ক) AI-তৈরি নমুনা উত্তর
ধরি, x = i 3 x = \sqrt[3]{i} x = 3 i ⟹ x 3 = i = − i 3 \implies x^3 = i = -i^3 ⟹ x 3 = i = − i 3 i 3 = − i ⟹ i = − i 3 i^3 = -i \implies i = -i^3 i 3 = − i ⟹ i = − i 3 ⟹ x 3 + i 3 = 0 \implies x^3 + i^3 = 0 ⟹ x 3 + i 3 = 0 ⟹ ( x + i ) ( x 2 − x i + i 2 ) = 0 \implies (x + i)(x^2 - xi + i^2) = 0 ⟹ ( x + i ) ( x 2 − x i + i 2 ) = 0 ⟹ ( x + i ) ( x 2 − i x − 1 ) = 0 \implies (x + i)(x^2 - ix - 1) = 0 ⟹ ( x + i ) ( x 2 − i x − 1 ) = 0
হয়, x + i = 0 ⟹ x = − i x + i = 0 \implies x = -i x + i = 0 ⟹ x = − i x 2 − i x − 1 = 0 x^2 - ix - 1 = 0 x 2 − i x − 1 = 0 ⟹ x = − ( − i ) ± ( − i ) 2 − 4 ( 1 ) ( − 1 ) 2 ( 1 ) = i ± − 1 + 4 2 = i ± 3 2 = ± 3 + i 2 \implies x = \frac{-(-i) \pm \sqrt{(-i)^2 - 4(1)(-1)}}{2(1)} = \frac{i \pm \sqrt{-1 + 4}}{2} = \frac{i \pm \sqrt{3}}{2} = \frac{\pm \sqrt{3} + i}{2} ⟹ x = 2 ( 1 ) − ( − i ) ± ( − i ) 2 − 4 ( 1 ) ( − 1 ) = 2 i ± − 1 + 4 = 2 i ± 3 = 2 ± 3 + i
অতএব, i 3 = − i , 3 + i 2 , − 3 + i 2 \sqrt[3]{i} = -i, \frac{\sqrt{3} + i}{2}, \frac{-\sqrt{3} + i}{2} 3 i = − i , 2 3 + i , 2 − 3 + i
খ) AI-তৈরি নমুনা উত্তর
দৃশ্যকল্প-১ এ প্রদত্ত সমীকরণ: 8 x 2 − 6 x + 1 = 0 8x^2 - 6x + 1 = 0 8 x 2 − 6 x + 1 = 0 8 x 2 − 4 x − 2 x + 1 = 0 8x^2 - 4x - 2x + 1 = 0 8 x 2 − 4 x − 2 x + 1 = 0 4 x ( 2 x − 1 ) − 1 ( 2 x − 1 ) = 0 4x(2x - 1) - 1(2x - 1) = 0 4 x ( 2 x − 1 ) − 1 ( 2 x − 1 ) = 0 ( 2 x − 1 ) ( 4 x − 1 ) = 0 (2x - 1)(4x - 1) = 0 ( 2 x − 1 ) ( 4 x − 1 ) = 0 ∴ x = 1 2 , 1 4 \therefore x = \frac{1}{2}, \frac{1}{4} ∴ x = 2 1 , 4 1
অতএব, মূলদ্বয় a = 1 2 a = \frac{1}{2} a = 2 1 b = 1 4 b = \frac{1}{4} b = 4 1
তাহলে নতুন মূলদ্বয় হবে:α = a + 1 b = 1 2 + 4 = 9 2 \alpha = a + \frac{1}{b} = \frac{1}{2} + 4 = \frac{9}{2} α = a + b 1 = 2 1 + 4 = 2 9 β = b + 1 a = 1 4 + 2 = 9 4 \beta = b + \frac{1}{a} = \frac{1}{4} + 2 = \frac{9}{4} β = b + a 1 = 4 1 + 2 = 4 9
নতুন সমীকরণের:α + β = 9 2 + 9 4 = 18 + 9 4 = 27 4 \alpha + \beta = \frac{9}{2} + \frac{9}{4} = \frac{18 + 9}{4} = \frac{27}{4} α + β = 2 9 + 4 9 = 4 18 + 9 = 4 27 α ⋅ β = 9 2 ⋅ 9 4 = 81 8 \alpha \cdot \beta = \frac{9}{2} \cdot \frac{9}{4} = \frac{81}{8} α ⋅ β = 2 9 ⋅ 4 9 = 8 81
সুতরাং নির্ণেয় দ্বিঘাত সমীকরণটি হলো:x 2 − ( α + β ) x + α β = 0 x^2 - (\alpha + \beta)x + \alpha\beta = 0 x 2 − ( α + β ) x + α β = 0 x 2 − 27 4 x + 81 8 = 0 x^2 - \frac{27}{4}x + \frac{81}{8} = 0 x 2 − 4 27 x + 8 81 = 0 8 8 8 8 x 2 − 54 x + 81 = 0 8x^2 - 54x + 81 = 0 8 x 2 − 54 x + 81 = 0
