Question

Let α > 0, β > 0 be such that α³ + β² = 4. If the maximum value of the term independent of x in the binomial expansion of (αx^1/9 + βx^–1/6)¹⁰ is 10k, then k is equal to ______.

विकल्प

• 176

• 336

• 352

• 84

Answer 1

Let α > 0, β > 0 be such that α³ + β² = 4. If the maximum value of the term independent of x in the binomial expansion of (αx^1/9 + βx^–1/6)¹⁰ is 10k, then k is equal to 336.

Explanation:

T_r+1 = ¹⁰C_r(αx^1/9)^10–r(βx^–1/6)^r

= `""^10C_r α^(10-r)x (10 - r)/9 β^r x (-r)/6`

For independent of x

`(10 - r)/9 - r/6` = 0

⇒ `(10 - r)/9 = r/6`

20 – 2r = 3r

r = 4

T₄₊₁ = ¹⁰C₄α⁶β⁴ should be maximum

 Let P = ¹⁰C₄α⁶β⁴ (α³ + β² = 4 given)

P = ¹⁰C₄(4 – β²)²β⁴α³ = (4 – β²)

`(dP)/(dβ)` = 2(4 – β²)(–2β)β⁴ + (4 – β²)²4b³

= 4(4 – β²)³(–β² + 4 – β²)

= 4(4 – β²)β²(4 – 2β²)

For maxima and minima

`(dP)/(dβ)` = 0

⇒ β = `+-2, 0, +- sqrt(2)`

But for β = ±2

⇒ α³ = 4 – β² = 0

So it is not possible equal to 10k

β = 0     Also not possible

(So only possible value β  = `+- sqrt(2)`)

We have to check second derivative for this but P = α⁶β⁴ so there is no need to check it because it is always +ve 

∴ Maximum value of P = ¹⁰C₄α⁶β⁴ 

= 210(4 – β²)²β⁴

= 210(4 – 2)² × 2²

= 210 × 4 × 4

= 3360

In equal to 10k given

10k = 3360

k = 336