Given that A={x2−x−cx2−c−1:x∈R,x2−c−1≠0} where c is a real constant. Also B={x:y=sin−1(3x−1),y∈⌈−π2,π2⌉} . If A∩B=ϕ , then c can be equal to

−1≤3x−1≤1⇒0≤x≤23⇒b=[0,23]Set A is range of y=x2−x−cx2−c−1A∩B=ϕ means range of y does not contains any value of [0,23]∴(y−1)x2+x−((c+1)y−c)=0If y∈[0,23], then the above QE must not posses any solution∴D<0∀y∈[0,23]⇒1+4(y−1)[c(y−1)+y]<0∀y∈[0,23]⇒1+4c(y−1)2+4y(y−1)<0∀y∈[0,23]⇒1+4ct2+4(1−t)(−t)<0∀t∈[13,1] (Here y=1−t)⇒4c+4<4t−1t2∀t∈[13,1]Range of 4t−1t2 is [3,4]⇒4c+4<3⇒C<−14

Given that A={x2−x−cx2−c−1:x∈R,x2−c−1≠0} where c is a real constant. Also B={x:y=sin−1(3x−1),y∈⌈−π2,π2⌉} . If A∩B=ϕ , then c can be equal to

−1≤3x−1≤1⇒0≤x≤23⇒b=[0,23]Set A is range of y=x2−x−cx2−c−1A∩B=ϕ means range of y does not contains any value of [0,23]∴(y−1)x2+x−((c+1)y−c)=0If y∈[0,23], then the above QE must not posses any solution∴D<0∀y∈[0,23]⇒1+4(y−1)[c(y−1)+y]<0∀y∈[0,23]⇒1+4c(y−1)2+4y(y−1)<0∀y∈[0,23]⇒1+4ct2+4(1−t)(−t)<0∀t∈[13,1] (Here y=1−t)⇒4c+4<4t−1t2∀t∈[13,1]Range of 4t−1t2 is [3,4]⇒4c+4<3⇒C<−14

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Given that $A = {\frac{x^{2} - x - c}{x^{2} - c - 1} : x \in R, x^{2} - c - 1 \neq 0}$ where c is a real constant. Also $B = {x : y = \sin^{- 1} (3 x - 1), y \in ⌈ - \frac{π}{2}, \frac{π}{2} ⌉}$ . If $A \cap B = ϕ$ , then c can be equal to

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$\frac{- 1}{\sqrt{15}}$

$\frac{- 1}{\sqrt{17}}$

$\frac{- 1}{\sqrt{14}}$

$\frac{- 1}{\sqrt{18}}$

answer is B, D.

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Detailed Solution

$- 1 \leq 3 x - 1 \leq 1 \Rightarrow 0 \leq x \leq \frac{2}{3} \Rightarrow b = [0, \frac{2}{3}]$
Set A is range of $y = \frac{x^{2} - x - c}{x^{2} - c - 1}$
$A \cap B = ϕ$ means range of y does not contains any value of $[0, \frac{2}{3}]$
$∴ (y - 1) x^{2} + x - ((c + 1) y - c) = 0$
If $y \in [0, \frac{2}{3}],$ then the above QE must not posses any solution
$∴ D < 0 \forall y \in [0, \frac{2}{3}]$
$\Rightarrow 1 + 4 (y - 1) [c (y - 1) + y] < 0 \forall y \in [0, \frac{2}{3}]$
$\Rightarrow 1 + 4 c {(y - 1)}^{2} + 4 y (y - 1) < 0 \forall y \in [0, \frac{2}{3}]$
$\Rightarrow 1 + 4 c t^{2} + 4 (1 - t) (- t) < 0 \forall t \in [\frac{1}{3}, 1] (Here y = 1 - t)$ $\Rightarrow 4 c + 4 < \frac{4 t - 1}{t^{2}} \forall t \in [\frac{1}{3}, 1]$

Range of $\frac{4 t - 1}{t^{2}}$ is [3,4]
$\Rightarrow 4 c + 4 < 3 \Rightarrow C < \frac{- 1}{4}$