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JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications
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Page 1 JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications Q1: If ? ? ?? ?? =?? ???? ?? ?? ?? +?? ?? ?? +?? = ?? ?? , ?????? ( ?? , ?? )= ?? , then ?? - ?? is equal to ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 2035 Solution: ( 1 + ?? ) 11 = 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ? 0 1 ?( 1 + ?? ) 11 ???? = ? 0 1 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 - ?? ) 12 12 ] 0 1 = 11 ?? 0?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 + ? + 11 ?? 11 ?? 12 12 ] 0 1 2 12 - 1 12 = ?? 0 + ?? 1 2 + ?? 2 3 + ? + ?? 11 12 ? ( 1) Now, ? -1 0 ?( 1 + ?? ) 11 ???? = ? -1 0 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 + ?? ) 12 12 ] -1 0 = 11 ?? 0 ?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 3 + ? + 11 ?? 11 ?? 12 12 ] -1 0 ( 2) 1 12 = ?? 0 - ?? 1 2 + ?? 2 3 . (1) -( 2) = 2 12 - 2 12 = 2 [ ?? 1 2 + ?? 3 4 + ? ] ? ? ?? =0 5 ? ?? 2?? +1 2?? + 2 = 2 11 - 1 12 = 2047 12 = ?? ?? = 2047- 12 = 2035 Q2: If ? ?? =?? ???? ? ?? ?? ( ???? ?? ?? ) ?? ???? ?? ?? -?? = ?? × ?? ???? , then ?? is equal to ____ . Ans: 465 Solution: Page 2 JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications Q1: If ? ? ?? ?? =?? ???? ?? ?? ?? +?? ?? ?? +?? = ?? ?? , ?????? ( ?? , ?? )= ?? , then ?? - ?? is equal to ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 2035 Solution: ( 1 + ?? ) 11 = 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ? 0 1 ?( 1 + ?? ) 11 ???? = ? 0 1 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 - ?? ) 12 12 ] 0 1 = 11 ?? 0?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 + ? + 11 ?? 11 ?? 12 12 ] 0 1 2 12 - 1 12 = ?? 0 + ?? 1 2 + ?? 2 3 + ? + ?? 11 12 ? ( 1) Now, ? -1 0 ?( 1 + ?? ) 11 ???? = ? -1 0 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 + ?? ) 12 12 ] -1 0 = 11 ?? 0 ?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 3 + ? + 11 ?? 11 ?? 12 12 ] -1 0 ( 2) 1 12 = ?? 0 - ?? 1 2 + ?? 2 3 . (1) -( 2) = 2 12 - 2 12 = 2 [ ?? 1 2 + ?? 3 4 + ? ] ? ? ?? =0 5 ? ?? 2?? +1 2?? + 2 = 2 11 - 1 12 = 2047 12 = ?? ?? = 2047- 12 = 2035 Q2: If ? ?? =?? ???? ? ?? ?? ( ???? ?? ?? ) ?? ???? ?? ?? -?? = ?? × ?? ???? , then ?? is equal to ____ . Ans: 465 Solution: ? r=1 30 ? r 2 ( 30 C r ) 2 30 C r-1 = ? r=1 30 ?r 2 ( 31 - r r )· 30! r! ( 30 - r) ! (? 30 C r 30 C r-1 = 30 - r + 1 r = 31 - r r ) = ? r=1 30 ? ( 31 - r) 30! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ? ( 31 - r) 29! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ?( 30 - r + 1) 29 C 30-r = 30( ? r=1 30 ?( 31 - r) 29 C 30-r + ? r=1 30 ? 