## If ac > b^2 then the sum of the coefficients in the expansion of (a α^2 x^2 + 2b α x + c)^n ; (a, b, c, α ∈ R, n ∈ N ) is

Q: If ac > b2 then the sum of the coefficients in the expansion of (a α2x2 + 2b α x + c)n; (a, b, c, α ∈ R, n ∈ N ) is

(A) positive if a > 0

(B) positive if c > 0

(C) negative if a < 0, n is odd

(D) positive if c < 0, n is even

Sol. In the expansion of (a α2x2 + 2b α x + c)n

the sum of the coefficients = (a α2 + 2b α + c)n

Let f(α) = (a α2 + 2b α + c)n

Its discriminant = 4b2 – 4ac = 4(b2 –ac ) < 0

Hence, f(α) < 0 or f(α) > 0 for all α ∈ R

If a > 0 then f(α) > 0 ⇒ (a α2 + 2b α + c)n > 0

If c > 0 i.e. f(0) > 0 ⇒ f(α) > 0 ⇒ (a α2 + 2b α + c)n > 0

If a < 0 then f(α) < 0 ⇒ (a α2 + 2b α + c)n < 0 if n is odd

If c < 0 i.e. f(0) < 0 ⇒ f(α) < 0 ⇒ (a α2 + 2b α + c)n > 0 if n is even.

Hence (A), (B), (C) and (D) are the correct answer.

## The largest coefficient in the expansion of (4+3x)^25 is

Q: The largest coefficient in the expansion of (4+3x)25 is

(A) 25C11.325 (4/3)11

(B) 25C11.425 (3/4)11

(C) 25C14.414 (3)11

(D) None of these

Ans: (B) & (C)

Sol: $\large (4+3x)^{25}= 4^{25} (1+\frac{3}{4}x)^{25}$

Let (r + 1)th term will have greatest coefficient,

$\large \frac{Coefficient \; of \; T_{r+1}}{Coefficient \; of \; T_r} \ge 1$

$\large \frac{25_{C_r}(\frac{3}{4})^r}{25_{C_{r-1}}(\frac{3}{4})^{r-1}} \ge 1$

$\large (\frac{25-r+1}{r}) \frac{3}{4} \ge 1$

$\large r \le \frac{78}{7}$

Greatest possible value of r is 11

Coefficient of T12 = 425 x 25C11 (3/4)11

Hence (B) and ( C ) are the correct answer.

## The value of nC0 + n+1C1 + n+2C2 + ….+ n+kCk is equal to

Q: The value of nC0 + n+1C1 + n+2C2 + ….+ n+kCk is equal to

(A) n+k+1Ck

(B) n+k+1Cn+1

(C) n+kCn+1

(D) None of these

Sol. We have (1+x)n + (1+x)n+1 + ….+ (1+x)n+k

$\large = (1+x)^n \frac{(1+x)^{k+1}-1}{x}$

$\large = \frac{(1+x)^{n+k+1}-(1+x)^n}{x}$

Equating coefficients of xn

nCn + n+1Cn + ….+n+kCn = n+k+1Cn+1

nC0 + n+1C1 + n+2C2 + ….+ n+kCk

= n+k+1Ck = n+k+1Cn+1

Hence (A), (B) are the correct answer.

## If (1 + x)2n = a0 + a1x + a2x2 + …..+ a2nx2n then

Q: If (1 + x)2n = a0 + a1x + a2x2 + …..+ a2nx2n then

(A) a0 + a2 + a4 + …. = (1/2)( a0 + a1 + a2 + a3 + ….)

(B) an + 1 < an

(C) an–3 = an+3

(D) none of these

Sol. a0 + a1 + a2 + ….= 22n

and a0 + a2 + a4 + ….= 22n–1

an = 2nCn = the greatest coefficient, being the middle coefficient

an–3 = 2nCn–3 = 2nC2n–(n –3) = 2nCn+3 = an+3

Hence (A), (B) and ( C ) are the correct answer.

