ক) AI-তৈরি নমুনা উত্তর
ধরি, y = x x y = x^x y = x x ln y = x ln x \ln y = x \ln x ln y = x ln x x x x 1 y d y d x = 1 ⋅ ln x + x ⋅ 1 x = 1 + ln x \frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = 1 + \ln x y 1 d x d y = 1 ⋅ ln x + x ⋅ x 1 = 1 + ln x ⟹ d y d x = x x ( 1 + ln x ) \implies \frac{dy}{dx} = x^x (1 + \ln x) ⟹ d x d y = x x ( 1 + ln x )
চরম মানের জন্য, d y d x = 0 \frac{dy}{dx} = 0 d x d y = 0 ⟹ x x ( 1 + ln x ) = 0 \implies x^x (1 + \ln x) = 0 ⟹ x x ( 1 + ln x ) = 0 x > 0 x > 0 x > 0 x x ≠ 0 x^x \neq 0 x x = 0 ∴ 1 + ln x = 0 ⟹ ln x = − 1 ⟹ x = e − 1 = 1 e \therefore 1 + \ln x = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e} ∴ 1 + ln x = 0 ⟹ ln x = − 1 ⟹ x = e − 1 = e 1
আবার, d 2 y d x 2 = d d x [ x x ( 1 + ln x ) ] = x x ( 1 + ln x ) 2 + x x ⋅ 1 x \frac{d^2y}{dx^2} = \frac{d}{dx} [x^x (1 + \ln x)] = x^x (1 + \ln x)^2 + x^x \cdot \frac{1}{x} d x 2 d 2 y = d x d [ x x ( 1 + ln x )] = x x ( 1 + ln x ) 2 + x x ⋅ x 1 x = 1 e x = \frac{1}{e} x = e 1 d 2 y d x 2 = 0 + ( 1 e ) 1 e ⋅ e > 0 \frac{d^2y}{dx^2} = 0 + \left(\frac{1}{e}\right)^{\frac{1}{e}} \cdot e > 0 d x 2 d 2 y = 0 + ( e 1 ) e 1 ⋅ e > 0 x = 1 e x = \frac{1}{e} x = e 1
অতএব, লঘিষ্ঠ মান = ( 1 e ) 1 e = e − 1 / e = \left(\frac{1}{e}\right)^{\frac{1}{e}} = e^{-1/e} = ( e 1 ) e 1 = e − 1/ e
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( x ) = x 3 − 9 x 2 + 24 x − 12 f(x) = x^3 - 9x^2 + 24x - 12 f ( x ) = x 3 − 9 x 2 + 24 x − 12 ∴ f ′ ( x ) = 3 x 2 − 18 x + 24 = 3 ( x 2 − 6 x + 8 ) = 3 ( x − 2 ) ( x − 4 ) \therefore f'(x) = 3x^2 - 18x + 24 = 3(x^2 - 6x + 8) = 3(x - 2)(x - 4) ∴ f ′ ( x ) = 3 x 2 − 18 x + 24 = 3 ( x 2 − 6 x + 8 ) = 3 ( x − 2 ) ( x − 4 ) f ′ ′ ( x ) = 6 x − 18 f''(x) = 6x - 18 f ′′ ( x ) = 6 x − 18
চরম মানের জন্য, f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 ⟹ 3 ( x − 2 ) ( x − 4 ) = 0 \implies 3(x - 2)(x - 4) = 0 ⟹ 3 ( x − 2 ) ( x − 4 ) = 0 ⟹ x = 2 \implies x = 2 ⟹ x = 2 x = 4 x = 4 x = 4
এখন,x = 2 x = 2 x = 2 f ′ ′ ( 2 ) = 6 ( 2 ) − 18 = − 6 < 0 f''(2) = 6(2) - 18 = -6 < 0 f ′′ ( 2 ) = 6 ( 2 ) − 18 = − 6 < 0 x = 2 x = 2 x = 2 ∴ \therefore ∴ = f ( 2 ) = 2 3 − 9 ( 2 ) 2 + 24 ( 2 ) − 12 = 8 − 36 + 48 − 12 = 8 = f(2) = 2^3 - 9(2)^2 + 24(2) - 12 = 8 - 36 + 48 - 12 = 8 = f ( 2 ) = 2 3 − 9 ( 2 ) 2 + 24 ( 2 ) − 12 = 8 − 36 + 48 − 12 = 8
