Column-IColumn-II(A) In R^2,if the magnitude | vedclass

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(A) The magnitude of the projection of $\vec{u} = \alpha \hat{i} + \beta \hat{j}$ on $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$ is $\frac{|\vec{u} \cdot \

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$(A) \rightarrow (P, Q), (B) \rightarrow (P, Q), (C) \rightarrow (P, Q, S, T), (D) \rightarrow (Q, T)$

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(A) The magnitude of the projection of $\vec{u} = \alpha \hat{i} + \beta \hat{j}$ on $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$ is $\frac{|\vec{u} \cdot \vec{v}|}{|\vec{v}|} = \frac{|\sqrt{3}\alpha + \beta|}{\sqrt{3+1}} = \frac{|\sqrt{3}\alpha + \beta|}{2} = \sqrt{3}$.So,$|\sqrt{3}\alpha + \beta| = 2\sqrt{3}$.Given $\alpha = 2 + \sqrt{3}\beta$,substitute $\beta = \frac{\alpha - 2}{\sqrt{3}}$:$|\sqrt{3}\alpha + \frac{\alpha - 2}{\sqrt{3}}| = 2\sqrt{3} \Rightarrow |3\alpha + \alpha - 2| = 6 \Rightarrow |4\alpha - 2| = 6$.$4\alpha - 2 = 6 \Rightarrow 4\alpha = 8 \Rightarrow \alpha = 2$,or $4\alpha - 2 = -6 \Rightarrow 4\alpha = -4 \Rightarrow \alpha = -1$.Thus,$|\alpha| = 2$ or $1$.$(B)$ For continuity at $x=1$: $-3a(1)^2 - 2 = b(1) + a^2 \Rightarrow -3a - 2 = b + a^2 \Rightarrow b = -a^2 - 3a - 2$.For differentiability at $x=1$: $f'(x) = -6ax$ for $x  1$.So,$-6a(1) = b \Rightarrow b = -6a$.Equating $b$: $-a^2 - 3a - 2 = -6a \Rightarrow a^2 - 3a + 2 = 0 \Rightarrow (a-1)(a-2) = 0$.So $a = 1$ or $2$.$(C)$ Let $X = 3-3\omega+2\omega^2$. Note that $2+3\omega-3\omega^2 = \omega(2\omega^2+3-3\omega) = \omega X$ and $-3+2\omega+3\omega^2 = \omega^2(2\omega^2+3-3\omega) = \omega^2 X$.The equation becomes $X^{4n+3} + (\omega X)^{4n+3} + (\omega^2 X)^{4n+3} = 0$.$X^{4n+3}(1 + \omega^{4n+3} + \omega^{8n+6}) = 0$.Since $X 
eq 0$,$1 + \omega^{4n} \cdot \omega^3 + \omega^{8n} \cdot \omega^6 = 0 \Rightarrow 1 + \omega^{4n} + \omega^{8n} = 0$.This holds if $4n$ is not a multiple of $3$,i.e.,$n$ is not a multiple of $3$.$(D)$ $a, 5, q, b$ are in $AP$. Let common difference be $d$. $a = 5-d, q = 5+d, b = 5+2d$.$HM = \frac{2ab}{a+b} = 4 \Rightarrow ab = 2(a+b)$.$(5-d)(5+2d) = 2(5-d+5+2d) = 2(10+d) = 20+2d$.$25 + 10d - 5d - 2d^2 = 20 + 2d \Rightarrow 2d^2 - 3d - 5 = 0$.$(2d-5)(d+1) = 0 \Rightarrow d = 2.5$ or $d = -1$.If $d = 2.5, |q-a| = |(5+2.5) - (5-2.5)| = |5| = 5$.If $d = -1, |q-a| = |(5-1) - (5+1)| = |-2| = 2$.

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

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

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