ক) AI-তৈরি নমুনা উত্তর
প্রদত্ত রাশি: ( 2 x 2 − 3 x ) 12 \left(2x^2 - \frac{3}{x}\right)^{12} ( 2 x 2 − x 3 ) 12 n = 12 n = 12 n = 12 ( 12 2 + 1 ) = ( 6 + 1 ) \left(\frac{12}{2} + 1\right) = (6 + 1) ( 2 12 + 1 ) = ( 6 + 1 )
∴ \therefore ∴ T 6 + 1 = 12 C 6 ( 2 x 2 ) 12 − 6 ( − 3 x ) 6 T_{6+1} = {}^{12}C_6 (2x^2)^{12-6} \left(-\frac{3}{x}\right)^6 T 6 + 1 = 12 C 6 ( 2 x 2 ) 12 − 6 ( − x 3 ) 6 = 12 C 6 ( 2 x 2 ) 6 ( 3 6 x 6 ) = {}^{12}C_6 (2x^2)^6 \left(\frac{3^6}{x^6}\right) = 12 C 6 ( 2 x 2 ) 6 ( x 6 3 6 ) = 924 × 2 6 ⋅ x 12 × 729 x 6 = 924 \times 2^6 \cdot x^{12} \times \frac{729}{x^6} = 924 × 2 6 ⋅ x 12 × x 6 729 = 924 × 64 × 729 × x 6 = 924 \times 64 \times 729 \times x^6 = 924 × 64 × 729 × x 6 = 43110144 x 6 = 43110144 x^6 = 43110144 x 6 924 ⋅ 64 ⋅ 729 x 6 924 \cdot 64 \cdot 729 x^6 924 ⋅ 64 ⋅ 729 x 6
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপক হতে, P = 4 x + 3 = 3 + 4 x P = 4x + 3 = 3 + 4x P = 4 x + 3 = 3 + 4 x ∴ P 34 = ( 3 + 4 x ) 34 \therefore P^{34} = (3 + 4x)^{34} ∴ P 34 = ( 3 + 4 x ) 34
এর বিস্তৃতিতে ( r + 1 ) (r+1) ( r + 1 ) T r + 1 = 34 C r ( 3 ) 34 − r ( 4 x ) r = 34 C r 3 34 − r 4 r x r T_{r+1} = {}^{34}C_r (3)^{34-r} (4x)^r = {}^{34}C_r 3^{34-r} 4^r x^r T r + 1 = 34 C r ( 3 ) 34 − r ( 4 x ) r = 34 C r 3 34 − r 4 r x r
অতএব, T r + 1 T_{r+1} T r + 1 = 34 C r 3 34 − r 4 r = {}^{34}C_r 3^{34-r} 4^r = 34 C r 3 34 − r 4 r r r r = 34 C r − 1 3 35 − r 4 r − 1 = {}^{34}C_{r-1} 3^{35-r} 4^{r-1} = 34 C r − 1 3 35 − r 4 r − 1
প্রশ্নমতে, এই দুটি ক্রমিক পদের সহগ সমান:34 C r − 1 3 35 − r 4 r − 1 = 34 C r 3 34 − r 4 r {}^{34}C_{r-1} 3^{35-r} 4^{r-1} = {}^{34}C_r 3^{34-r} 4^r 34 C r − 1 3 35 − r 4 r − 1 = 34 C r 3 34 − r 4 r ⇒ 34 C r 34 C r − 1 = 3 35 − r 4 r − 1 3 34 − r 4 r = 3 4 \Rightarrow \frac{{}^{34}C_r}{{}^{34}C_{r-1}} = \frac{3^{35-r} 4^{r-1}}{3^{34-r} 4^r} = \frac{3}{4} ⇒ 34 C r − 1 34 C r = 3 34 − r 4 r 3 35 − r 4 r − 1 = 4 3 ⇒ 34 − r + 1 r = 3 4 \Rightarrow \frac{34 - r + 1}{r} = \frac{3}{4} ⇒ r 34 − r + 1 = 4 3 ⇒ 35 − r r = 3 4 \Rightarrow \frac{35 - r}{r} = \frac{3}{4} ⇒ r 35 − r = 4 3 ⇒ 4 ( 35 − r ) = 3 r \Rightarrow 4(35 - r) = 3r ⇒ 4 ( 35 − r ) = 3 r ⇒ 140 − 4 r = 3 r \Rightarrow 140 - 4r = 3r ⇒ 140 − 4 r = 3 r ⇒ 7 r = 140 \Rightarrow 7r = 140 ⇒ 7 r = 140 ⇒ r = 20 \Rightarrow r = 20 ⇒ r = 20
সুতরাং পদ দুটি হলো r = 20 r = 20 r = 20 T 20 T_{20} T 20 ( r + 1 ) = 21 (r+1) = 21 ( r + 1 ) = 21 T 21 T_{21} T 21
T 20 T_{20} T 20 x x x = r − 1 = 20 − 1 = 19 = r - 1 = 20 - 1 = 19 = r − 1 = 20 − 1 = 19 T 21 T_{21} T 21 x x x = r = 20 = r = 20 = r = 20
অতএব, পদ দুইটিতে x x x 19 19 19 20 20 20
গ) AI-তৈরি নমুনা উত্তর
