AP EAMCET · Maths · Functions
A real valued function \(f: A \rightarrow B\) defined by \(f(x)=\frac{4-x^2}{4+x^2} \forall x \in A\) is a bijection. If \(-4 \in A\) then \(A \cap B=\)
- A \((-1,1]\)
- B \([0,1]\)
- C \([0, \infty)\)
- D \((-1,0]\)
Answer & Solution
Correct Answer
(D) \((-1,0]\)
Step-by-step Solution
Detailed explanation
\(y = \frac{4-x^2}{4+x^2} \implies y(4+x^2) = 4-x^2 \implies 4y+yx^2 = 4-x^2 \implies x^2(y+1) = 4(1-y)\) \(x^2 = \frac{4(1-y)}{y+1}\) Since \(x^2 \ge 0\), \(\frac{1-y}{y+1} \ge 0 \implies -1 \(f(x)\) is an even function. For injectivity and \(-4 \in A\),…
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