CBSE Class 11 Maths Exemplar Chapter 8: Binomial Theorem NCERT Solutions

Solution:

The given binomial expansion is $\backslash left(\backslash frac\{3x^2\}\{2\}\; -\; \backslash frac\{1\}\{3x\}\backslash right)^\{15\}$. We use the general term formula for the binomial expansion of $(a+b)^n$, which is $T\_\{r+1\}\; =\; \{\}^n\; C\_r\; a^\{n-r\}\; b^r$.

Here, $n\; =\; 15$, $a\; =\; \backslash frac\{3x^2\}\{2\}$, and $b\; =\; -\backslash frac\{1\}\{3x\}$.

The general term $T\_\{r+1\}$ is:

$T\_\{r+1\}\; =\; \{\}^\{15\}C\_r\; \backslash left(\backslash frac\{3x^2\}\{2\}\backslash right)^\{15-r\}\; \backslash left(-\backslash frac\{1\}\{3x\}\backslash right)^r$

Let's separate the variable part (x) and the constant part:

$T\_\{r+1\}\; =\; \{\}^\{15\}C\_r\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{15-r\}\; (x^2)^\{15-r\}\; \backslash left(-\backslash frac\{1\}\{3\}\backslash right)^r\; (x^\{-1\})^r$

$T\_\{r+1\}\; =\; \{\}^\{15\}C\_r\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{15-r\}\; \backslash left(-\backslash frac\{1\}\{3\}\backslash right)^r\; x^\{2(15-r)\}\; x^\{-r\}$

$T\_\{r+1\}\; =\; \{\}^\{15\}C\_r\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{15-r\}\; \backslash left(-\backslash frac\{1\}\{3\}\backslash right)^r\; x^\{30-2r-r\}$

$T\_\{r+1\}\; =\; \{\}^\{15\}C\_r\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{15-r\}\; \backslash left(-\backslash frac\{1\}\{3\}\backslash right)^r\; x^\{30-3r\}$

For the term to be independent of x, the exponent of x must be zero:

$30\; -\; 3r\; =\; 0$

$3r\; =\; 30$

$r\; =\; 10$

So, the 11th term ($T\_\{10+1\}$) is independent of x. Now, we substitute $r\; =\; 10$ back into the general term formula:

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{15-10\}\; \backslash left(-\backslash frac\{1\}\{3\}\backslash right)^\{10\}$

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{5\}\; \backslash left(\backslash frac\{1\}\{3\}\backslash right)^\{10\}$

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash frac\{3^5\}\{2^5\}\; \backslash frac\{1\}\{3^\{10\}\}$

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash frac\{1\}\{2^5\; \backslash cdot\; 3^5\}$

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash frac\{1\}\{(2\; \backslash cdot\; 3)^5\}$

$T\_\{11\}\; =\; \{\}^\{15\}C\_\{10\}\; \backslash frac\{1\}\{6^5\}$

We know that $\{\}^\{15\}C\_\{10\}\; =\; \{\}^\{15\}C\_\{15-10\}\; =\; \{\}^\{15\}C\_5$.

$\{\}^\{15\}C\_5\; =\; \backslash frac\{15!\}\{5!(15-5)!\}\; =\; \backslash frac\{15!\}\{5!10!\}\; =\; \backslash frac\{15\; \backslash times\; 14\; \backslash times\; 13\; \backslash times\; 12\; \backslash times\; 11\}\{5\; \backslash times\; 4\; \backslash times\; 3\; \backslash times\; 2\; \backslash times\; 1\}$

$\{\}^\{15\}C\_5\; =\; 3\; \backslash times\; 7\; \backslash times\; 13\; \backslash times\; 1\; \backslash times\; 11\; =\; 3003$

Therefore, the term independent of x is:

$T\_\{11\}\; =\; 3003\; \backslash times\; \backslash frac\{1\}\{6^5\}\; =\; \backslash frac\{3003\}\{7776\}$

Answer: The term independent of x is $\backslash frac\{3003\}\{7776\}$.

Solution:

The given binomial expansion is $\backslash left(\backslash sqrt\{x\}\; -\; \backslash frac\{k\}\{x^2\}\backslash right)^\{10\}$. We use the general term formula $T\_\{r+1\}\; =\; \{\}^n\; C\_r\; a^\{n-r\}\; b^r$.

Here, $n\; =\; 10$, $a\; =\; \backslash sqrt\{x\}\; =\; x^\{1/2\}$, and $b\; =\; -\backslash frac\{k\}\{x^2\}\; =\; -k\; x^\{-2\}$.

The general term $T\_\{r+1\}$ is:

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; (x^\{1/2\})^\{10-r\}\; (-k\; x^\{-2\})^r$

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; x^\{\backslash frac\{1\}\{2\}(10-r)\}\; (-k)^r\; (x^\{-2\})^r$

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; (-k)^r\; x^\{\backslash frac\{10-r\}\{2\}\}\; x^\{-2r\}$

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; (-k)^r\; x^\{\backslash frac\{10-r\}\{2\}\; -\; 2r\}$

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; (-k)^r\; x^\{\backslash frac\{10-r-4r\}\{2\}\}$

$T\_\{r+1\}\; =\; \{\}^\{10\}C\_r\; (-k)^r\; x^\{\backslash frac\{10-5r\}\{2\}\}$

For the term to be free from x (independent of x), the exponent of x must be zero:

$\backslash frac\{10-5r\}\{2\}\; =\; 0$

$10\; -\; 5r\; =\; 0$

$5r\; =\; 10$

$r\; =\; 2$

So, the 3rd term ($T\_\{2+1\}$) is free from x. We are given that this term is 405.