x = 4 x = 4 x = 4 f ′ ′ ( 4 ) = 6 ( 4 ) − 18 = 6 > 0 f''(4) = 6(4) - 18 = 6 > 0 f ′′ ( 4 ) = 6 ( 4 ) − 18 = 6 > 0 x = 4 x = 4 x = 4 ∴ \therefore ∴ = f ( 4 ) = 4 3 − 9 ( 4 ) 2 + 24 ( 4 ) − 12 = 64 − 144 + 96 − 12 = 4 = f(4) = 4^3 - 9(4)^2 + 24(4) - 12 = 64 - 144 + 96 - 12 = 4 = f ( 4 ) = 4 3 − 9 ( 4 ) 2 + 24 ( 4 ) − 12 = 64 − 144 + 96 − 12 = 4
অতএব, গরিষ্ঠ মান 8 8 8 4 4 4
গ) AI-তৈরি নমুনা উত্তর
(i) ∫ φ ( x ) d x = ∫ 1 12 − 16 x 2 d x \int \varphi(x)\,dx = \int \frac{1}{\sqrt{12 - 16x^2}}\,dx ∫ φ ( x ) d x = ∫ 12 − 16 x 2 1 d x = ∫ 1 16 ( 12 16 − x 2 ) d x = 1 4 ∫ 1 ( 3 2 ) 2 − x 2 d x = \int \frac{1}{\sqrt{16\left(\frac{12}{16} - x^2\right)}}\,dx = \frac{1}{4} \int \frac{1}{\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 - x^2}}\,dx = ∫ 16 ( 16 12 − x 2 ) 1 d x = 4 1 ∫ ( 2 3 ) 2 − x 2 1 d x = 1 4 sin − 1 ( x 3 2 ) + c = 1 4 sin − 1 ( 2 x 3 ) + c = \frac{1}{4} \sin^{-1}\left(\frac{x}{\frac{\sqrt{3}}{2}}\right) + c = \frac{1}{4} \sin^{-1}\left(\frac{2x}{\sqrt{3}}\right) + c = 4 1 sin − 1 ( 2 3 x ) + c = 4 1 sin − 1 ( 3 2 x ) + c c c c
(ii) ∫ ψ ( x ) d x = ∫ tan − 1 ( x 5 ) d x \int \psi(x)\,dx = \int \tan^{-1}\left(\frac{x}{5}\right)\,dx ∫ ψ ( x ) d x = ∫ tan − 1 ( 5 x ) d x = tan − 1 ( x 5 ) ∫ 1 d x − ∫ [ d d x ( tan − 1 ( x 5 ) ) ∫ 1 d x ] d x = \tan^{-1}\left(\frac{x}{5}\right) \int 1\,dx - \int \left[ \frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{5}\right)\right) \int 1\,dx \right] dx = tan − 1 ( 5 x ) ∫ 1 d x − ∫ [ d x d ( tan − 1 ( 5 x ) ) ∫ 1 d x ] d x = x tan − 1 ( x 5 ) − ∫ 1 5 1 + ( x 5 ) 2 ⋅ x d x = x \tan^{-1}\left(\frac{x}{5}\right) - \int \frac{\frac{1}{5}}{1 + \left(\frac{x}{5}\right)^2} \cdot x\,dx = x tan − 1 ( 5 x ) − ∫ 1 + ( 5 x ) 2 5 1 ⋅ x d x = x tan − 1 ( x 5 ) − ∫ 5 x 25 + x 2 d x = x \tan^{-1}\left(\frac{x}{5}\right) - \int \frac{5x}{25 + x^2}\,dx = x tan − 1 ( 5 x ) − ∫ 25 + x 2 5 x d x = x tan − 1 ( x 5 ) − 5 2 ∫ 2 x 25 + x 2 d x = x \tan^{-1}\left(\frac{x}{5}\right) - \frac{5}{2} \int \frac{2x}{25 + x^2}\,dx = x tan − 1 ( 5 x ) − 2 5 ∫ 25 + x 2 2 x d x = x tan − 1 ( x 5 ) − 5 2 ln ∣ 25 + x 2 ∣ + c = x \tan^{-1}\left(\frac{x}{5}\right) - \frac{5}{2} \ln|25 + x^2| + c = x tan − 1 ( 5 x ) − 2 5 ln ∣25 + x 2 ∣ + c c c c