29 C 30-r ) = 30( 29 × 2 28 + 2 29 )= 30( 29 + 2) 2 28 = 15 × 31 × 2 29 = 465 ( 2 29 ) ?? = 465 Q3: The sum of all rational terms in the expansion of ( ?? + ?? ?? /?? + ?? ?? /?? ) ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 612 Solution: ( 1 + 2 1 3 + 3 1 2 ) 6 = 6 ?? 1 ?? 2 ?? 3 ( 1) ?? 1 ( 2) ?? 2 3 ( 3) ?? 3 2 ?? ?? ?? ?? ?? ?? 6 0 0 4 0 2 2 0 4 0 0 6 3 3 0 1 3 2 0 6 0 = 60 600 + 6 402 ( 3)+ 6 20 | 4 ( 3) 2 + 6 0 | 0 | 6 ( 3) 3 + 6 |3|3 | 0 ( 2)+ 6 1|3|2 ( 2) 1 ( 3) 1 + 6 0|6|0 ( 2) 2 = 1 + 45 + 135+ 27 + 40 + 360+ 4 = 612 Page 3 JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications Q1: If ? ? ?? ?? =?? ???? ?? ?? ?? +?? ?? ?? +?? = ?? ?? , ?????? ( ?? , ?? )= ?? , then ?? - ?? is equal to ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 2035 Solution: ( 1 + ?? ) 11 = 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ? 0 1 ?( 1 + ?? ) 11 ???? = ? 0 1 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 - ?? ) 12 12 ] 0 1 = 11 ?? 0?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 + ? + 11 ?? 11 ?? 12 12 ] 0 1 2 12 - 1 12 = ?? 0 + ?? 1 2 + ?? 2 3 + ? + ?? 11 12 ? ( 1) Now, ? -1 0 ?( 1 + ?? ) 11 ???? = ? -1 0 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 + ?? ) 12 12 ] -1 0 = 11 ?? 0 ?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 3 + ? + 11 ?? 11 ?? 12 12 ] -1 0 ( 2) 1 12 = ?? 0 - ?? 1 2 + ?? 2 3 . (1) -( 2) = 2 12 - 2 12 = 2 [ ?? 1 2 + ?? 3 4 + ? ] ? ? ?? =0 5 ? ?? 2?? +1 2?? + 2 = 2 11 - 1 12 = 2047 12 = ?? ?? = 2047- 12 = 2035 Q2: If ? ?? =?? ???? ? ?? ?? ( ???? ?? ?? ) ?? ???? ?? ?? -?? = ?? × ?? ???? , then ?? is equal to ____ . Ans: 465 Solution: ? r=1 30 ? r 2 ( 30 C r ) 2 30 C r-1 = ? r=1 30 ?r 2 ( 31 - r r )· 30! r! ( 30 - r) ! (? 30 C r 30 C r-1 = 30 - r + 1 r = 31 - r r ) = ? r=1 30 ? ( 31 - r) 30! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ? ( 31 - r) 29! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ?( 30 - r + 1) 29 C 30-r = 30( ? r=1 30 ?( 31 - r) 29 C 30-r + ? r=1 30 ? 29 C 30-r ) = 30( 29 × 2 28 + 2 29 )= 30( 29 + 2) 2 28 = 15 × 31 × 2 29 = 465 ( 2 29 ) ?? = 465 Q3: The sum of all rational terms in the expansion of ( ?? + ?? ?? /?? + ?? ?? /?? ) ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 612 Solution: ( 1 + 2 1 3 + 3 1 2 ) 6 = 6 ?? 1 ?? 2 ?? 3 ( 1) ?? 1 ( 2) ?? 2 3 ( 3) ?? 3 2 ?? ?? ?? ?? ?? ?? 6 0 0 4 0 2 2 0 4 0 0 6 3 3 0 1 3 2 0 6 0 = 60 600 + 6 402 ( 3)+ 6 20 | 4 ( 3) 2 + 6 0 | 0 | 6 ( 3) 3 + 6 |3|3 | 0 ( 2)+ 6 1|3|2 ( 2) 1 ( 3) 1 + 6 0|6|0 ( 2) 2 = 1 + 45 + 135+ 27 + 40 + 360+ 4 = 612 Q4: If ?? = ?? + ? ?? =?? ?? ?( -?? ) ?? -?? ???? ?? ?? ?? -?? , then the distance of the point ( ???? , v?? ) from the line ???? - v?? ?? + ?? = ?? is ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 5 Solution: ?? = 1 + ? r=1 6 ?( -1) r-112 C 2r-1 3 r-1 ?? = 1 + ? r=1 6 ? 12 C 2r-1 ( v 3i) 2t-1 v 3i i = iota, let v3i = x ?? = 1 + 1 v3i ( 12 C 1 x + 12 C 3 x 3 + ? . 