If fx-1-sin2 x3 cos2 x - x -2- then fx is continuous at x-2- if

F(x) is continuous at x=π2 So, f(π2^ )=f (π2)=f (π2^+) ⇒lim_x →π2^ 1 sin^2x3 cos^2x=a=lim_x →π2^+b(1 sin x)(π 2x)^2 ⇒13=a=lim_x →π2^+b( cos x)2 (π 2x)( 2) a=13 ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

If fx=1-sin2 ...

Question

If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{2} x}{3 {cos}^{2} x} & , & x < \frac{π}{2} a & , & x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}} & , & x > \frac{π}{2} \end{matrix}$ then $f (x)$ is continuous at $x = \frac{π}{2}$ , if

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Solution

$f (x)$ is continuous at $x = \frac{π}{2}$
So, $f ({\frac{π}{2}}^{-}) = f (\frac{π}{2}) = f ({\frac{π}{2}}^{+})$
$\Rightarrow lim x \to {\frac{π}{2}}^{-} \frac{1 - {sin}^{2} x}{3 {cos}^{2} x} = a = lim x \to {\frac{π}{2}}^{+} \frac{b (1 - sin x)}{(π - 2 x)^{2}}$
$\Rightarrow \frac{1}{3} = a = lim x \to {\frac{π}{2}}^{+} \frac{b (- cos x)}{2 (π - 2 x) (- 2)}$
$a = \frac{1}{3} = lim x \to {\frac{π}{2}}^{+} \frac{b (sin x)}{- 4 (- 2)} = \frac{b}{8}$
So, $a = \frac{1}{3}$ and $b = \frac{8}{3}$

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If f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1−sin2x3cos2x,x<π2a,x=π2b(1−sinx)(π−2x)2,x>π2 then f(x) is continuous at x=π2, if

f(x) is continuous at x=π2 So, f(π2−)=f(π2)=f(π2+) ⇒limx→π2−1−sin2x3cos2x=a=limx→π2+b(1−sinx)(π−2x)2 ⇒13=a=limx→π2+b(−cosx)2(π−2x)(−2) a=13=limx→π2+b(sinx)−4(−2)=b8 So, a=13 and b=83