Algebra of real functions

((f-g)(x)=x^3-x^2=x^2(x-1)), so the zeroes are (x=0) and (x=1). Factorisation is the fastest method at expert level.

Function addition is performed pointwise: \((f+g)(x)=f(x)+g(x)=(x^2+2x)+(x^2-2x)=2x^2\). Therefore, \(\left(\frac{f+g}{2}\right)(x)=\frac{2x^2}{2}=x^2\), so option A is correct. Options C and D are simply \(f\) and \(g\), respectively, while option B omits the squared term. Exam tip: when adding symmetric expressions, first check which opposite terms cancel.

(\frac{f}{g}=\frac{x^2}{4}) and (x^2\ge 0), so the range is ([0,\infty)). Division by a positive constant does not change sign.

(f−g)(x)=f(x)−g(x)=(x²+1)−(x²−1)=x²+1−x²+1=2. Thus, the difference is the constant function 2 for every x. Option D results from mishandling the signs while subtracting g(x). Exam tip: when subtracting a function, change the sign of every term in the function being subtracted.

((f-g)(x)=\frac{x-2}{x-2}=1), but (x=2) is outside the original domain. Keep restrictions even after simplification.

Since \(x^2+1>0\) and \(x^2-1<x^2+1\), the ratio is always less than (1). With a positive denominator, the inequality direction does not change.

\((f+g)(x)=x^2-6x+10+3=x^2-6x+13=(x-3)^2+4\). Since \((x-3)^2\geq 0\), its minimum is 0, attained at \(x=3\). Therefore, the minimum value of \((f+g)(x)\) is 4. Option 1 is only the minimum of \(f(x)\) and incorrectly ignores the added constant \(g(x)=3\). Exam tip: For a quadratic function, completing the square quickly reveals its minimum value.

The domain of (f) is (x\ne 0), and (fg=\frac{1}{x}\cdot x=1). The product may simplify, but the domain comes from the original functions.

For the sum of two functions, (f + g)(x) = f(x) + g(x) = x² + (−x² + 4) = 4. The like terms x² and −x² cancel, so the result is 4 for every x, making it a constant function. The quadratic option is incorrect because the x² term vanishes. Exam tip: combine the coefficients of like powers before identifying the function type.

\((f-g)(x)=f(x)-g(x)=(x^2+px+1)-(x^2-px+1)\). The like terms \(x^2\) and the constants \(1\) cancel, while \(px-(-px)=2px\). Hence, the correct answer is \(2px\). Option D results from mishandling the minus sign before \(-px\). Exam tip: when subtracting functions, distribute the minus sign to every term of the second function.

(f=(x-2)^2), so (\frac{f}{g}=x-2), but (x=2) is excluded. Cancelled factors still leave domain restrictions.

(f(1)=1) and (g(1)=4), so (\lambda+4=9) and (\lambda=5). In a linear combination, first find the function values.

((fg)(x)=2x^3), whose range is all real numbers. An odd-degree polynomial often covers (\mathbb{R}).

(\lfloor x\rfloor+{x}=\lfloor x\rfloor+x-\lfloor x\rfloor=x). Substituting the definition directly makes the question simple.

(\sqrt{x}) needs (x\ge 0), and the denominator needs (x-2\ne 0). In a quotient, both square-root and denominator restrictions apply.

The denominator has (\sqrt{x-2}), so (x-2>0), meaning (x>2). A square root in the denominator cannot be zero.

For the product of functions, \((fg)(1)=f(1)g(1)\). Here, \(f(1)=1+a\) and \(g(1)=1-a\), so \((1+a)(1-a)=0\). By the zero-product property, \(a=-1\) or \(a=1\), hence \(a=\pm1\). Option B is incorrect because substituting \(a=0\) gives a product equal to \(1\). Exam tip: Evaluate the functions at the specified point first, then apply the zero-product property.

The (\frac{1}{x}) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.

(f(x)) simplifies to (x+1), but (f) is not defined at (x=1). Equal functions need both the same rule and the same domain.

((f+g)(x)=x^2-x=x(x-1)), so the zeroes are (0) and (1). For a zero set, set the function equal to (0).

(x^2-4=(x-2)(x+2)), so (x=2) and (x=-2) are forbidden. For a sum, take the intersection of the domains.

(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}), so the value stays below (1) and approaches (1). The actual maximum is not attained here.

First, \((f+g)(x)=f(x)+g(x)=(x^2-1)+(2x+3)=x^2+2x+2\). Completing the square gives \(x^2+2x+2=(x^2+2x+1)+1=(x+1)^2+1\), so option A is correct. Option C becomes \(x^2+2x\), which misses the constant term \(+2\). Exam tip: for \(x^2+2ax\), add and subtract \(a^2\) to form \((x+a)^2-a^2\).

(\frac{1}{x-1}-\frac{1}{x+1}=\frac{x+1-(x-1)}{x^2-1}=\frac{2}{x^2-1}). Watch numerator signs while taking a common denominator.

\((f-g)(x)=f(x)-g(x)=x^2+4-4x=x^2-4x+4=(x-2)^2\). A perfect square satisfies \((x-2)^2\ge 0\) for every real \(x\), so the inequality is true for all real numbers. At \(x=2\), the expression is merely equal to zero; the inequality is not restricted to that value. Exam tip: rewrite a quadratic as a perfect square whenever possible to determine its sign quickly.