# CBSE Solutions for Class 11 English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

If in the expansion of (1 + y)20, the coefficients of rth and (r + 4)th terms are equal, then r is equal to

જવાબ : 9

The term without y in the expansion of (2y − 1/2y2)12 is

જવાબ : 7920

The middle term in the expansion of (2y2/3 + 3/2y2)10 is

જવાબ : 252

If an the expansion of (1+y)15, the coefficients of (2r+3)th and (r−1)th terms are equal, then the value of r is

જવાબ : 5

In the expansion of (y2−1/3y)9, the term without y  is equal to

જવાબ : 28/243

The coefficient of x−17 in the expansion of (x4−1/x3)15 is

જવાબ : −1365

The number of irrational terms in the expansion of (41/5+71/10)45 is

જવાબ : 41

If A and B are the sums of odd and even terms respectively in the expansion of (y + a)n, then (y + a)2n − (y − a)2n is equal to

જવાબ : 4 AB

If in the expansion of (x + y)n and (x + y)n + 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, then n is

જવાબ : 5

The coefficient of 1/y in the expansion of (1+y)n (1+1/y)n is

જવાબ : 2n ! / [(n-1) ! (n+1) !]

If T2/T3 in the expansion of (a+b)n  and T3/T4 in the expansion of (a+b)n+3 are equal, then n =

જવાબ : 5

The total number of terms in the expansion of (y+a)100 + (y−a)100 after simplification is

જવાબ : 51

The coefficient of x4 in (x/2 – 3/x2)10 is

જવાબ : 405/256

If the coefficient of y in (y2+λ/y)5 is 270, then λ=

જવાબ : 3

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of (x+a)n are A and B respectively, then the value of (x2−a2)n is

જવાબ : A2 − B2​​​​​​​

In the expansion of ( ½ y1/3 + y−1/5)8, the term independent of y is

જવાબ : T6

If in the expansion of (1 + x)n, the coefficients of 5th, 6th and 7th terms are in A.P., then n is equal to

જવાબ : 7, 14

In the expansion of (y − 1/3y2)9, the term independent of y is

જવાબ : T4

If in the expansion of (y4 − 1/y3)15, x−17 occurs in rth term, then

જવાબ : r = 12

The general term of the expansion (p + q)n is

જવાબ : Tr+1 = nCr × pn-r × qr

If a and b are the roots of the equation y² – y + 1 = 0 then the value of a2009 + b2009 is

જવાબ : 1

The value of n in the expansion of (p + q)n if the first three terms of the expansion are 729, 7290 and 30375, respectively is

જવાબ : 6

The coefficient of yn in the expansion of (1 – 2y + 3y² – 4y³ + ……..)-n is

જવાબ : (2n)!/(n!)²

The greatest coefficient in the expansion of (1 + y)10 is

જવાબ : 10!/(5!)²

If the third term in the binomial expansion of (1 + y)n is (-1/8)y² then the rational value of n is

જવાબ : 1/2

If m is a positive integer, then (√7+1)2m+1 + (√7−1)2m+1 is

જવાબ : an irrational number

The fourth term in the expansion (a – 2b)12 is

જવાબ : -1760 a9 × b³

(1.2)10000 is _____ 1000

જવાબ : greater than

The coefficient of x in the expansion of (x² + a/x)5 is

જવાબ : 10a3

The number of ordered triplets of positive integers which are solution of the equation a + b + c = 100 is

જવાબ : 4851

The greatest coefficient in the expansion of (1 + y)10 is

જવાબ : 10!/(5!)²

if n is a positive ineger then 33nn – 3n  is divisible by

જવાબ : 49

In the expansion of (p + q)n, if n is odd then the number of middle term is/are

જવાબ : 2

In the expansion of (p + q)n, if n is even then the middle term is

જવાબ : (n/2 + 1)th term

If n is a positive integer, then (√3+1)2n + 1 − (√3−1)2n + 1 is

જવાબ : not an integer

The coefficient of yn in the expansion (1 + y + y² + …..)-n is

જવાબ : (-1)n

If the coefficients of y2 and y3 in the expansion of (3 + ay)9 are the same, then the value of a is

જવાબ : 97

If the sum of the binomial coefficients of the expansion (2y + 1/y)n is equal to 256, then the term independent of y is

જવાબ : 1120

If the fifth term of the expansion ( p2/3 + p−1)n does not contain 'p'. Then n is equal to

જવાબ : 10

The coefficient of y−3 in the expansion of (y − n/y)11 is

જવાબ : −330 n7

The coefficient of the term independent of x in the expansion of (ax+bx)14 is

જવાબ : 14!/(7!)2 a7 b7

The coefficient of y5 in the expansion of (1+y)21 + (1+y)22 + ... + (1+y)30

જવાબ : 31C6 − 21C6

The coefficient of p8 q10 in the expansion of (p + q)18 is

જવાબ : 18C8

If the coefficients of the (m + 1)th term and the (m + 3)th term in the expansion of (1 + y)20 are equal, then the value of m is

જવાબ : 9

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1+y)n, n ∈ N are in A.P., then n =

જવાબ : 7

The middle term in the expansion of (2y/3 – 3/2y2)2n is

જવાબ : (−1)n 2nCn y−n

If rth term is the middle term in the expansion of (y2−1/2y)20, then (r+3)th term is

જવાબ : −20C7 y, 2−13

The number of terms with integral coefficients in the expansion of (171/3 + 351/2 y)600 is

જવાબ : 101

Constant term in the expansion of (y−1/y)10 is

જવાબ : -252

If rth term in the expansion of (2y2−1/y)12 is without y, then r is equal to

જવાબ : 9

Find the middle terms(s) in the expansion of : (x/a – a/x)10

જવાબ : (x/a + a/x)10

Here, n is an even number.