P = 3 + 4 x P = 3 + 4x P = 3 + 4 x ∴ P − 1 2 = ( 3 + 4 x ) − 1 2 = 3 − 1 2 ( 1 + 4 x 3 ) − 1 2 = 1 3 ( 1 + 4 x 3 ) − 1 2 \therefore P^{-\frac{1}{2}} = (3 + 4x)^{-\frac{1}{2}} = 3^{-\frac{1}{2}} \left(1 + \frac{4x}{3}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{3}} \left(1 + \frac{4x}{3}\right)^{-\frac{1}{2}} ∴ P − 2 1 = ( 3 + 4 x ) − 2 1 = 3 − 2 1 ( 1 + 3 4 x ) − 2 1 = 3 1 ( 1 + 3 4 x ) − 2 1
আমরা জানি, ( 1 + y ) − n = 1 + ( − n ) y + ( − n ) ( − n − 1 ) 2 ! y 2 + ⋯ + ( − n ) ( − n − 1 ) … ( − n − r + 1 ) r ! y r + … (1 + y)^{-n} = 1 + (-n)y + \frac{(-n)(-n-1)}{2!} y^2 + \dots + \frac{(-n)(-n-1)\dots(-n-r+1)}{r!} y^r + \dots ( 1 + y ) − n = 1 + ( − n ) y + 2 ! ( − n ) ( − n − 1 ) y 2 + ⋯ + r ! ( − n ) ( − n − 1 ) … ( − n − r + 1 ) y r + … y = 4 x 3 y = \frac{4x}{3} y = 3 4 x = − 1 2 = -\frac{1}{2} = − 2 1
( 1 + 4 x 3 ) − 1 2 \left(1 + \frac{4x}{3}\right)^{-\frac{1}{2}} ( 1 + 3 4 x ) − 2 1 ( r + 1 ) (r+1) ( r + 1 ) x r x^r x r = ( − 1 2 ) ( − 3 2 ) ( − 5 2 ) … ( − 2 r − 1 2 ) r ! ( 4 x 3 ) r = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\dots\left(-\frac{2r-1}{2}\right)}{r!} \left(\frac{4x}{3}\right)^r = r ! ( − 2 1 ) ( − 2 3 ) ( − 2 5 ) … ( − 2 2 r − 1 ) ( 3 4 x ) r = ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) 2 r ⋅ r ! ⋅ 4 r 3 r x r = \frac{(-1)^r \cdot 1 \cdot 3 \cdot 5 \dots (2r-1)}{2^r \cdot r!} \cdot \frac{4^r}{3^r} x^r = 2 r ⋅ r ! ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 3 r 4 r x r = ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r r ! ⋅ 3 r x r = \frac{(-1)^r \cdot 1 \cdot 3 \cdot 5 \dots (2r-1) \cdot 2^r}{r! \cdot 3^r} x^r = r ! ⋅ 3 r ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r x r
অতএব, P − 1 2 P^{-\frac{1}{2}} P − 2 1 x r x^r x r = 1 3 ⋅ ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r 3 r ⋅ r ! = \frac{1}{\sqrt{3}} \cdot \frac{(-1)^r \cdot 1 \cdot 3 \cdot 5 \dots (2r-1) \cdot 2^r}{3^r \cdot r!} = 3 1 ⋅ 3 r ⋅ r ! ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r = ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r 3 r + 1 2 ⋅ r ! = \frac{(-1)^r \cdot 1 \cdot 3 \cdot 5 \dots (2r-1) \cdot 2^r}{3^{r + \frac{1}{2}} \cdot r!} = 3 r + 2 1 ⋅ r ! ( − 1 ) r ⋅ 1 ⋅ 3 ⋅ 5 … ( 2 r − 1 ) ⋅ 2 r
পঞ্চম পদ (r = 4 r = 4 r = 4 T 5 = 1 3 ⋅ ( − 1 ) 4 ⋅ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 2 4 3 4 ⋅ 4 ! x 4 T_5 = \frac{1}{\sqrt{3}} \cdot \frac{(-1)^4 \cdot 1 \cdot 3 \cdot 5 \cdot 7 \cdot 2^4}{3^4 \cdot 4!} x^4 T 5 = 3 1 ⋅ 3 4 ⋅ 4 ! ( − 1 ) 4 ⋅ 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 2 4 x 4 = 1 3 ⋅ 105 × 16 81 × 24 x 4 = \frac{1}{\sqrt{3}} \cdot \frac{105 \times 16}{81 \times 24} x^4 = 3 1 ⋅ 81 × 24 105 × 16 x 4 = 1 3 ⋅ 1680 1944 x 4 = 1 3 ⋅ 35 27 x 4 = 35 27 3 x 4 = \frac{1}{\sqrt{3}} \cdot \frac{1680}{1944} x^4 = \frac{1}{\sqrt{3}} \cdot \frac{35}{27} x^4 = \frac{35}{27\sqrt{3}} x^4 = 3 1 ⋅ 1944 1680 x 4 = 3 1 ⋅ 27 35 x 4 = 27 3 35 x 4 35 3 81 x 4 \frac{35\sqrt{3}}{81} x^4 81 35 3 x 4