গ) AI-তৈরি নমুনা উত্তর
দৃশ্যকল্প-২ এ প্রদত্ত রাশি: ( 1 + 3 y ) 2 n (1 + 3y)^{2n} ( 1 + 3 y ) 2 n 2 n 2n 2 n ( 2 n + 1 ) (2n + 1) ( 2 n + 1 ) ( 2 n 2 + 1 ) \left(\frac{2n}{2} + 1\right) ( 2 2 n + 1 ) ( n + 1 ) (n + 1) ( n + 1 )
( n + 1 ) (n + 1) ( n + 1 ) T n + 1 = 2 n C n ( 1 ) 2 n − n ( 3 y ) n T_{n+1} = {}^{2n}C_n (1)^{2n-n} (3y)^n T n + 1 = 2 n C n ( 1 ) 2 n − n ( 3 y ) n = ( 2 n ) ! n ! ( 2 n − n ) ! ⋅ 3 n y n = \frac{(2n)!}{n!(2n - n)!} \cdot 3^n y^n = n ! ( 2 n − n )! ( 2 n )! ⋅ 3 n y n = ( 2 n ) ! n ! n ! ⋅ 3 n y n = \frac{(2n)!}{n! \, n!} \cdot 3^n y^n = n ! n ! ( 2 n )! ⋅ 3 n y n = 1 ⋅ 2 ⋅ 3 ⋅ 4 … ( 2 n − 1 ) ( 2 n ) n ! n ! ⋅ 3 n y n = \frac{1 \cdot 2 \cdot 3 \cdot 4 \dots (2n - 1)(2n)}{n! \, n!} \cdot 3^n y^n = n ! n ! 1 ⋅ 2 ⋅ 3 ⋅ 4 … ( 2 n − 1 ) ( 2 n ) ⋅ 3 n y n = { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) } { 2 ⋅ 4 ⋅ 6 … ( 2 n ) } n ! n ! ⋅ 3 n y n = \frac{\{1 \cdot 3 \cdot 5 \dots (2n - 1)\} \{2 \cdot 4 \cdot 6 \dots (2n)\}}{n! \, n!} \cdot 3^n y^n = n ! n ! { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 )} { 2 ⋅ 4 ⋅ 6 … ( 2 n )} ⋅ 3 n y n = { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) } ⋅ 2 n { 1 ⋅ 2 ⋅ 3 … n } n ! n ! ⋅ 3 n y n = \frac{\{1 \cdot 3 \cdot 5 \dots (2n - 1)\} \cdot 2^n \{1 \cdot 2 \cdot 3 \dots n\}}{n! \, n!} \cdot 3^n y^n = n ! n ! { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 )} ⋅ 2 n { 1 ⋅ 2 ⋅ 3 … n } ⋅ 3 n y n = { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) } ⋅ 2 n ⋅ n ! n ! n ! ⋅ 3 n y n = \frac{\{1 \cdot 3 \cdot 5 \dots (2n - 1)\} \cdot 2^n \cdot n!}{n! \, n!} \cdot 3^n y^n = n ! n ! { 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 )} ⋅ 2 n ⋅ n ! ⋅ 3 n y n = 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) n ! ⋅ ( 2 ⋅ 3 ) n y n = \frac{1 \cdot 3 \cdot 5 \dots (2n - 1)}{n!} \cdot (2 \cdot 3)^n y^n = n ! 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) ⋅ ( 2 ⋅ 3 ) n y n = 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) n ! 6 n y n = \frac{1 \cdot 3 \cdot 5 \dots (2n - 1)}{n!} 6^n y^n = n ! 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 6 n y n