SATHEE: Binomial Theorem 1 Question 21

Binomial Theorem 1 Question 21

23.

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are $3$ and $- 6$ respectively, then $m$ is euqal to

(1999, 2M)

(a) $6$

(b) $9$

(d) $24$

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Answer:

Correct Answer: 23. (c)

Solution:

$(1 + x)^{m} (1 - x)^{n} = [1 + m x + \frac{m (m - 1)}{2} x^{2} + \dots]$

$[1 - n x + \frac{n (n - 1)}{2} x^{2} - \dots]$

$= 1 + (m - n) x + [\frac{m (m - 1)}{2} + \frac{n (n - 1)}{2} - m n] x^{2} + \dots$

term containing power of $x \geq 3$ .

Now, $m - n = 3$ …….(i)

$[∵$ coefficient of $x = 3$ , given]

and $\frac{1}{2} m (m - 1) + \frac{1}{2} n (n - 1) - m n = - 6$

$\begin{aligned} \Rightarrow m (m - 1) + n (n - 1) - 2 m n = - 12 \\ \Rightarrow m^{2} - m + n^{2} - n - 2 m n = - 12 \\ \Rightarrow (m - n)^{2} - (m + n) = - 12 \\ \Rightarrow m + n = 9 + 12 = 21 \dots \dots . (i i) \end{aligned}$

On solving Eqs. (i) and (ii), we get $m = 12$

Binomial Theorem 1 Question 21

23.

Answer:

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Binomial Theorem 1 Question 21

23.

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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