If - r -1 k cos -1- r - k-2 for any k - 1 and A - r -1 k - r r- then lim x - A 1- x 1 - 3-1-2 x 1 - 4- x - x 2 is equal toA- 0B- 1-2C- -2D- 5-6

The correct option is D ∵Σ _r=1^kcos^ 1β_r=k π/2 Which is possible by when cos^ 1β_1=cos^ 1β_2=cos^ 1β_3=……=cos^ 1β_k=π/2 ∴β_1=β_2=β_3=……=β_k=0 Hence, A=Σ ...

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Standard XII

Mathematics

Angle between Two Vectors

If Σ r =1 k c...

Question

If $Σ_{r = 1}^{k} c o s^{- 1} β_{r} = \frac{k π}{2}$ for any $k \leq 1$ and $A = Σ_{r = 1}^{k} (β_{r})^{r}$ , then $l i m x \to A \frac{(1 + x)^{1 / 3} - (1 - 2 x)^{1 / 4}}{x + x^{2}}$ is equal to

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Solution

The correct option is D
$∵ Σ_{r = 1}^{k} c o s^{- 1} β_{r} = \frac{k π}{2} W h i c h i s p o s s i b l e b y w h e n c o s^{- 1} β_{1} = c o s^{- 1} β_{2} = c o s^{- 1} β_{3} = \dots \dots = c o s^{- 1} β_{k} = \frac{π}{2} ∴ β_{1} = β_{2} = β_{3} = \dots \dots = β_{k} = 0 H e n c e, A = Σ_{r = 1}^{k} (β_{r})^{r} = 0 ∴ l i m x \to 0 \frac{(1 + x)^{1 / 3} - (1 - 2 x)^{1 / 4}}{x + x^{2}} \Rightarrow l i m h \to 0 \frac{(\frac{(1 + x)^{1 / 3} - (1)^{1 / 3}}{(1 + x) - 1}) (\frac{(1 - 2 x)^{1 / 4} - (1)^{1 / 4}}{(1 - 2 x) - 1} \times (- 2))}{(1 + x)} \Rightarrow \frac{\frac{1}{2} + 2 \times \frac{1}{4}}{1 + 0} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$

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