📝 Arithmetic Operations on Functions (15 MCQs)

📖 Explanation: The quotient requires both numerator and denominator to be defined, and the denominator must not be zero. Therefore we start with the intersection of the individual domains and then remove any x for which \g(x)=0\. This yields option C, which correctly describes the domain.

📖 Explanation: The sum \f+g\ is defined wherever both f and g are defined. Since g is defined for all real numbers, the domain of the sum is limited only by the domain of f, which is \[2,\\infty)\. Hence option A correctly states the domain.

📖 Explanation: Both f and g are only defined for \x\\ge0\, so their product \fg\ has domain \[0,\\infty)\. The function \h(x)=x\ is defined for every real number, giving a larger domain \( -\\infty,\\infty)\. Thus the domain of \fg\ is a proper subset of the domain of \h\, which matches option B.

📖 Explanation: Multiplying a constant function by a scalar multiplies its constant value. Here \f(x)=2\, so \5f(x)=5\\times2=10\. The resulting function is the constant \10\, which corresponds to option B. The other choices either forget the multiplication or incorrectly introduce a variable factor.

📖 Explanation: Both f and g require \x-2>0\ for the square root, and f additionally requires \x\\neq2\. The intersection of these conditions is \x>2\. Since \g(x)\ never equals zero on this interval, no further points are removed. Hence the domain is \(2,\\infty)\, which is option A.

📖 Explanation: From the derived formula \(f+g)(x)=x-2+\\sqrt{x-2}\, substitute \x=4\. This gives \4-2+\\sqrt{4-2}=2+\\sqrt{2}\. Therefore the correct value is \2+\\sqrt{2}\, which matches option A. The other options differ by either adding or subtracting extra constants.

📖 Explanation: First evaluate \f(5)=1+\\sqrt{5-2}=1+\\sqrt{3}\. Multiplying by 7 gives \7\\times(1+\\sqrt{3})=7+7\\sqrt{3}\. This corresponds to option A. The other choices either use incorrect radicands or omit the multiplication by 7.

📖 Explanation: The quotient requires both functions to be defined and the denominator non‑zero. \k(x)=x^{2}\ is zero only at \x=0\. Since \h\ needs \x\\ge0\, the intersection is \[0,\\infty)\. Removing the point where the denominator vanishes leaves \(0,\\infty)\, described as \[0,\\infty)\ excluding \x=0\. Hence option A is correct.

📖 Explanation: The difference \f-g\ is defined only where both functions exist, i.e., the intersection \[0,\\infty)\\cap( -\\infty,2)=[0,2)\. Among the choices, \-1\ and \3\ lie outside this interval, \2\ is excluded because the interval is half‑open, and \0\ lies within. Therefore option B correctly identifies a valid point.

📖 Explanation: The product of two functions is obtained by multiplying their formulas: \(f\\cdot g)(x)=\\frac{1}{x-1}\\cdot(x+2)=\\frac{x+2}{x-1}\. This matches option A. The other options either invert the fraction or multiply incorrectly, so they do not represent the correct product.

📖 Explanation: Statement B suggests that a point where both f and g are undefined could still belong to the domain of the quotient, which contradicts the definition requiring each function to be defined at that point. Therefore B is false. The other statements correctly describe the domain restrictions.

📖 Explanation: The product is defined where both functions exist. The intersection of \[1,\\infty)\ with \( -\\infty,1)\\cup(1,\\infty)\ removes the point \x=1\ because g is undefined there, leaving \(1,\\infty)\. Thus the domain of \fg\ is \(1,\\infty)\, which corresponds to option B.

📖 Explanation: Multiplying a function by a non‑zero constant does not introduce any new restrictions; the resulting function is defined exactly wherever the original function is defined. Therefore the domain remains \D\. Option C correctly states this, while the other options add unnecessary exclusions or expansions.

📖 Explanation: The function \h\ is precisely the quotient \f/g\; its domain consists of points where both \f\ and \g\ are defined and the denominator is non‑zero. This is the intersection of the domains of \f\ and \g\ with the point \x=1\ removed, yielding \(1,\\infty)\. Hence option C accurately describes the domain.

📖 Explanation: Both functions require \x-2>0\ for the square root and \x\\neq2\ for the denominator. Their individual domains are \(2,\\infty)\; the point \x=2\ is excluded by both conditions. Consequently, the sum \f+g\ is defined exactly on \(2,\\infty)\, which matches option B.