यहाँ (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1)। वर्गों को जोड़ते समय सर्वसमिका पहचानें। / Here (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1). Recognise the identity while adding squares.
Mathematics Answer, Explanation and Revision Hints
यदि (f(x)=\frac{x-2-1}{x-2+1}) और (g(x)=\frac{2x}{x-2+1}) हैं तो (\(f^2+g^2\)(x)) का मान क्या है? / If (f(x)=\frac{x-2-1}{x-2+1}) and (g(x)=\frac{2x}{x-2+1}) then what is the value of (\(f^2+g^2\)(x))?
Correct Answer: A. (1). Explanation: यहाँ (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1)। वर्गों को जोड़ते समय सर्वसमिका पहचानें। / Here (f-2(x)+g-2(x)=\frac{\(x^2-1\)2+(2x)2}{\(x^2+1\)2}=\frac{\(x^2+1\)2}{\(x^2+1\)2}=1). Recognise the identity while adding squares.
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