b→ and c→ are non-collinear ifa→×(b→×c→)+(a→.b→)b→=(4−2x−siny)b→+(x2−1)c→ and (c→.c→)a→=c→ . Then

a→×(b→×c→)+(a→.b→)b→ =(4−2x−siny)b→+(x2−1)c→ (a→.c→)b→−(a→.b→)c→+(a→.b→)c→Now, (c→.c→)a→=c→. Therefore, (c→.c→)(a→.c→)=(c→.c→) or a→.c→=1 ⇒ 1+a→.b→=4−2x−siny,x2−1=−(a→.b→)Or 1=4−2x−siny+x2−1Or siny=x2−2x+2=(x−1)2+1But siny≤1⇒x=1,siny=1 ⇒ y=(4n+1)π2,n∈I

b→ and c→ are non-collinear ifa→×(b→×c→)+(a→.b→)b→=(4−2x−siny)b→+(x2−1)c→ and (c→.c→)a→=c→ . Then

a→×(b→×c→)+(a→.b→)b→ =(4−2x−siny)b→+(x2−1)c→ (a→.c→)b→−(a→.b→)c→+(a→.b→)c→Now, (c→.c→)a→=c→. Therefore, (c→.c→)(a→.c→)=(c→.c→) or a→.c→=1 ⇒ 1+a→.b→=4−2x−siny,x2−1=−(a→.b→)Or 1=4−2x−siny+x2−1Or siny=x2−2x+2=(x−1)2+1But siny≤1⇒x=1,siny=1 ⇒ y=(4n+1)π2,n∈I

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$\vec{b}$ and $\vec{c}$ are non-collinear if $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} . \vec{b}) \vec{b} = (4 - 2 x - \sin y) \vec{b} + (x^{2} - 1) \vec{c}$ and $(\vec{c} . \vec{c}) \vec{a} = \vec{c}$ . Then

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$x = 1$

$x = - 1$

$y = (4 n + 1) \frac{π}{2}, n \in I$

$y = (2 n + 1) \frac{π}{2}, n \in I$

answer is A, C.

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Detailed Solution

$\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} . \vec{b}) \vec{b} = (4 - 2 x - \sin y) \vec{b} + (x^{2} - 1) \vec{c} (\vec{a} . \vec{c}) \vec{b} - (\vec{a} . \vec{b}) \vec{c} + (\vec{a} . \vec{b}) \vec{c}$

Now, $(\vec{c} . \vec{c}) \vec{a} = \vec{c}$ . Therefore,
$(\vec{c} . \vec{c}) (\vec{a} . \vec{c}) = (\vec{c} . \vec{c})$ or $\vec{a} . \vec{c} = 1$
$\Rightarrow 1 + \vec{a} . \vec{b} = 4 - 2 x - \sin y, x^{2} - 1 = - (\vec{a} . \vec{b})$
Or $1 = 4 - 2 x - \sin y + x^{2} - 1$
Or $\sin y = x^{2} - 2 x + 2 = {(x - 1)}^{2} + 1$
But $\sin y \leq 1 \Rightarrow x = 1, \sin y = 1$
$\Rightarrow y = (4 n + 1) \frac{π}{2}, n \in I$