Question

Column Matching:

$$\begin{array}{cc}\text{Column (I)}& \text{Column (II)}\\ \begin{array}{cc}\text{(A)}& \text{In}{R}^2,\text{if the magnitude of the projection}\\ & \text{vector of the vector}\alpha \hat{i}+\beta \hat{j}\text{on}\sqrt{3}\hat{i}+\hat{j}\\ & \text{is}\sqrt{3}\text{and if}\alpha =2+\sqrt{3}\beta ,\text{then possible}\\ & \text{value(s) of}|\alpha |\text{is (are)}\end{array}& \text{(P) 1}\\ & \\ \begin{array}{cc}\text{(B)}& \text{Let}a\text{and}b\text{be real numbers such that}\\ & \text{the function}\\ & f\left(x\right)=\{\begin{array}{c}-3a{x}^2-2,x<1\\ bx+a^2,x\ge 1\end{array}\\ & \text{is differentiable for all}x\in R.\text{Then}\\ & \text{possible value(s) of}a\text{is (are)}\end{array}& \text{(Q) 2}\\ & \\ \begin{array}{cc}\text{(C)}& \text{Let}\omega \ne 1\text{be a complex cube root of}\\ & \text{unity. If}(3-3\omega +2{\omega }^2{)}^{4n+3}+\\ & (2+3\omega -3{\omega }^2{)}^{4n+3}+(-3+2\omega +3{\omega }^2{)}^{4n+3}=0,\\ & \text{then possible value(s) of}n\text{is (are)}\end{array}& \text{(R) 3}\\ & \\ \begin{array}{cc}\text{(D)}& \text{Let the harmonic mean of two positive real}\\ & \text{numbers}a\text{and}b\text{be}4.\text{If}q\text{is a positive real}\\ & \text{number such that}a,5,q,b\text{is an arithmetic}\\ & \text{progression, then the value(s) of}|q-a|\text{is (are)}\end{array}& \text{(S) 4}\\ & \\ & \text{(T) 5}\\ & \end{array}$$
Option $\text{(D)}$ matches with which of the elements of right hand column?

• A. $P$
• B. $Q$
• C. $R$
• D. $S$
• E. $T$

Answer 1

Answer: (A), (B)

$Q$

Option $\left(A\right)$:
The projection vector of the vector $\alpha \hat{i}+\beta \hat{j}$ on $\sqrt{3}\hat{i}+\hat{j}$ is $(\alpha \hat{i}+\beta \hat{j})\cdot \frac{\sqrt{3}\hat{i}+\hat{j}}{2}\Rightarrow |\alpha \hat{i}+\beta \hat{j})\cdot \frac{\sqrt{3}\hat{i}+\hat{j}}{2}|=\sqrt{3}\Rightarrow \sqrt{3}\alpha +\beta =\pm 2\sqrt{3}\cdots \left(1\right)\alpha =2+\sqrt{3}\beta \cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$ we get
$\Rightarrow \alpha =2,-1|\alpha |=2,1$