SATHEE: Binomial Theorem 1 Question 11

Binomial Theorem 1 Question 11

12.

The positive value of $λ$ for which the coefficient of $x^{2}$ in the expression $x^{2} (\sqrt{x} + \frac{λ}{x^{2}})^{10}$ is $720$ , is

(a) 3

(b) $\sqrt{5}$

(d) 4

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Answer:

Correct Answer: 12. (d)

Solution:

The general term in the expansion of binomial expression $(a + b)^{n}$ is $T_{r + 1} =^{n} C_{r} a^{n \to r} b^{r}$ , so the general term in the expansion of binomial expression $x^{2} (\sqrt{x} + \frac{λ}{x^{2}})^{10}$ is

$\begin{aligned} T_{r + 1} & = x^{2} (^{10} C_{r} (\sqrt{x})^{10 \to r} (\frac{λ}{x^{2}})^{r}) =^{10} C_{r} x^{2} \cdot x^{\frac{10 \to r}{2}} λ^{r} x^{- 2 r} \\ =^{10} C_{r} λ^{r} x^{2 + \frac{10 \to}{2} - 2 r} \end{aligned}$

Now, for the coefficient of $x^{2}$ , put $2 + \frac{10 - r}{2} - 2 r = 2$

$\begin{aligned} \Rightarrow & \frac{10 - r}{2} - 2 r = 0 \\ \Rightarrow & 10 - r = 4 r \Rightarrow r = 2 \end{aligned}$

So, the coefficient of $x^{2}$ is $^{10} C_{2} λ^{2} = 720$ [given]

$\begin{array}{lc} \Rightarrow & \frac{10!}{2! 8!} λ^{2} = 720 \Rightarrow \frac{10 \cdot 9 \cdot 8!}{2 \cdot 8!} λ^{2} = 720 \\ \Rightarrow & 45 λ^{2} = 720 \\ \Rightarrow & λ^{2} = 16 \Rightarrow λ = \pm 4 \\ ∴ & λ = 4 \end{array}$

Binomial Theorem 1 Question 11

12.

Answer:

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Binomial Theorem 1 Question 11

12.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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