Question

For an integer \(k \ge 0\), let \(P_k\) denote the vector space of all real polynomials in one variable of degree less than or equal to \(k\). Define a linear transformation \(T : P_2 \to P_3\) by \[ T(f(x)) = f''(x) + x f(x). \] Which one of the following polynomials is not in the range of \(T\)?

• A. \(x + x^2\)
• B. \(x^2 + x^3 + 2\)
• C. \(x + x^3 + 2\)
• D. \(x + 1\)

Hint:

When checking the range of a linear operator, compare coefficients using the mapping’s structure to find consistency constraints.

Answer 1

The Correct Option is D

Step 1: Represent \(f(x) = a + bx + cx^2.\)
Then \(f''(x) = 2c.\) So \[ T(f(x)) = 2c + x(a + bx + cx^2) = a x + b x^2 + c x^3 + 2c. \] Hence, \[ T(f(x)) = c x^3 + b x^2 + a x + 2c. \]
Step 2: Compare with each option.
The coefficient of \(x^3\) equals the constant term divided by 2, i.e. \[ \text{Coeff}(x^3) = \frac{\text{Const}}{2}. \] In option (C), the constant term = 2, coefficient of \(x^3 = 1\). Since \(1 \ne 2/2 = 1\)? Wait that satisfies; check carefully: for option (C) we have const=2, coeff(x^3)=1, so condition holds. But we must check all degrees. For \(x+x^3+2\): coefficient pattern (x^3=1, x^2=0, x=1, const=2). We can’t find \(a,b,c\) satisfying simultaneously: \[ a=1, \ b=0, \ c=1, \text{ but const term }=2c=2 \Rightarrow c=1, \text{ OK.} \] Wait, that fits—so check again other options? Actually for option (C), we can check linear dependence. For (B), \(x^2 + x^3 + 2\): \(a=0,b=1,c=1\Rightarrow T(f)=x^3+x^2+2\) works. For (A), \(x+x^2\): \(a=1,b=1,c=0\Rightarrow T(f)=x^2+x\) works. For (C), we need \(b=0,a=1,c=1\Rightarrow T(f)=x^3+x+2\) works. Hmm—all work. Actually, the polynomial not in range must violate the constraint that const term = 2×coeff(x^3). For (C), constant = 2, coeff(x^3)=1 → valid. But for (B), const=2, coeff(x^3)=1 → also valid. For (A), const=0, coeff(x^3)=0 → valid. For (D), const=1, coeff(x^3)=0 → violates const=2×coeff(x^3). Hence, correct answer is (D).
Step 3: Conclusion.
Polynomial \(x + 1\) is not in the range of \(T.\)