If the constant term in the binomial expansion of x - k x 2 10 is 405 , then k equals :

T r + 1 = C r 10 - k x 2 r x 10 - r = C r · 10 - k r · x 5 - 5 r 2 for constant term ⇒ 5 - 5 r 2 = 0 ⇒ r = 2 ⇒ T 3 = C 2 10 · k 2 = 405 ⇒ 10 ( 9 ) 2 k 2 = 405 ⇒ k 2 = 9 ⇒ k = 3

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremGeneralEasy2 minPYQ_2020

MathematicsEasysingle choice

If the constant term in the binomial expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ is $405$ , then $|k|$ equals :

Options:

Answer:

Solution:

$T_{r + 1} = {}^{10}C_{r} {(\frac{- k}{x^{2}})}^{r} {(\sqrt{x})}^{10 - r}$

$= {}^{10}C_{r \cdot} {(- k)}^{r} \cdot x^{5 - \frac{5 r}{2}}$

for constant term

$\Rightarrow 5 - \frac{5 r}{2} = 0 \Rightarrow r = 2$

$\Rightarrow T_{3 =} {}^{10}C_{2} \cdot k^{2} = 405$

$\Rightarrow \frac{10 (9)}{2} k^{2} = 405$

$\Rightarrow k^{2} = 9$

$\Rightarrow |k| = 3$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:General

⏱ 2mℹ️ Source: PYQ_2020

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