Phrases Previous Year Questions (PYQs) – 2023 – Page 1 of 12

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:

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

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

Quick DL Method Solution

Given:

$$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}=\underset{x\to k}{lim}\frac{x^3-k^2}{x^2-k^2}$$

LHS using derivative:

$$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}={\frac{d}{dx}(x^4)|}_{x=1}=4x^3{|}_{x=1}=4$$

RHS using DL logic:

$$\underset{x\to k}{lim}\frac{x^3-k^2}{x^2-k^2}\approx \frac{3k^2(x-k)}{2k(x-k)}=\frac{3k}{2}$$

Equating both sides:

$$\frac{3k}{2}=4\Rightarrow k=\frac{8}{3}$$

$$\boxed{{\displaystyle k=\frac{8}{3}}}$$

Probability — No Black Ball is Chosen

Given:

• Yellow balls = 5
• Black balls = 4
• Green balls = 3
• Total balls = 5 + 4 + 3 = 12

We are to find:

Probability that no black ball is selected when 3 balls are drawn at random.

Step 1: Total number of ways to choose any 3 balls from 12:

$$\text{Total ways}=(\genfrac{}{}{0ex}{}{12}{3})=220$$

Step 2: Ways to choose 3 balls such that no black ball is chosen:

Only yellow and green balls are allowed ⇒ Total = 5 (yellow) + 3 (green) = 8
$$\text{Favorable ways}=(\genfrac{}{}{0ex}{}{8}{3})=56$$

Step 3: Probability

$$P(\text{no black ball})=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=\frac{56}{220}=\frac{14}{55}$$

$$\boxed{{\displaystyle \text{Probability}=\frac{14}{55}}}$$