\( \Large f \left(x\right)=\frac{1}{2}-\tan \frac{ \pi x}{2},\ -1 < x < 1\ and\ g \left(x\right) \)  \( \Large =\sqrt{ \left(3+4x-4x^{2}\right) } \) then dom \( \Large \left(f + g\right) \) is given by:


A) \( \Large \left[ \frac{1}{2}, 1 \right] \)

B) \( \Large \left[ \frac{1}{2}, -1 \right] \)

C) \( \Large \left[ -\frac{1}{2}, 1 \right] \)

D) \( \Large \left[ -\frac{1}{2}, -1 \right] \)

Correct Answer:
C) \( \Large \left[ -\frac{1}{2}, 1 \right] \)

Description for Correct answer:

Given that \( \Large f \left(x\right)=\frac{1}{2}- \tan \left(\frac{ \pi x}{2}\right),\ -1 < x < 1 \)
br /> Here domain of \( \Large f \left(x\right)\ is\ d_{1} = \left(-1,\ 1\right) \)
br /> and \( \Large g \left(x\right)=\sqrt{3+4x-4x^{2}}=\sqrt{-4x^{2}-6x+2x-3} \)
=> \( \Large \sqrt{- \left(2x-3\right) \left(2x+1\right) } \)

=> \( \Large - \left(2x-3\right) \left(2x+1\right)\ge 0 \)

=> \( \Large \left(2x-3\right) \left(2x+1\right) \le 0 \)

=> \( \Large -\frac{1}{2} \le x \le \frac{3}{2} \)

Therefore, Domain of \( \Large g \left(x\right) \) is \( \Large d_{2} = \left[ -\frac{1}{2},\ \frac{3}{2} \right] \)

Therefore, Domain of \( \Large \left(f+g\right)=d_{1} \cap d_{2} = \left[ -\frac{1}{2,\ 1} \right] \)


Part of solved Set theory questions and answers : >> Elementary Mathematics >> Set theory








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