∴ Middle term = (10/2 +1 )th = 6th term

Now, we have T6 = T5+1 = 10C5 (x/a)10−5 (a/x)5 = 10×9×8×7×6/5×4×3×2×1 = 252

Find the middle terms(s) in the expansion of : (p/x + x/p)9

જવાબ : (p/x - x/p)9

Here, n is an odd number.

Therefore, the middle terms are (9+1/2)th and [(9+1)/2 + 1]th, i.e., 5th and 6th terms.

Now, we have T5 = T4+1 = 9C4 (p/x)9−4 (-x/p)4 = (9×8×7×6)/(4×3×2×1) × (p/x) = 126 p/x And,T6 = T5+1 = 9C5 (p/x)9−5 (-x/p)5 = (9×8×7×6)/(4×3×2×1) × (-x/p) = -126 x/p

Find the middle terms(s) in the expansion of : (2ax−b/x2)12

જવાબ : (2ax + b/x2)12

Here, n is an even number.

∴ Middle term = (12/2 +1)th = 7th term

Now, we have T7 = T6 + 1 = 12C6  (2ax)12−6 (b/x2)6 = (12×11×10×9×8×7)/(6×5×4×3×2×1) × (2ab/x)6 = 59136 a6b6/x6

Find the middle terms(s) in the expansion of : (3 − x3/6)7

જવાબ : (3 + x3/6)7

Here, n is an odd number.

Therefore, the middle terms are (7+1/2)th and [(7+1)/2 + 1]th, i.e., 4th and 5th terms.

Now, we have T4 = T3+1 = 7C3 (3)7−3 (x3/6)3

= 105/8 x9

And, T5 = T4+1 = 9C4 (3)9−4 (x3/6)4 = (7×6×5)/(3×2) × 35 × 1/64 x12 = 35/48 x12

Find the middle terms(s) in the expansion of : (x/3 + 9y)10

જવાબ : (x/3 - 9y)10

Here, n is an even number.

Therefore, the middle term is (10/2 + 1 )th, i.e., 6th term.

Now, we have T6 = T5+1 = 10C5 (x/3)10−5 (-9y)5

= -(10×9×8×7×6)/(5×4×3×2) × 1/35 × 95 × x5 y5

= -61236 x5 y5

Find the middle terms(s) in the expansion of : (x – 1/x)2n+1

જવાબ : (x + 1/x)2n+1

Here, (2n+1)  is an odd number.

Therefore, the middle terms are (2n+1+1/2)th and [(2n+1+1)/2 + 1]th i.e. (n+1)th and (n+2)th terms.

Now, we have: Tn+1 = 2n+1Cn x2n+1−n  × (1)n/xn = (1)n 2n+1Cn x

And,Tn+2 = Tn+1+1 = 2n+1Cn+1 x2n+1−n−1   (1)n+1/xn+1 = (1)n+1 2n+1Cn+1 × 1/x

Find the middle terms(s) in the expansion of : (2x − x2/4)9

જવાબ : (2x + x2/4)9

Here, n is an odd number.

Therefore, the middle terms are [(n+1)/2]th and [(n+1)/2 + 1]th, i.e. 5th and 6th terms.

Now, we haveT5 = T4+1 = 9C4 (2x)9−4 (x2/4)4 = (9×8×7×6)/(4×3×2) × 25 1/44 x5+8 =63/4 x13

And,T6 = T5+1 = 9C5 (2x)9−5 (x2/4)5 = (9×8×7×6)/(4×3×2) × 24 1/45 x4 + 10 = 63/32 x14

Find the middle terms(s) in the expansion of : (1 + 3x + 3x2 + x3)2n

જવાબ : (1 + 3x - 3x2 + x3)2n =(1 - x)6n

Here, n  is an even number.

∴ Middle term = (6n/2 + 1 )th = (3n+1)th term

Now, we have T3n+1 = 6nC3n x3n = (6n)!/(3n!)2 x3n

Find the middle terms(s) in the expansion of : (1 + 2x + x2)n

જવાબ : (1 + 2x + x2)n

= (1+x)2n

n is an even number.

∴ Middle term = (2n/2 + 1)th = (n+1)th term

Now, we haveTn+1 = 2nCn (1)n (x)n = (2n)!/(n!)2 (1)n xn

Find the middle terms(s) in the expansion of : (x + 1/x)10

જવાબ : (x + 1/x)10

Here, n is an even number.

Middle term = (10/2 + 1)th= 6th term

Now, we haveT6 = T5 + 1 =10C5 x10−5 (1/x)5 = (10×9×8×7×6)/(5×4×3×2) = 252