12 C 11 x 11 ) = 1 + 1 v3i ( ( 1 + v3i) 12 - ( 1 - v3i) 12 2 ) = 1 + 1 v3i ( ( -2?? 2 ) 12 - ( 2?? ) 12 2 ) = 1 so distance of ( 12, v3) from ?? - v3?? + 1 = 0 is 12 - 3 + 1 2 = 5 Q5: Let ( ?? + ?? + ?? ?? ) ???? = ?? ?? + ?? ?? ?? + ?? ?? ?? ?? + ? + ?? ???? ?? ???? . If ( ?? ?? + ?? ?? + ?? ?? + ? + ?? ???? )- ???? ?? ?? = ?????? ?? , then ?? is equal to ____ . Ans: 239 Solution: Let ?? ( ?? )= ( 1 + ?? + ?? 2 ) 10 = ? ?? =0 20 ??? ?? ?? ?? The sum of odd coefficients: ?? odd = ?? 1 + ?? 3 + ?? 5 + ? + ?? 19 Subtracting 11?? 2 from above will give the answer ?? odd = ?? ( 1)- ?? ( -1) 2 ?? ( 1)= ( 1 + 1 + 1) 10 = 3 10 ?? ( -1)= ( 1 - 1 + 1) 10 = ( 1) 10 = 1 ?? odd = ? odd ??? ?? = 3 10 - 1 2 Now for ?? 2 1 + ?? + ?? 2 = 1 - ?? 3 1 - ?? ? ?? ( ?? )= ( 1 - ?? 3 1 - ?? ) = ( 1 - ?? 3 ) 10 ( 1 - ?? ) 10 Now use: ( 1 - ?? 3 ) 10 = ? ?? =0 10 ?( -1) ?? ( 10 ?? )?? 3?? ( 1 - ?? ) -10 = ? ?? =0 8 ?( ?? + 9 9 )?? ?? So Page 4 JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications Q1: If ? ? ?? ?? =?? ???? ?? ?? ?? +?? ?? ?? +?? = ?? ?? , ?????? ( ?? , ?? )= ?? , then ?? - ?? is equal to ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 2035 Solution: ( 1 + ?? ) 11 = 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ? 0 1 ?( 1 + ?? ) 11 ???? = ? 0 1 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 - ?? ) 12 12 ] 0 1 = 11 ?? 0?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 + ? + 11 ?? 11 ?? 12 12 ] 0 1 2 12 - 1 12 = ?? 0 + ?? 1 2 + ?? 2 3 + ? + ?? 11 12 ? ( 1) Now, ? -1 0 ?( 1 + ?? ) 11 ???? = ? -1 0 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 + ?? ) 12 12 ] -1 0 = 11 ?? 0 ?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 3 + ? + 11 ?? 11 ?? 12 12 ] -1 0 ( 2) 1 12 = ?? 0 - ?? 1 2 + ?? 2 3 . (1) -( 2) = 2 12 - 2 12 = 2 [ ?? 1 2 + ?? 3 4 + ? ] ? ? ?? =0 5 ? ?? 2?? +1 2?? + 2 = 2 11 - 1 12 = 2047 12 = ?? ?? = 2047- 12 = 2035 Q2: If ? ?? =?? ???? ? ?? ?? ( ???? ?? ?? ) ?? ???? ?? ?? -?? = ?? × ?? ???? , then ?? is equal to ____ . Ans: 465 Solution: ? r=1 30 ? r 2 ( 30 C r ) 2 30 C r-1 = ? r=1 30 ?r 2 ( 31 - r r )· 30! r! ( 30 - r) ! (? 30 C r 30 C r-1 = 30 - r + 1 r = 31 - r r ) = ? r=1 30 ? ( 31 - r) 30! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ? ( 31 - r) 29! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ?( 30 - r + 1) 29 C 30-r = 30( ? r=1 30 ?( 31 - r) 29 C 30-r + ? r=1 30 ? 29 C 30-r ) = 30( 29 × 2 28 + 2 29 )= 30( 29 + 2) 2 28 = 15 × 31 × 2 29 = 465 ( 2 29 ) ?? = 465 Q3: The sum of all rational terms in the expansion of ( ?? + ?? ?? /?? + ?? ?? /?? ) ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 612 Solution: ( 1 + 2 1 3 + 3 1 2 ) 6 = 6 ?? 1 ?? 2 ?? 3 ( 1) ?? 1 ( 2) ?? 2 3 ( 3) ?? 3 2 ?? ?? ?? ?? ?? ?? 6 0 0 4 0 2 2 0 4 0 0 6 3 3 0 1 3 2 0 6 0 = 60 600 + 6 402 ( 3)+ 6 20 | 4 ( 3) 2 + 6 0 | 0 | 6 ( 3) 3 + 6 |3|3 | 0 ( 2)+ 6 1|3|2 ( 2) 1 ( 3) 1 + 6 0|6|0 ( 2) 2 = 1 + 45 + 135+ 27 + 40 + 360+ 4 = 612 Q4: If ?? = ?? + ? ?? =?? ?? ?