Substitute $r\; =\; 2$ into the general term formula:

$T\_3\; =\; \{\}^\{10\}C\_2\; (-k)^2\; x^\{\backslash frac\{10-5(2)\}\{2\}\}$

$T\_3\; =\; \{\}^\{10\}C\_2\; (-k)^2\; x^0$

$T\_3\; =\; \{\}^\{10\}C\_2\; k^2$

Now, calculate $\{\}^\{10\}C\_2$:

$\{\}^\{10\}C\_2\; =\; \backslash frac\{10!\}\{2!(10-2)!\}\; =\; \backslash frac\{10!\}\{2!8!\}\; =\; \backslash frac\{10\; \backslash times\; 9\}\{2\; \backslash times\; 1\}\; =\; 45$

So, $T\_3\; =\; 45k^2$.

We are given that $T\_3\; =\; 405$.

$45k^2\; =\; 405$

$k^2\; =\; \backslash frac\{405\}\{45\}$

$k^2\; =\; 9$

$k\; =\; \backslash pm\; \backslash sqrt\{9\}$

$k\; =\; \backslash pm\; 3$

Answer: The value of k is $\backslash pm\; 3$.

Solution:

We need to find the coefficient of x in the expansion of $(1-3x+7x^2)(1-x)^\{16\}$. Let's first expand $(1-x)^\{16\}$ up to the term containing x.

Using the binomial theorem, $(1-x)^\{16\}\; =\; \{\}^\{16\}C\_0\; (1)^\{16\}\; (-x)^0\; +\; \{\}^\{16\}C\_1\; (1)^\{15\}\; (-x)^1\; +\; \{\}^\{16\}C\_2\; (1)^\{14\}\; (-x)^2\; +\; ...$

$(1-x)^\{16\}\; =\; 1\; \backslash cdot\; 1\; \backslash cdot\; 1\; +\; 16\; \backslash cdot\; 1\; \backslash cdot\; (-x)\; +\; \backslash frac\{16\; \backslash times\; 15\}\{2\; \backslash times\; 1\}\; \backslash cdot\; 1\; \backslash cdot\; x^2\; +\; ...$

$(1-x)^\{16\}\; =\; 1\; -\; 16x\; +\; 120x^2\; +\; ...$

Now, we multiply this expansion by $(1-3x+7x^2)$:

$(1-3x+7x^2)(1\; -\; 16x\; +\; 120x^2\; +\; ...)$

To find the coefficient of x, we need to identify the terms that result in x when multiplied:

• Multiply the constant term of the first factor (1) by the x term of the second factor (-16x): $1\; \backslash times\; (-16x)\; =\; -16x$.
• Multiply the x term of the first factor (-3x) by the constant term of the second factor (1): $(-3x)\; \backslash times\; 1\; =\; -3x$.
• The $x^2$ term in the first factor ($7x^2$) multiplied by any term in the second factor will result in a power of x greater than 1, so we don't consider it for the coefficient of x.

The terms containing x are $-16x$ and $-3x$.

The coefficient of x is the sum of the coefficients of these terms: $-16\; +\; (-3)\; =\; -19$.

Answer: The coefficient of x is -19.

Solution:

The given binomial expansion is $\backslash left(3x\; -\; \backslash frac\{2\}\{x^2\}\backslash right)^\{13\}$. We use the general term formula $T\_\{r+1\}\; =\; \{\}^n\; C\_r\; a^\{n-r\}\; b^r$.

Here, $n\; =\; 13$, $a\; =\; 3x$, and $b\; =\; -\backslash frac\{2\}\{x^2\}\; =\; -2x^\{-2\}$.

The general term $T\_\{r+1\}$ is:

$T\_\{r+1\}\; =\; \{\}^\{13\}C\_r\; (3x)^\{13-r\}\; \backslash left(-\backslash frac\{2\}\{x^2\}\backslash right)^r$

$T\_\{r+1\}\; =\; \{\}^\{13\}C\_r\; (3)^\{13-r\}\; x^\{13-r\}\; (-2)^r\; (x^\{-2\})^r$

$T\_\{r+1\}\; =\; \{\}^\{13\}C\_r\; (3)^\{13-r\}\; (-2)^r\; x^\{13-r\}\; x^\{-2r\}$

$T\_\{r+1\}\; =\; \{\}^\{13\}C\_r\; (3)^\{13-r\}\; (-2)^r\; x^\{13-r-2r\}$

$T\_\{r+1\}\; =\; \{\}^\{13\}C\_r\; (3)^\{13-r\}\; (-2)^r\; x^\{13-3r\}$

For the term to be independent of x, the exponent of x must be zero:

$13\; -\; 3r\; =\; 0$

$3r\; =\; 13$

$r\; =\; \backslash frac\{13\}\{3\}$

Since 'r' must be a non-negative integer, there is no integer value of 'r' that satisfies this condition. This means there is no term independent of x in the expansion of $\backslash left(3x\; -\; \backslash frac\{2\}\{x^2\}\backslash right)^\{13\}$.

Answer: There is no term independent of x in this expansion.