BlogIIT-JEEJEE Binomial Theorem Previous Year Question Paper With Solutions

JEE Binomial Theorem Previous Year Question Paper With Solutions

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Question 1: If the total coefficients of all equal forces of x in a product (1 + x + x2 + … + x2n) (1 – x + x2 – x3 +… + x2n) are 61, get the value n.

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    Solution:

    Allow (1 + x + x2 +… + x2n) (1 – x + x2 – x3 +… + x2n) = a0 + ax + a2x2 + a3x3 + ……

    Set x = 1 => 2n + 1 = a0 + a1 + a2 + a3 + ……… (i)

    Put x = -1 => 2n + 1 = a0 – a1 + a2 – a3 + ……… .. (ii)

    Add (i) and (ii), we get

    2n + 1 = 61

    or n = 30

    Question 2: If α and β are coefficients of x4 and x2 respectively in addition (x + √ (x2 – 1))) 6 + (x – √ (x2 – 1) 6), then -ke:

    (a) α + β = -30

    (b) α− β = −132

    (c) α− β = 60

    (d) α + β = 60

    Solution: (x + √ (x2 – 1)) 6 + (x – √ (x2 – 1) 6

    = 2 [6C0 x6 + 6C2x4 (x2 – 1) + 6C4x2 (x2 – 1) 2 + 6C6 (x2 – 1) 3]

    = 2 [32x6 – 48x4 + 18x2 – 1]

    => Now, α=- 96 and Β=-36

    => α – β = -132

    Question 3: Find the coefficient of x4 in addition to (1 + x + x2) 10.

    Solution:

    The general term for a given speech is,

    {10!} \ {P! Q! R!} x {q + 2r}

     

    Here, q + 2r = 4

    At p = 6, q = 4, r = 0, coefficient = 10! / [6! x 4!] = 210

    At p = 7, q = 2, r = 1, coefficient = 10! / [7! x2! x 1!] = 360

    At p = 8, q = 0, r = 2, coefficient = 10! / [8! x2!] = 45

    Therefore, total = 615

    Question 4: If x is positive, the first negative word in addition to

    (1 + x) 27/5 by

    (a) term 5 (b) term 6 (c) term 7 (d) term 8

    Answer: (d)

    Solution:

    We know, T{r + 1} = {n (n-1)… (n-r + 1)} \ {r!} X r

    Therefore, n – r + 1 <0

    => 27/5 + 1 <r

    or r> 6

    Here r = 7

    Thus, Tr + 1 <0

    So Tr + 1 = T7 + 1 = T8 = term 8 is negative.

    Question 5: Deposit number greater than (1 + 0.0001) 10000

    (a) 4 (b) 5 (c) 2 (d) 3

    Answer: (d)

    Solution:

    (1 + 0.0001) 10000 similar form (1 + 1 / n) n where n = 10000.

    Using binomial extensions, we have

    (1 + 0.0001) 10000 = (1 + 1 / n) n

    = 1 + n x 1 / n + n (n-1) / 2! x 1 / n2 + [n (n-1) (n-2)] / 3! X 1 / n3 + …… ..

    = 1 + 1 + 1/2! (1 – 1 / n) + 1/3! (1 – 1 / n) + (1 – 2 / n) + ……

    <1 + 1/1! + 1/2! + 1/3! + ….. + 1 / (9999)!

    = 1 + 1/1! + 1/2! + 1/3! + …… .∞

    = e <3

    Question 6: The coefficients of xp and xq in the expression (1 + x) p + q

    (a) Equal

    (b) It is equal to the opposite sign

    (c) Reconciliation with each other

    (d) None of this

    Answer: (a)

    Solution:

    The coefficients of xp and xq in the expression (1 + x) p + q

    We know, yp + 1 = p + qCp xp and yq + 1 = p + qCq xq

    Coefficient of xp = p + qCp = [p + q]! / P! Q!

    and

    Coefficient of xq = p + qCq = [p + q]! / P! Q!

    Therefore, the coefficients of xp and xq in the expression (1 + x) p + q are equal.

    Question 7: If an=√7+(√7+(√7+……
    having n radical signs then in the form of true mathematical inventions

    (a) <7 for all n ≥ 1

    (b) an> 7 for all n ≥ 1

    (c) 3 for all n ≥ 1

    (d) <4 for all n ≥ 1

    Answer: (c)

    Solution: an=√7+(√7+(√7+……

    => i = √ (7 + an)

    => an2 – an – 7 = 0

    Solving above the quadratic equation, we find

    i = [1 ± √29] / 2

    But> 0, therefore, is = [1 + √29] / 2> 3

    => i> 3

    Question 8: If the sum of the coefficients in the extension (a + b) n is 4096, then the largest coefficient in the extension is

    1. 1594
    2. 924
    3. 792
    4. 2924

    Answer: (b)

    Solution:

    Binomial Expansion: (x + y) n = C0 xn + C1 xn-1 y + C2 xn-2y2 + …. + Cnyn

    setting x=y=1,then

    2n = C0 + C1 + C2 +…. + Cn = 4096 = 212

    => n = 12

    As n is present even here, so the coefficient of the largest term states

    nCn / 2 = 12C12 / 2 = 12C6

    = 12/6 x 11/5 x 10/4 x 9/3 x8 / 2 x 7/1

    = 924

    Question 9: The co-efficient amount of all the unusual term terms in the extension of

    (x + √ (x3-1) 5 + (x – √ (x3-1) 5, (x> 1)

    (a) 0 (b) 2 (c) 1 (d) -1

    Answer:

    Solution:

    Allow y = √ (x3-1)

    Thus, the given expression is reduced to (x + y) 5 + (x – y) 5

    Using binomial extensions, we have

    (x + y) 5 + (x – y) 5 = x5 + 5C1 x4y + 5C2 x3y2 + 5C3 x2y3 + 5C4 xy4 + 5C5 y5 + x55C1 x4y + 5C2 x3y25C3 x2y3 + 5C5

    = 2 (x5 + 10 x3y2 + 5xy4)

    = 2 (x5 + 10 x3 (x3 – 1) + 5x (x3 – 1) 2)

    = 2 {10x6 + 5x7 + x5 – 10x3 – 10x4 + 5x}

    Total coefficient of unconventional x: 2 {5 + 1 – 10 + 5} = 2

    Question 10: Extend quote (2x-3) 6 using the binomial theorem.

    Solution:

    Presentation: (2x-3) 6

    Using the binomial theorem, expression (2x-3) 6 can be extended as follows:

    (2x-3) 6 = 6C0 (2x) 66C1 (2x) 5 (3) + 6C2 (2x) 4 (3) 2 6C3 (2x) 3 (3) 3 + 6C4 (2x) 2 (3) 46C5 (2x) (3) 5 + 6C6 (3) 6

    (2x-3) 6 = 64x6 – 6 (32x5) (3) +15 (16x4) (9) – 20 (8x3) (27) +15 (4x2) (81) – 6 (2x) (243) + 729

    (2x-3) 6 = 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

    Thus, the binomial extension of the given expression (2x-3) 6 is 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.

    FAQs

    Is binomial theorem easy for JEE?

    Some basic introduction to the Binomial Theorem introduction is included in the eighth-grade syllabus as well. The concept of the Binomial Theorem becomes easier to understand once students become familiar with its output.

    Is the binomial theorem important to JEE?

    Binomial Theorem is one of the most important chapters of Algebra in the JEE syllabus. !

     

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