( -?? ) ?? -?? ???? ?? ?? ?? -?? , then the distance of the point ( ???? , v?? ) from the line ???? - v?? ?? + ?? = ?? is ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 5 Solution: ?? = 1 + ? r=1 6 ?( -1) r-112 C 2r-1 3 r-1 ?? = 1 + ? r=1 6 ? 12 C 2r-1 ( v 3i) 2t-1 v 3i i = iota, let v3i = x ?? = 1 + 1 v3i ( 12 C 1 x + 12 C 3 x 3 + ? . 12 C 11 x 11 ) = 1 + 1 v3i ( ( 1 + v3i) 12 - ( 1 - v3i) 12 2 ) = 1 + 1 v3i ( ( -2?? 2 ) 12 - ( 2?? ) 12 2 ) = 1 so distance of ( 12, v3) from ?? - v3?? + 1 = 0 is 12 - 3 + 1 2 = 5 Q5: Let ( ?? + ?? + ?? ?? ) ???? = ?? ?? + ?? ?? ?? + ?? ?? ?? ?? + ? + ?? ???? ?? ???? . If ( ?? ?? + ?? ?? + ?? ?? + ? + ?? ???? )- ???? ?? ?? = ?????? ?? , then ?? is equal to ____ . Ans: 239 Solution: Let ?? ( ?? )= ( 1 + ?? + ?? 2 ) 10 = ? ?? =0 20 ??? ?? ?? ?? The sum of odd coefficients: ?? odd = ?? 1 + ?? 3 + ?? 5 + ? + ?? 19 Subtracting 11?? 2 from above will give the answer ?? odd = ?? ( 1)- ?? ( -1) 2 ?? ( 1)= ( 1 + 1 + 1) 10 = 3 10 ?? ( -1)= ( 1 - 1 + 1) 10 = ( 1) 10 = 1 ?? odd = ? odd ??? ?? = 3 10 - 1 2 Now for ?? 2 1 + ?? + ?? 2 = 1 - ?? 3 1 - ?? ? ?? ( ?? )= ( 1 - ?? 3 1 - ?? ) = ( 1 - ?? 3 ) 10 ( 1 - ?? ) 10 Now use: ( 1 - ?? 3 ) 10 = ? ?? =0 10 ?( -1) ?? ( 10 ?? )?? 3?? ( 1 - ?? ) -10 = ? ?? =0 8 ?( ?? + 9 9 )?? ?? So ?? ( ?? )= ( ? ?? =0 10 ?( -1) ?? ( 10 ?? )?? 3?? )· ( ? ?? =0 8 ?( ?? + 9 9 )?? ?? ) Only the term with ?? 0 from the first sum (i.e., ?? = 0 ) can contribute to ?? 2 , since all other ?? = 1 gives ?? 3?? = ?? 3 From ( 1 - ?? 3 ) 10 : the ?? 0 term is ( 10 0 )= 1 From ( 1 - ?? ) -10 : the coefficient of ?? 2 is ( 2 + 9 9 )= ( 11 9 )= 55 Hence, ?? 2 = 1.55 = 55 Now, ?? odd - 11?? 2 = 3 10 -1 2 - 11 · 55 = 121 ?? 3 10 = 59049 So: ?? = 59049- 1 2 - 605 = 59048 2 - 605 = 29524- 605 = 28919 So: 121 ?? = 28919 ? ?? = 28919 121 = 239 Q6: The sum of the series ?? × ?? × ???? ?? ?? - ?? × ?? × ???? ?? ?? + ?? × ?? × ???? ?? ?? - ?? × ?? × ???? ?? ?? + ? ? + ???? × ???? × ???? ?? ???? , is equal to ____ . JEE Main 2025 (Online) 7th April Evening Shift Ans: 34 Solution: ( 1 - x) 20 = 20 C 0 - 20 C 1 x + 20 C 2 x 2 … … + 20 C 20 x 20 ( 1 - x) 20 x 2 = 20 C 0 x 2 - 20 C 1 x + 20 C 2 - 20 C 3 x + 20 C 4 x 2 … Diff twice and put x = 1 = 6 - 20 C 1 ( 2)+ A A = 40 - 6 = 34 Q7: The product of the last two digits of (1919) ???????? is JEE Main 2025 (Online) 8th April Evening Shift Ans: 63 Solution: ( 1919) 1919 = ( 1920- 1) 1919 = 1919 C 0 ( 1920 ) 1919 - 1919 C 1 ( 1920 ) 1918 + ?. + 1919 C 1918 ( 1920 ) 1 - 1919 C 1919 = 100 ?? + 1919 × 1920- 1 Page 5 JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications Q1: If ? ? ?? ?? =?? ???? ?? ?? ?? +?? ?? ?? +?? = ?? ?? , ?????? ( ?? , ?? )= ?? , then ?? - ?? is equal to ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 2035 Solution: ( 1 + ?? ) 11 = 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ? 