Question

If $\sum_{r = 1}^{k}{}$ cos^–1β_r = $\frac{k\pi}{2}$ for any k ≥ 1 and A =$\sum_{r = 1}^{k}{}$ (β_r)^r. Then $\operatorname { Lim } _ { \mathrm { x } \rightarrow \mathrm { A } }$ $\frac{(1 + x^{2})^{1/3} - (1 - 2x)^{1/4}}{x + x^{2}}$ =

• A. ½
• B. 0
• C. A/2
• D. π/2

Answer 1

Answer: (A)

½

Answer 2

For k = 1, cos^–1 β₁ = $\frac { \pi } { 2 }$ ⇒ β₁ = cos $\frac { \pi } { 2 }$ = 0

k = 2, cos^–1 β₁ + cos^–1 β₂ = π

⇒ cos^–1 β₂ = $\frac { \pi } { 2 }$ = β₂ = 0

By induction β_j = 0, j = 1, …… k.

Hence A = (β_r)^r = 0

Required limit $\lim _ { x \rightarrow 0 }$

$\lim _ { x \rightarrow 0 }$ $\frac { \frac { 2 x } { 3 } \left( 1 + x ^ { 2 } \right) ^ { 2 / 3 } + \frac { 2 } { 4 } ( 1 - 2 x ) ^ { 3 / 4 } } { 1 + 2 x }$ = $\frac { 1 } { 2 }$