Question

If a = min {x² + 4x + 5, x ∈ R} and

b = $( - \infty,1\rbrack \cup \lbrack 5,\infty)$ then the value of $\lbrack 0,\infty)$ is-

• A. $y(x)$
• B. 2^{n + 1} – 1
• C. $2^{x} + 2^{y} = 2$
• D. None of these

Answer 1

Answer: (B)

2^{n + 1} – 1

Answer 2

a = min{x² + 4x + 5} = $\frac { 4.1 .5 - 4 ^ { 2 } } { 4 \times 1 }$ = 1

b = $\lim _ { \theta \rightarrow 0 } \frac { 1 - \cos 2 \theta } { \theta ^ { 2 } }$ = 2

= $\sum _ { \mathrm { r } = 0 } ^ { \mathrm { n } } 1.2 ^ { \mathrm { n } - \mathrm { r } }$

= 2ⁿ(1 + 2^–1 + 2^–2 + .... + 2^–n)

= 2ⁿ $\left( 1 \frac { \left[ 1 - ( 1 / 2 ) ^ { n + 1 } \right] } { 1 - ( 1 / 2 ) } \right)$ = 2ⁿ⁺¹ – 1