma) = \sin^2 \gamma ⟹ cos 2 α − 2 cos α cos β cos γ + cos 2 β ( cos 2 γ + sin 2 γ ) = sin 2 γ ⟹ cos 2 α + cos 2 β − 2 cos α cos β cos γ = sin 2 γ \implies \cos^2 \alpha + \cos^2 \beta - 2\cos \alpha \cos \beta \cos \gamma = \sin^2 \gamma ⟹ cos 2 α + cos 2 β − 2 cos α cos β cos γ = sin 2 γ
এখন, cos α = x p \cos \alpha = \frac{x}{p} cos α = p x cos β = y q \cos \beta = \frac{y}{q} cos β = q y x 2 p 2 + y 2 q 2 − 2 ⋅ x p ⋅ y q cos γ = sin 2 γ \frac{x^2}{p^2} + \frac{y^2}{q^2} - 2 \cdot \frac{x}{p} \cdot \frac{y}{q} \cos \gamma = \sin^2 \gamma p 2 x 2 + q 2 y 2 − 2 ⋅ p x ⋅ q y cos γ = sin 2 γ ⟹ x 2 p 2 + y 2 q 2 − 2 x y p q cos γ = sin 2 γ \implies \frac{x^2}{p^2} + \frac{y^2}{q^2} - \frac{2xy}{pq}\cos \gamma = \sin^2 \gamma ⟹ p 2 x 2 + q 2 y 2 − pq 2 x y cos γ = sin 2 γ
গ) AI-তৈরি নমুনা উত্তর
উদ্দীপক-২ অনুযায়ী, f ( x ) = sec x f(x) = \sec x f ( x ) = sec x
প্রদত্ত সমীকরণ:f ( x ) ⋅ f ( 3 x ) + 2 = 0 f(x) \cdot f(3x) + 2 = 0 f ( x ) ⋅ f ( 3 x ) + 2 = 0 ⟹ sec x ⋅ sec 3 x + 2 = 0 \implies \sec x \cdot \sec 3x + 2 = 0 ⟹ sec x ⋅ sec 3 x + 2 = 0 ⟹ 1 cos x cos 3 x + 2 = 0 \implies \frac{1}{\cos x \cos 3x} + 2 = 0 ⟹ c o s x c o s 3 x 1 + 2 = 0 ⟹ 1 + 2 cos 3 x cos x = 0 \implies 1 + 2\cos 3x \cos x = 0 ⟹ 1 + 2 cos 3 x cos x = 0 cos x ≠ 0 \cos x \neq 0 cos x = 0 cos 3 x ≠ 0 \cos 3x \neq 0 cos 3 x = 0 ⟹ 1 + cos ( 3 x + x ) + cos ( 3 x − x ) = 0 \implies 1 + \cos(3x + x) + \cos(3x - x) = 0 ⟹ 1 + cos ( 3 x + x ) + cos ( 3 x − x ) = 0 ⟹ 1 + cos 4 x + cos 2 x = 0 \implies 1 + \cos 4x + \cos 2x = 0 ⟹ 1 + cos 4 x + cos 2 x = 0 ⟹ 2 cos 2 2 x + cos 2 x = 0 \implies 2\cos^2 2x + \cos 2x = 0 ⟹ 2 cos 2 2 x + cos 2 x = 0 ⟹ cos 2 x ( 2 cos 2 x + 1 ) = 0 \implies \cos 2x(2\cos 2x + 1) = 0 ⟹ cos 2 x ( 2 cos 2 x + 1 ) = 0
হয়, cos 2 x = 0 \cos 2x = 0 cos 2 x = 0 ⟹ 2 x = ( 2 n + 1 ) π 2 \implies 2x = (2n + 1)\frac{\pi}{2} ⟹ 2 x = ( 2 n + 1 ) 2 π ⟹ x = ( 2 n + 1 ) π 4 \implies x = (2n + 1)\frac{\pi}{4} ⟹ x = ( 2 n + 1 ) 4 π n ∈ Z n \in \mathbb{Z} n ∈ Z
অথবা,2 cos 2 x + 1 = 0 2\cos 2x + 1 = 0 2 cos 2 x + 1 = 0 ⟹ cos 2 x = − 1 2 \implies \cos 2x = -\frac{1}{2} ⟹ cos 2 x = − 2 1 ⟹ cos 2 x = cos ( π − π 3 ) = cos ( 2 π 3 ) \implies \cos 2x = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) ⟹ cos 2 x = cos ( π − 3 π ) = cos ( 3 2 π ) ⟹ 2 x = 2 n π ± 2 π 3 \implies 2x = 2n\pi \pm \frac{2\pi}{3} ⟹ 2 x = 2 nπ ± 3 2 π ⟹ x = n π ± π 3 \implies x = n\pi \pm \frac{\pi}{3} ⟹ x = nπ ± 3 π n ∈ Z n \in \mathbb{Z} n ∈ Z
কিন্তু cos x ≠ 0 \cos x \neq 0 cos x = 0 cos 3 x ≠ 0 \cos 3x \neq 0 cos 3 x = 0 x = ( 2 n + 1 ) π 4 x = (2n + 1)\frac{\pi}{4} x = ( 2 n + 1 ) 4 π cos x = cos ( ( 2 n + 1 ) π 4 ) ≠ 0 \cos x = \cos\left((2n + 1)\frac{\pi}{4}\right) \neq 0 cos x = cos ( ( 2 n + 1 ) 4 π ) = 0 cos 3 x = cos ( 3 ( 2 n + 1 ) π 4 ) ≠ 0 \cos 3x = \cos\left(3(2n + 1)\frac{\pi}{4}\right) \neq 0 cos 3 x = cos ( 3 ( 2 n + 1 ) 4 π ) = 0 x = n π ± π 3 x = n\pi \pm \frac{\pi}{3} x = nπ ± 3 π cos x ≠ 0 \cos x \neq 0 cos x = 0 cos 3 x = cos ( 3 n π ± π ) = ± 1 ≠ 0 \cos 3x = \cos(3n\pi \pm \pi) = \pm 1 \neq 0 cos 3 x = cos ( 3 nπ ± π ) = ± 1 = 0
অতএব, নির্ণেয় সাধারণ সমাধান:x = ( 2 n + 1 ) π 4 x = (2n + 1)\frac{\pi}{4} x = ( 2 n + 1 ) 4 π x = n π ± π 3 x = n\pi \pm \frac{\pi}{3} x = nπ ± 3 π n ∈ Z n \in \mathbb{Z} n ∈ Z