f(x)=12−tan(πx2), −1<x<1 and g(x)=3+4x−4x2 then domain of (f+g) is

f(x)=12−Tan(πx2)Domain of f(x) is (−1, 1)and g(x)=3+4x−4x2 is defined iff3+4x−4x2≥0⇒ 4x2−4x−3≤0⇒ 4x2−6x+2x−3≤0⇒ (2x+1)(2x−3)≤0∴ x∈ [−12,32]Domain of (f+g) is Domain of f ∩ DOmain of gx∈ [−12, 1]

f(x)=12−tan(πx2), −1<x<1 and g(x)=3+4x−4x2 then domain of (f+g) is

f(x)=12−Tan(πx2)Domain of f(x) is (−1, 1)and g(x)=3+4x−4x2 is defined iff3+4x−4x2≥0⇒ 4x2−4x−3≤0⇒ 4x2−6x+2x−3≤0⇒ (2x+1)(2x−3)≤0∴ x∈ [−12,32]Domain of (f+g) is Domain of f ∩ DOmain of gx∈ [−12, 1]

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$f (x) = \frac{1}{2} - \tan (\frac{π x}{2}), - 1 < x < 1 a n d g (x) = \sqrt{3 + 4 x - 4 x^{2}} t h e n d o m a i n o f (f + g) i s$

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$[\frac{1}{2}, 1]$

$[- 1, \frac{1}{2}]$

$[- \frac{1}{2}, 1)$

$[- 1, \frac{- 1}{2}]$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{2} - Tan (\frac{π x}{2}) \\ D o m a i n o f f (x) i s (- 1, 1) \\ a n d g (x) = \sqrt{3 + 4 x - 4 x^{2}} i s d e f i n e d i f f \\ 3 + 4 x - 4 x^{2} \geq 0 \\ \Rightarrow 4 x^{2} - 4 x - 3 \leq 0 \\ \Rightarrow 4 x^{2} - 6 x + 2 x - 3 \leq 0 \\ \Rightarrow (2 x + 1) (2 x - 3) \leq 0 \\ ∴ x \in [\frac{- 1}{2}, \frac{3}{2}] \\ D o m a i n o f (f + g) i s D o m a i n o f f ∩ D O m a i n o f g \\ x \in [\frac{- 1}{2}, 1] \end{array}$