0 1 ?( 1 + ?? ) 11 ???? = ? 0 1 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 - ?? ) 12 12 ] 0 1 = 11 ?? 0?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 + ? + 11 ?? 11 ?? 12 12 ] 0 1 2 12 - 1 12 = ?? 0 + ?? 1 2 + ?? 2 3 + ? + ?? 11 12 ? ( 1) Now, ? -1 0 ?( 1 + ?? ) 11 ???? = ? -1 0 ?( 11 ?? 0 + 11 ?? 1 ?? + 11 ?? 2 ?? 2 + ? + 11 ?? 11 ?? 11 ) ???? ( 1 + ?? ) 12 12 ] -1 0 = 11 ?? 0 ?? + 11 ?? 1 ?? 2 2 + 11 ?? 2 ?? 3 3 + ? + 11 ?? 11 ?? 12 12 ] -1 0 ( 2) 1 12 = ?? 0 - ?? 1 2 + ?? 2 3 . (1) -( 2) = 2 12 - 2 12 = 2 [ ?? 1 2 + ?? 3 4 + ? ] ? ? ?? =0 5 ? ?? 2?? +1 2?? + 2 = 2 11 - 1 12 = 2047 12 = ?? ?? = 2047- 12 = 2035 Q2: If ? ?? =?? ???? ? ?? ?? ( ???? ?? ?? ) ?? ???? ?? ?? -?? = ?? × ?? ???? , then ?? is equal to ____ . Ans: 465 Solution: ? r=1 30 ? r 2 ( 30 C r ) 2 30 C r-1 = ? r=1 30 ?r 2 ( 31 - r r )· 30! r! ( 30 - r) ! (? 30 C r 30 C r-1 = 30 - r + 1 r = 31 - r r ) = ? r=1 30 ? ( 31 - r) 30! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ? ( 31 - r) 29! ( r - 1) ! ( 30 - r) ! = 30? r=1 30 ?( 30 - r + 1) 29 C 30-r = 30( ? r=1 30 ?( 31 - r) 29 C 30-r + ? r=1 30 ? 29 C 30-r ) = 30( 29 × 2 28 + 2 29 )= 30( 29 + 2) 2 28 = 15 × 31 × 2 29 = 465 ( 2 29 ) ?? = 465 Q3: The sum of all rational terms in the expansion of ( ?? + ?? ?? /?? + ?? ?? /?? ) ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 612 Solution: ( 1 + 2 1 3 + 3 1 2 ) 6 = 6 ?? 1 ?? 2 ?? 3 ( 1) ?? 1 ( 2) ?? 2 3 ( 3) ?? 3 2 ?? ?? ?? ?? ?? ?? 6 0 0 4 0 2 2 0 4 0 0 6 3 3 0 1 3 2 0 6 0 = 60 600 + 6 402 ( 3)+ 6 20 | 4 ( 3) 2 + 6 0 | 0 | 6 ( 3) 3 + 6 |3|3 | 0 ( 2)+ 6 1|3|2 ( 2) 1 ( 3) 1 + 6 0|6|0 ( 2) 2 = 1 + 45 + 135+ 27 + 40 + 360+ 4 = 612 Q4: If ?? = ?? + ? ?? =?? ?? ?( -?? ) ?? -?? ???? ?? ?? ?? -?? , then the distance of the point ( ???? , v?? ) from the line ???? - v?? ?? + ?? = ?? is ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 5 Solution: ?? = 1 + ? r=1 6 ?( -1) r-112 C 2r-1 3 r-1 ?? = 1 + ? r=1 6 ? 12 C 2r-1 ( v 3i) 2t-1 v 3i i = iota, let v3i = x ?? = 1 + 1 v3i ( 12 C 1 x + 12 C 3 x 3 + ? . 12 C 11 x 11 ) = 1 + 1 v3i ( ( 1 + v3i) 12 - ( 1 - v3i) 12 2 ) = 1 + 1 v3i ( ( -2?? 2 ) 12 - ( 2?? ) 12 2 ) = 1 so distance of ( 12, v3) from ?? - v3?? + 1 = 0 is 12 - 3 + 1 2 = 5 Q5: Let ( ?? + ?? + ?? ?? ) ???? = ?? ?? + ?? ?? ?? + ?? ?? ?? ?? + ? + ?? ???? ?? ???? . If ( ?? ?? + ?? ?? + ?? ?? + ? + ?? ???? )- ???? ?? ?? = ?????? ?? , then ?? is equal to ____ . Ans: 239 Solution: Let ?? ( ?? )= ( 1 + ?? + ?? 2 ) 10 = ? ?? =0 20 ??? ?? ?? ?? The sum of odd coefficients: ?? odd = ?? 1 + ?? 3 + ?? 5 + ? + ?? 19 Subtracting 11?? 2 from above will give the answer ?? odd = ?? ( 1)- ?? ( -1) 2 ?? ( 1)= ( 1 + 1 + 1) 10 = 3 10 ?? ( -1)= ( 1 - 1 + 1) 10 = ( 1) 10 = 1 ?? odd = ? odd ??? ?? = 3 10 - 1 2 Now for ?? 