Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

if the constant terms in the expansion of |√x - k/x²|¹⁰ is 405 then we have to find value of |k|

solution : using formula,

$T_{(r+1)}=^nC_r\left(\frac{-K}{x^2}\right)^r(\sqrt{x})^{(n-r)}$

n = 10

so, $T_{(r+1)}=^{10}C_r\left(\frac{-K}{x^2}\right)^{r}(\sqrt{x})^{(10-r)}$

= $^{10}C_r(-K)^{r}\frac{(\sqrt{x})^{(10-r)}}{(x^2)^{r}}$

= $^{10}C_r(-K)^rx^{\frac{(10-r)}{2}-2r}$

= $^{10}C_r(-K)^rx^{(5-5r/2)}$

for constant term

power of x = 0

⇒5 - 5r/2 = 0

⇒r = 2

now $T_{(2+1)}=T_3=^{10}C_2(-K)^2x^0$

⇒405 = ¹⁰C₂ K²

⇒405 = 10!/(2! × 8!) × K²

⇒405 = 10 × 9/2 × K²

⇒405/45 = K²

⇒ = K² = 9

⇒K = ±3

Therefore the value of k is ±3

Answer 2

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The value of K=+3

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