## Which of the following is/are true in the expansion of (x + 2y + 3z)12

Q: Which of the following is/are true in the expansion of (x + 2y + 3z)12

(A) The number of terms in the expansion is 91

(B) The sum of all the coefficient in the expansion is 126

(C) The number of terms in the expansion is 36

(D) The sum of all the coefficient in the expansion is 612

Sol. (x + 2y + 3z)12

= 12C0 (x + 2y)12 (3z)0 + 12C1 (x + 2y)11 (3z)1 + ….+ 12C12 (x + 2y)0 (3z)12

= 13 + 12 + 11 + …….1 = 91

Sum of coefficients = (1 + 2 + 3)12 = 612

Hence (A) and (D) are the correct answer.

## $\frac{1}{1!(n-1)!} – \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} – …is \; equal \; to$

Q: $\large \frac{1}{1!(n-1)!} – \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} – …is \; equal \; to$

(A) $\large \frac{2^{n-1}}{n!}$ for all n ∈ N

(B) $\large \frac{2^{n-1}}{n!}$ for odd values of n only

(C) $\large \frac{2^{n-1}}{n!}$ for even values of n only

(D) none of these

Sol. Given expression $\large = \frac{1}{n!}(n_{C_1} – n_{C_3} + n_{C_5} -…..) = \frac{2^{n/2}sin\frac{n\pi}{4}}{n!}$

## Given the integers r > 1, n > 2 , and coefficients of (3r)th and (r+2)nd term in the binomial expansion of (1+x)2n are equal, then

Q: Given the integers r > 1, n > 2 , and coefficients of (3r)th and (r+2)nd term in the binomial expansion of (1+x)2n are equal, then

(A) n = 2r

(B) n =3r

(C) n = 2r+1

(D) None of these

Sol. Coefficients of (3r)th and (r+2)nd terms will be 2nC3r-1 and 2nCr+1

These are equal ⇒ (3r – 1) + (r + 1) = 2n ⇒ n = 2r

Hence (A) is the correct answer.

## The value of (n + 2) C0 2n+1 – (n + 1) C12n + n. C2 2n-1 – … is equal to

Q: The value of (n + 2) C0 2n+1 – (n + 1) C12n + n. C2 2n-1 – … is equal to

(A) 4 (1 + n)

(B) 4n

(C) 2n

(D) 2n + 4

Sol. (x – 1)n = C0 xn – C1 xn-1 + Cnxn-2 – ….

x2 (x – 1)n = C0 xn+2 – C1 xn+1 + C2 xn -….

differentiating with respect to x , we get

2x (x – 1)n + x2n (x – 1)n-1 = (n + 2) C0xn+1 – (n + 1) C1 xn + nC2xn-1 – ………

put, x = 2, we get

(n + 2) C0 xn+1 – (n + 1) C12n + n. C22n-1 – …. = 4 + 4n

Hence (A) is the correct answer.

## If the coefficients of x^7 and x^8 in $(2 + \frac{x}{3})^n$ are equal, then n is

Q: If the coefficients of x7 and x8 in $(2 + \frac{x}{3})^n$ are equal, then n is

(A) 56

(B) 55

(C) 45

(D) 15

Sol. $\large n_{C_7} \frac{2^{n-7}}{3^7} = n_{C_8} \frac{2^{n-8}}{3^8}$

⇒ n = 55

Hence (B) is the correct answer.

## The term independent of x in $( \sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} )^{10}$ is

Q: The term independent of x in $\large (\sqrt{\frac{x}{3}} + \sqrt{\frac{3}{2x^2}} )^{10}$ is

(A) 1

(B) 5/12

(C) 10C1

(D) None of these

Sol. General term in the expansion is

$\large = 10_{C_r} (\frac{x}{3})^{r/2} (\frac{3}{2x^2})^{\frac{10-r}{2}}$

$\large = 10_{C_r} x^{\frac{3r}{2}-10} . \frac{3^{5-r}}{2^{(10-r)/2}}$

For constant term, 3r/2 = 10  ⇒ r = 20/3

which is not an integer. Therefore, there will be no constant term.

Hence (D) is the correct answer.