Phrases Previous Year Questions (PYQs) – Mathematics – Page 1 of 61

Given:

$\overrightarrow{a}=\hat{i}-\hat{k},$ $\overrightarrow{b}=x\hat{i}+\hat{j}+(1-x)\hat{k},$ $\overrightarrow{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$

Form the matrix:

$M=\left[\begin{array}{ccc}1& x& y\\ 0& 1& x\\ -1& 1-x& 1+x-y\end{array}\right]$

Find the determinant:

$det(M)=\left|\begin{array}{ccc}1& x& y\\ 0& 1& x\\ -1& 1-x& 1+x-y\end{array}\right|=1$

Since the determinant is constant and non-zero, the vectors are linearly independent.

$\boxed{{\displaystyle \text{The matrix does not depend on }x\text{ or }y}}$

Quick Solution

Given: $\overrightarrow{a},\overrightarrow{b}$ are unit vectors and

$2\overrightarrow{a}+\overrightarrow{b}=3$

Take magnitude on both sides:

$|2\overrightarrow{a}+\overrightarrow{b}|=3\Rightarrow |2\overrightarrow{a}+\overrightarrow{b}{|}^2=9$

Use identity:

$$|2\overrightarrow{a}+\overrightarrow{b}{|}^2=4|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+4(\overrightarrow{a}\cdot \overrightarrow{b})=4+1+4(\overrightarrow{a}\cdot \overrightarrow{b})=5+4(\overrightarrow{a}\cdot \overrightarrow{b})$$

Set equal to 9:

$$5+4(\overrightarrow{a}\cdot \overrightarrow{b})=9\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=1\Rightarrow \cos \theta =1\Rightarrow \theta =0^{\circ }$$

Quick Solution

Given:

$\int f(x)dx=g(x)$

Required: $\int x^{5}f(x^3)dx$

Use substitution:

Let $u=x^3\Rightarrow du=3x^{2}dx\Rightarrow dx=\frac{du}{3x^2}$

Now rewrite the integral:

$$\int x^{5}f(x^3)dx=\int x^{5}f(u)\cdot \frac{du}{3x^2}=\frac{1}{3}\int x^{3}f(u)du$$

But $x^3=u$, so:

$$\frac{1}{3}\int uf(u)du$$

Now integrate by parts or use the identity:

$$\int uf(u)du=ug(u)-\int g(u)du$$

Final answer:

$$\int x^{5}f(x^3)dx=\frac{1}{3}[x^{3}g(x^3)-\int g(x^3)\cdot 3x^{2}dx]=x^{3}g(x^3)-\int x^{2}g(x^3)dx$$

$$\boxed{{\displaystyle \int x^{5}f(x^3)dx=x^{3}g(x^3)-\int x^{2}g(x^3)dx}}$$