⨍ x=1-sin3x3cos2xifx π2 isis continuous at x=π2 then p, q=

fx=1-sin3 x3cos2 xif x π2 is continuous at x=π2limx→π-2 fx=fπ2=limx→π+2fxlimx→π-21-sin3 x3cos2 x=ρ=limx→π+2q(1-sinx)π-2x2limx→π-2 (1-sinx)(1+sinx+sin2 x)3(1-sinx)(1+sinx)=ρ= limx→π+2 q (1-cos π2-x22π2-x2 332=ρ=q4×12⇒ρ=12, q=4Option 4 is correct

⨍ x=1-sin3x3cos2xifx π2 isis continuous at x=π2 then p, q=

fx=1-sin3 x3cos2 xif x π2 is continuous at x=π2limx→π-2 fx=fπ2=limx→π+2fxlimx→π-21-sin3 x3cos2 x=ρ=limx→π+2q(1-sinx)π-2x2limx→π-2 (1-sinx)(1+sinx+sin2 x)3(1-sinx)(1+sinx)=ρ= limx→π+2 q (1-cos π2-x22π2-x2 332=ρ=q4×12⇒ρ=12, q=4Option 4 is correct

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$⨍ (x) = \{\begin{matrix} \frac{1 - \sin^{3} x}{3 \cos^{2} x} & if & x < \frac{π}{2} \\ p & if & x = \frac{π}{2} \\ \frac{q (1 - sinx)}{{(π - 2 x)}^{2}} & if & x > \frac{π}{2} \end{matrix}$ is
is continuous at $x = \frac{π}{2}$ then $(p, q) =$

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$(4, \frac{1}{2})$

$(- \frac{1}{2}, 4)$

$(4, - \frac{1}{2})$

$(\frac{1}{2}, 4)$

answer is D.

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Detailed Solution

$f (x) = \{\begin{matrix} \frac{1 - \sin^{3} x}{3 \cos^{2} x} if x < \frac{π}{2} \\ p if x = \frac{π}{2} \\ \frac{q (1 - sinx)}{{(π - 2 x)}^{2}} if x > \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$

$\lim_{x \to \frac{π^{-}}{2}} f (x) = f (\frac{π}{2}) = \lim_{x \to \frac{π +}{2}} f (x)$

$\lim_{x \to \frac{π^{-}}{2}} \frac{1 - \sin^{3} x}{3 \cos^{2} x} = ρ = \lim_{x \to \frac{π +}{2}} q \frac{(1 - sinx)}{{(π - 2 x)}^{2}}$

$\lim_{x \to \frac{π^{-}}{2}} \frac{(1 - sinx) (1 + sinx + \sin^{2} x)}{3 (1 - sinx) (1 + sinx)} = ρ = \lim_{x \to \frac{π^{+}}{2}} q \frac{(1 - \cos (\frac{π}{2} - x)}{2^{2} {(\frac{π}{2} - x)}^{2}}$