2 1 + ?? + ?? 2 = 1 - ?? 3 1 - ?? ? ?? ( ?? )= ( 1 - ?? 3 1 - ?? ) = ( 1 - ?? 3 ) 10 ( 1 - ?? ) 10 Now use: ( 1 - ?? 3 ) 10 = ? ?? =0 10 ?( -1) ?? ( 10 ?? )?? 3?? ( 1 - ?? ) -10 = ? ?? =0 8 ?( ?? + 9 9 )?? ?? So ?? ( ?? )= ( ? ?? =0 10 ?( -1) ?? ( 10 ?? )?? 3?? )· ( ? ?? =0 8 ?( ?? + 9 9 )?? ?? ) Only the term with ?? 0 from the first sum (i.e., ?? = 0 ) can contribute to ?? 2 , since all other ?? = 1 gives ?? 3?? = ?? 3 From ( 1 - ?? 3 ) 10 : the ?? 0 term is ( 10 0 )= 1 From ( 1 - ?? ) -10 : the coefficient of ?? 2 is ( 2 + 9 9 )= ( 11 9 )= 55 Hence, ?? 2 = 1.55 = 55 Now, ?? odd - 11?? 2 = 3 10 -1 2 - 11 · 55 = 121 ?? 3 10 = 59049 So: ?? = 59049- 1 2 - 605 = 59048 2 - 605 = 29524- 605 = 28919 So: 121 ?? = 28919 ? ?? = 28919 121 = 239 Q6: The sum of the series ?? × ?? × ???? ?? ?? - ?? × ?? × ???? ?? ?? + ?? × ?? × ???? ?? ?? - ?? × ?? × ???? ?? ?? + ? ? + ???? × ???? × ???? ?? ???? , is equal to ____ . JEE Main 2025 (Online) 7th April Evening Shift Ans: 34 Solution: ( 1 - x) 20 = 20 C 0 - 20 C 1 x + 20 C 2 x 2 … … + 20 C 20 x 20 ( 1 - x) 20 x 2 = 20 C 0 x 2 - 20 C 1 x + 20 C 2 - 20 C 3 x + 20 C 4 x 2 … Diff twice and put x = 1 = 6 - 20 C 1 ( 2)+ A A = 40 - 6 = 34 Q7: The product of the last two digits of (1919) ???????? is JEE Main 2025 (Online) 8th April Evening Shift Ans: 63 Solution: ( 1919) 1919 = ( 1920- 1) 1919 = 1919 C 0 ( 1920 ) 1919 - 1919 C 1 ( 1920 ) 1918 + ?. + 1919 C 1918 ( 1920 ) 1 - 1919 C 1919 = 100 ?? + 1919 × 1920- 1 = 100 ?? + 3684480- 1 = 100 ?? + ? … … . .79 (last two digit) ? Number having last two digit 79 ? Product of last two digit 63 Q8: Let ?? , ?? , ?? and ?? be the coefficients of ?? ?? , ?? ?? , ?? ?? and ?? respectively in the expansion of ( ?? + v?? ?? - ?? ) ?? + ( ?? - v?? ?? - ?? ) ?? , ?? > ?? . If ?? and ?? satisfy the equations ???? + ???? = ???? , ???? + ???? = ???? , then ?? + ?? equals : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 4 B. 3 C. 5 D. 8 Ans: C Solution: To find the sum of ?? and ?? , we first need to expand the expression: ( ?? + v ?? 3 - 1) 5 + ( ?? - v ?? 3 - 1) 5 Using the Binomial Theorem, the expansion yields: = 2( 5 ?? 0 · ?? 5 + 5 ?? 2 · ?? 3 ( ?? 3 - 1)+ 5 ?? 4 · ?? ( ?? 3 - 1) 2 ) Simplifying this, we obtain: = 2( 5?? 7 + 10?? 6 + ?? 5 - 10?? 4 - 10?? 3 + 5?? ) From this expansion, we can identify the coefficients: The coefficient of ?? 7 is ?? = 10 The coefficient of ?? 5 is ?? = 2 The coefficient of ?? 3 is ?? = -20 The coefficient of ?? is ?? = 10 Given the equations: ???? + ???? = 18 ???? + ???? = 20 Substituting in the coefficients: 10?? + 2?? = 18 -20?? + 10?? = 20 By solving these equations, we find: From 10?? + 2?? = 18, simplify to 5?? + ?? = 9. From -20?? + 10?? = 20, simplify to -2?? + ?? = 2. Solving these linear equations simultaneously, we find: Subtracting equation 2 from equation 1 : 5?? + ?? = 9 -( -2?? + ?? = 2)Read More
