In the Expansion of\( \left(2 x^{2}+r x+1\right)^{6} \), where \( r \)

1.	In the Expansion of\( \left(2 x^{2}+r x+1\right)^{6} \), where \( r \) is a number and\( x \) is a variable, the coefficient of \( x^{2} \) and\( x^{4} \) are 27 and \( -192 \) respectively findThe possible value of \( r \)
Answer» Given expression:- (2x² + rx + 1)⁶ The general term of this expression = 6!/ (p! q! r!) × 1^p × rx^q × (2x²)^r = 6!/ (p! q! r!) × r^q × 2^r × (x^{q + 2r}) Case1:- x^q^+2r= x² q + 2r = 2 Also, p + q + r = 6, The possible values of p, q, r are (4,2,0) and (5,0,1) The coefficient of x² = 27(given) [{6! / (4! 2! 1!)} × r² × 2⁰] + [{6! / (5! 0! 1!)} × r⁰ × 2¹] = 27 [{6! / (4! 2!)} × r²] + [{6! / (5! 1!)} × 2] = 27 15r² + 12 = 27 15r² = 15 ∴r = ±1 .... (1) Case2:- x^q^{+ 2r} = x¹¹ q + 2r = 11 Also, p + q + r = 6 The only possible value of p, q, r is (0,1,5) The coefficient of x¹¹ = -192 {6! / (0! 1! 5!)} × r¹ × 2⁵ = -192 {6! / 5!} × r × 32 = -192 6r = -6 r = -1 .... (2) From (1) and (2), r = -1

In the Expansion of\( \left(2 x^{2}+r x+1\right)^{6} \), where \( r \) is a number and\( x \) is a variable, the coefficient of \( x^{2} \) and\( x^{4} \) are 27 and \( -192 \) respectively findThe possible value of \( r \)

Answer»

Given expression:- (2x² + rx + 1)⁶

The general term of this expression = 6!/ (p! q! r!) × 1^p × rx^q × (2x²)^r

= 6!/ (p! q! r!) × r^q × 2^r × (x^{q + 2r})

Case1:-

x^q^+2r= x²

q + 2r = 2

Also, p + q + r = 6,

The possible values of p, q, r are (4,2,0) and (5,0,1)

The coefficient of x² = 27(given)

[{6! / (4! 2! 1!)} × r² × 2⁰] + [{6! / (5! 0! 1!)} × r⁰ × 2¹] = 27

[{6! / (4! 2!)} × r²] + [{6! / (5! 1!)} × 2] = 27

15r² + 12 = 27

15r² = 15 ∴r = ±1 .... (1)

Case2:-

x^q^{+ 2r} = x¹¹

q + 2r = 11

Also, p + q + r = 6

The only possible value of p, q, r is (0,1,5)

The coefficient of x¹¹ = -192

{6! / (0! 1! 5!)} × r¹ × 2⁵ = -192

{6! / 5!} × r × 32 = -192

6r = -6

r = -1 .... (2)

From (1) and (2), r = -1