ExamGOAL
Books
24
MCQ (Single Correct Answer)

If the middle term of $\left(\frac{1}{x}+x \sin x\right)^{10}$ is equal to $7 \frac{7}{8}$, then the value of $x$ is

A
$2 n \pi+\frac{\pi}{6}$
B
$n \pi+\frac{\pi}{6}$
C
$n \pi+(-1)^n \frac{\pi}{6}$
D
$n \pi+(-1)^n \frac{\pi}{3}$
25

The largest coefficient in the expansion of $(1+x)^{30}$ is .............. .

Explanation

Largest coefficient in the expansion of $(1+x)^{30}={ }^{30} \mathrm{C}_{30 / 2}={ }^{30} \mathrm{C}_{15}$

26

The number of terms in the expansion of $(x+y+z)^n$ ............ .

Explanation

Given expansion is $(x+y+z)^n=[x+(y+z)]^n$.

$$\begin{aligned} {[x+(y+z)]^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1}(} & (y) z) \\ & +{ }^n C_2 x^{n-2}(y+z)^2+\ldots+{ }^n C_n(y+z)^n \end{aligned}$$

$$\begin{aligned} \therefore \text { Number of terms } & =1+2+3+\ldots+n+(n+1) \\ & =\frac{(n+1)(n+2)}{2} \end{aligned}$$

27

In the expansion of $\left(x^2-\frac{1}{x^2}\right)^{16}$, the value of constant term is ............. .

Explanation

Let constant be $T_{t+1}$.

$$\begin{aligned} T_{r+1} & ={ }^{16} C_r\left(x^2\right)^{16-r}\left(-\frac{1}{x^2}\right)^r \\ & ={ }^{16} C_r x^{32-2 r}(-1)^r x^{-2 r} \\ & ={ }^{16} C_r x^{32-4 r}(-1)^r \end{aligned}$$

For constant term, $\quad 32-4 r=0 \Rightarrow r=8$

$$\therefore \quad T_{8+1}={ }^{16} C_8$$

28

If the seventh term from the beginning and the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ are equal, then $n$ equals to ............ .

Explanation

Given expansions is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$.

$$T_7=T_{6+1}={ }^n C_6(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6\quad \text{.... (i)}$$

Since, $T_7$ from end is same as the $T_7$ from beginning of $\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^n$.

Then, $$T_7={ }^n C_6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^6\quad \text{.... (ii)}$$

Given that,

$${ }^n C_6(2)^{\frac{n-6}{3}}(3)^{-6 / 3}={ }^n C_6(3)^{-\frac{(n-6)}{3}} 2^{6 / 3}$$

$\Rightarrow\quad (2)^{\frac{n-12}{3}}=\left(\frac{1}{3^{1 / 3}}\right)^{n-12}$

which is true, when $\frac{n-12}{3}=0$.

$$\Rightarrow \quad n-12=0 \Rightarrow n=12$$