ক) AI-তৈরি নমুনা উত্তর
প্রদত্ত সমীকরণ:2 x 2 − 2 ( d + p ) x + d 2 + p 2 = 0 2x^2 - 2(d+p)x + d^2 + p^2 = 0 2 x 2 − 2 ( d + p ) x + d 2 + p 2 = 0 D = B 2 − 4 A C D = B^2 - 4AC D = B 2 − 4 A C A = 2 A = 2 A = 2 B = − 2 ( d + p ) B = -2(d+p) B = − 2 ( d + p ) C = d 2 + p 2 C = d^2+p^2 C = d 2 + p 2 ∴ D = { − 2 ( d + p ) } 2 − 4 ⋅ 2 ⋅ ( d 2 + p 2 ) \therefore D = \{-2(d+p)\}^2 - 4 \cdot 2 \cdot (d^2+p^2) ∴ D = { − 2 ( d + p ) } 2 − 4 ⋅ 2 ⋅ ( d 2 + p 2 ) = 4 ( d 2 + 2 d p + p 2 ) − 8 ( d 2 + p 2 ) = 4(d^2 + 2dp + p^2) - 8(d^2+p^2) = 4 ( d 2 + 2 d p + p 2 ) − 8 ( d 2 + p 2 ) = 4 ( d 2 + 2 d p + p 2 − 2 d 2 − 2 p 2 ) = 4(d^2 + 2dp + p^2 - 2d^2 - 2p^2) = 4 ( d 2 + 2 d p + p 2 − 2 d 2 − 2 p 2 ) = 4 ( − d 2 + 2 d p − p 2 ) = 4(-d^2 + 2dp - p^2) = 4 ( − d 2 + 2 d p − p 2 ) = − 4 ( d 2 − 2 d p + p 2 ) = -4(d^2 - 2dp + p^2) = − 4 ( d 2 − 2 d p + p 2 ) = − 4 ( d − p ) 2 = -4(d-p)^2 = − 4 ( d − p ) 2
যেহেতু d ≠ p d \neq p d = p ( d − p ) 2 > 0 (d-p)^2 > 0 ( d − p ) 2 > 0 D = − 4 ( d − p ) 2 < 0 D = -4(d-p)^2 < 0 D = − 4 ( d − p ) 2 < 0 D < 0 D < 0 D < 0
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপক অনুসারে, f ( x ) = a x 2 + b x + c f(x) = ax^2 + bx + c f ( x ) = a x 2 + b x + c f ( x ) = 0 ⟹ a x 2 + b x + c = 0 f(x) = 0 \implies ax^2 + bx + c = 0 f ( x ) = 0 ⟹ a x 2 + b x + c = 0 b = c b = c b = c a x 2 + b x + b = 0 ax^2 + bx + b = 0 a x 2 + b x + b = 0 m m m n n n p : q p : q p : q m n = p q \frac{m}{n} = \frac{p}{q} n m = q p m + n = − b a m + n = -\frac{b}{a} m + n = − a b m n = b a mn = \frac{b}{a} mn = a b
বামপক্ষ = p q + q p + c a = \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = q p + p q + a c = m n + n m + b a = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \sqrt{\frac{b}{a}} = n m + m n + a b b = c b = c b = c = m + n m n + b a = \frac{m + n}{\sqrt{mn}} + \sqrt{\frac{b}{a}} = mn m + n + a b = − b a b a + b a = \frac{-\frac{b}{a}}{\sqrt{\frac{b}{a}}} + \sqrt{\frac{b}{a}} = a b − a b + a b = − b a + b a = -\sqrt{\frac{b}{a}} + \sqrt{\frac{b}{a}} = − a b + a b = 0 = = 0 = = 0 =
গ) AI-তৈরি নমুনা উত্তর
উদ্দীপক মতে, f ( x ) = a x 2 + b x + c = 0 f(x) = ax^2+bx+c=0 f ( x ) = a x 2 + b x + c = 0 α \alpha α β \beta β ∴ α + β = − b a \therefore \alpha + \beta = -\frac{b}{a} ∴ α + β = − a b α β = c a \alpha\beta = \frac{c}{a} α β = a c
নতুন সমীকরণের মূলদ্বয় ( α + 1 β ) \left(\alpha + \frac{1}{\beta}\right) ( α + β 1 ) ( β + 1 α ) \left(\beta + \frac{1}{\alpha}\right) ( β + α 1 )
মূলদ্বয়ের যোগফল:S = ( α + 1 β ) + ( β + 1 α ) S = \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) S = ( α + β 1 ) + ( β + α 1 ) = ( α + β ) + ( 1 α + 1 β ) = (\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = ( α + β ) + ( α 1 + β 1 ) = ( α + β ) + α + β α β = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = ( α + β ) + α β α + β = ( α + β ) ( 1 + 1 α β ) = (\alpha + \beta)\left(1 + \frac{1}{\alpha\beta}\right) = ( α + β ) ( 1 + α β 1 ) = ( − b a ) ( 1 + a c ) = \left(-\frac{b}{a}\right)\left(1 + \frac{a}{c}\right) = ( − a b ) ( 1 + c a ) = − b a ⋅ a + c c = − b ( a + c ) a c = -\frac{b}{a} \cdot \frac{a+c}{c} = -\frac{b(a+c)}{ac} = − a b ⋅ c a + c = − a c b ( a + c )
মূলদ্বয়ের গুণফল:P = ( α + 1 β ) ( β + 1 α ) P = \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) P = ( α + β 1 ) ( β + α 1 ) = α β + 1 + 1 + 1 α β = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = α β + 1 + 1 + α β 1 = α β + 1 α β + 2 = \alpha\beta + \frac{1}{\alpha\beta} + 2 = α β + α β 1 + 2 = c a + a c + 2 = \frac{c}{a} + \frac{a}{c} + 2 = a c + c a + 2 = c 2 + a 2 + 2 a c a c = ( a + c ) 2 a c = \frac{c^2 + a^2 + 2ac}{ac} = \frac{(a+c)^2}{ac} = a c c 2 + a 2 + 2 a c = a c ( a + c ) 2
নির্ণেয় দ্বিঘাত সমীকরণটি হলো:x 2 − S x + P = 0 x^2 - S x + P = 0 x 2 − S x + P = 0 ⟹ x 2 − ( − b ( a + c ) a c ) x + ( a + c ) 2 a c = 0 \implies x^2 - \left(-\frac{b(a+c)}{ac}\right)x + \frac{(a+c)^2}{ac} = 0 ⟹ x 2 − ( − a c b ( a + c ) ) x + a c ( a + c ) 2 = 0 ⟹ x 2 + b ( a + c ) a c x + ( a + c ) 2 a c = 0 \implies x^2 + \frac{b(a+c)}{ac}x + \frac{(a+c)^2}{ac} = 0 ⟹ x 2 + a c b ( a + c ) x + a c ( a + c ) 2 = 0 ⟹ a c x 2 + b ( a + c ) x + ( a + c ) 2 = 0 \implies acx^2 + b(a+c)x + (a+c)^2 = 0 ⟹ a c x 2 + b ( a + c ) x + ( a + c ) 2 = 0
উত্তর: a c x 2 + b ( a + c ) x + ( a + c ) 2 = 0 acx^2 + b(a+c)x + (a+c)^2 = 0 a c x 2 + b ( a + c ) x + ( a + c ) 2 = 0