Let f x -left begin matrix lfrac1-sin - 3 x3 cos - 2 x- - xfrac- p i2lend - matrix right-l If f x is continuous at x -2- p - q -A- 1-2- 2B- 1-4C- 1-2- 4D- I -2- 7

The correct option is C ( 1/2,4 )f [ ( π/2 )^ ]=h→ 0lim1 sin^3 [ ( π/2 ) h ]/3 cos^2 [ ( π/2 ) h ] =h→ 0lim1 cos^3h/3 sin^2 h=1/2 f [ ( π/2 )^+ ]=h→ ...

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Standard XII

Mathematics

In Radius

Let f x =left...

Question

Let $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - s i n^{3} x}{3 c o s^{2} x} : & x < \frac{π}{2} p : & x = \frac{π}{2} \frac{q (1 - s i n x)}{(x - 2 x)^{2}} : & x > \frac{π}{2} \end{matrix}$ If f(x) is continuous at $x = \frac{π}{2}, (p, q) =$

(1, 4)

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

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

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(\frac{1}{2}, 7)

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Solution

The correct option is C $(\frac{1}{2}, 4)$
$f [{(\frac{π}{2})}^{-}] = l i m h \to 0 \frac{1 - s i n^{3} [(\frac{π}{2}) - h]}{3 c o s^{2} [(\frac{π}{2}) - h]}$
$= l i m h \to 0 \frac{1 - c o s^{3} h}{3 s i n^{2} h} = \frac{1}{2}$
$f [{(\frac{π}{2})}^{+}] = l i m h \to 0 \frac{q [1 - s i n {(\frac{π}{2}) + h}]}{{[π - 2 {(\frac{π}{2}) + h}]}^{2}}$
$= l i m h \to 0 \frac{q (1 - c o s h)}{4 h^{2}} = \frac{q}{8}$
$∴ p = \frac{1}{2} = \frac{q}{8} \Rightarrow p = \frac{1}{2}, q = 4$

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