FAQs on JEE Main Previous Year Questions (2025): Binomial Theorem & its Applications
| 1. What is the Binomial Theorem and how is it applied in algebra? |  |
Ans. The Binomial Theorem provides a formula for expanding expressions of the form (a + b)ⁿ, where n is a non-negative integer. It states that (a + b)ⁿ = Σ (n choose k) * a^(n-k) * b^k, where k ranges from 0 to n. This theorem is widely used in algebra to simplify polynomial expressions, compute probabilities in statistics, and in combinatorial problems to find the number of ways to choose items from a set.
| 2. How can the Binomial Theorem be used to find coefficients in polynomial expansions? | |
Ans. The coefficients in the expansion of (a + b)ⁿ can be found using the Binomial Theorem, where each coefficient corresponds to "n choose k" for k = 0 to n. Specifically, the coefficient of a^(n-k) * b^k in the expansion is given by the binomial coefficient C(n, k) = n! / (k!(n-k)!), which denotes the number of ways to choose k successes in n trials, making it valuable in both algebra and probability.
| 3. Can the Binomial Theorem be applied to negative or fractional exponents? | |
Ans. Yes, the Binomial Theorem can be extended to negative and fractional exponents using the generalized Binomial series. For a negative integer exponent, (1 + x)⁻ⁿ can be expanded as Σ (n choose k) * (-1)^(k) * x^k for k = 0 to ∞. For fractional exponents, the series converges under certain conditions and can be expressed similarly, allowing for the expansion of (a + b)^(m/n) under specific restrictions.
| 4. What are some practical applications of the Binomial Theorem in real-life scenarios? | |
Ans. The Binomial Theorem has numerous practical applications, including in finance for calculating compound interest, in statistics for determining probabilities in binomial distributions, and in computer science for algorithm analysis. It is also used in engineering for modeling situations involving random variables and in various fields of science for computations involving series and expansions.
| 5. How do you solve problems involving the Binomial Theorem in competitive exams? | |
Ans. To solve problems involving the Binomial Theorem in competitive exams, first understand the given expression and identify the values of a, b, and n. Apply the Binomial Theorem formula to expand the expression and then extract the required coefficient or term. Practicing past exam questions helps in familiarizing with different types of problems, including finding specific terms and coefficients in expansions, which is crucial for quick and accurate problem-solving during exams.
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