KCET · Maths · Differential Equations
\[
f(x)=\frac{1}{2}-\tan \left(\frac{\Pi x}{2}\right)-1 < x < 1 \text { and } g(x)=\sqrt{\left(3+4 x-4 x^{2}\right)}
\]
Find domain of \( (f+g) \)
- A \( \left(-\frac{1}{2}, 1\right) \)
- B \( \left(-\frac{1}{2}, 1\right] \)
- C \( \left[-\frac{1}{2}, \frac{3}{2}\right] \)
- D \( (-1,1) \)
Answer & Solution
Correct Answer
(B) \( \left(-\frac{1}{2}, 1\right] \)
Step-by-step Solution
Detailed explanation
Given that,
\(f(x)=\frac{1}{2}-\tan ^{-1}\left(\frac{\Pi x}{2}\right)\)
and \(g(x)=\sqrt{3+4 x-4 x^{2}}\)
Since, \(f x()\) is defined in \((-1,1)\)
So, \(g(x)\) to defined, we have
\(3+4 x-4 x^{2} \geq 0\)
\(4 x^{2}-4 x-3 \leq 0\)
\(\Rightarrow 4 x^{2}-4 x+1-4 \leq 0\)
\(\Rightarrow(2 x-1)^{2}-4 \leq 0\)
\(\Rightarrow(2 x-1)^{2} \leq 4\)
\(\Rightarrow-2 \leq 2 x-1 \leq 2\)
\(\Rightarrow-1 \leq 2 x \leq 3\)
\(\Rightarrow-\frac{1}{2} \leq x \leq \frac{3}{2}\)
Therefore, domain of \((f+g)=\left(-\frac{1}{2}, 1\right)\)
\(f(x)=\frac{1}{2}-\tan ^{-1}\left(\frac{\Pi x}{2}\right)\)
and \(g(x)=\sqrt{3+4 x-4 x^{2}}\)
Since, \(f x()\) is defined in \((-1,1)\)
So, \(g(x)\) to defined, we have
\(3+4 x-4 x^{2} \geq 0\)
\(4 x^{2}-4 x-3 \leq 0\)
\(\Rightarrow 4 x^{2}-4 x+1-4 \leq 0\)
\(\Rightarrow(2 x-1)^{2}-4 \leq 0\)
\(\Rightarrow(2 x-1)^{2} \leq 4\)
\(\Rightarrow-2 \leq 2 x-1 \leq 2\)
\(\Rightarrow-1 \leq 2 x \leq 3\)
\(\Rightarrow-\frac{1}{2} \leq x \leq \frac{3}{2}\)
Therefore, domain of \((f+g)=\left(-\frac{1}{2}, 1\right)\)
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