(A)
In a plane, if the magnitude of the projection vector
Of the vector $α \hat{i} + β \hat{j}$ on $\sqrt{3} \hat{i} + \hat{j}$ is $\sqrt{3}$ and if $α = 2 + \sqrt{3} β$ ,
then possible value(s) of $|α|$ is (are)
(P) 1
(B)
Let a and b be real numbers such that the
function is $f (x) = \{\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ b x + a^{2}, & x \geq 1 \end{matrix}$
differentiable for all real values of x. Then possible
value(s) of a is(are)
(Q) 2
(C)
Let $ω \neq 1$ be a complex cube root of unity. If
${(3 - 3 ω + 2 ω^{2})}^{4 n + 3} + {(2 + 3 ω - 3 ω^{2})}^{4 n + 3} +$ ${(- 3 + 2 ω + 3 ω^{2})}^{4 n + 3} = 0$
, 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

Difficult

Solution

$\begin{array}{l} (A) The projection of α \hat{i} + β \hat{j} on the vector \sqrt{3} \hat{i} + \hat{j} is \sqrt{3} \\ Hence, \frac{\sqrt{3} α + β}{2} = \sqrt{3} \\ \sqrt{3} α + β = 2 \sqrt{3} \\ substitute α = 2 + \sqrt{3} β \\ \sqrt{3} (2 + \sqrt{3} β) + β = 2 \sqrt{3} \\ 2 \sqrt{3} + 3 β + β = 2 \sqrt{3} \\ β = 0 and α = 2 \end{array}$
$\begin{array}{l} (B) f (x) = \{\begin{matrix} - 3 a x^{2} - 2 & x < 1 \\ b x + a^{2} & x \geq 1 \end{matrix} \\ \underset{x \to 1^{-}}{L t} f (x) = \underset{x \to 1^{+}}{L t} f (x) \\ - 3 a - 2 = b + a^{2} \\ - a^{2} - 3 a - 2 = b \\ f^{1} (x) = \{\begin{matrix} - 6 a x & x < 1 \\ b & x > 1 \end{matrix} \\ f^{1} (1^{-}) = f^{1} (1^{+}) \\ - 6 a = b \\ - a^{2} - 3 a - 2 = - 6 a \\ a^{2} - 3 a + 2 = 0 \\ a = 1, a = 2 \end{array}$
$\begin{array}{l} (C) {(3 - 3 ω + 2 ω^{2})}^{4 n + 3} + {(2 + 3 ω - 3 ω^{2})}^{4 n + 3} + {(- 3 + 2 ω + 3 ω^{2})}^{4 n + 3} = 0 \\ 3 - 3 ω + 2 ω^{2} = 2 (1 + ω^{2}) + 1 - 3 ω = - 5 ω + 1 \\ 2 + 3 ω - 3 ω^{2} = 2 (1 + ω) + ω - 3 ω^{2} = ω - 5 ω^{2} = ω (- 5 ω + 1) \\ - 3 + 2 ω + 3 ω^{2} = - 3 + 2 (ω + ω^{2}) + ω^{2} = - 5 + ω^{2} = ω^{2} (- 5 ω + 1) \\ {(- 5 ω + 1)}^{4 n + 3} \{1 + ω^{4 n + 3} + {(ω^{2})}^{4 n + 3}\} = 0 \\ n = 1, n = 2, n = 4, n = 5 \end{array}$
$\begin{array}{l} (D) \frac{2 a b}{a + b} = 4 \Rightarrow a b = 2 (a + b) \\ \Rightarrow b = \frac{2 a}{a - 2} \\ a, 5, q, b are in AP \\ a + d = 5 \Rightarrow d = 5 - a \\ b = a + 3 d = \frac{2 a}{a - 2} \\ \Rightarrow a + 3 (5 - a) = \frac{2 a}{a - 2} \\ \Rightarrow 2 a^{2} - 17 a + 30 = 0 \\ \Rightarrow a = 6 or \frac{5}{2} \\ |q - a| = 2, 5 \end{array}$

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