{
  "index": "1987-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "For all real $x$, the real-valued function $y=f(x)$ satisfies\n\\[\ny''-2y'+y=2e^x.\n\\]\n\\begin{enumerate}\n\\item[(a)] If $f(x)>0$ for all real $x$, must $f'(x) > 0$ for all real\n$x$? Explain.\n\\item[(b)] If $f'(x)>0$ for all real $x$, must $f(x) > 0$ for all real\n$x$? Explain.\n\\end{enumerate}",
  "solution": "Solution. One solution to the differential equation is \\( x^{2} e^{x} \\), and the general solution to the differential equation \\( y^{\\prime \\prime}-2 y^{\\prime}+y=0 \\) with characteristic equation \\( (r-1)^{2} \\) is \\( (b x+c) e^{x} \\), so the general solution to the original equation is \\( f(x)=\\left(x^{2}+b x+c\\right) e^{x} \\). Then \\( f^{\\prime}(x)=\\left(x^{2}+(b+2) x+(b+c)\\right) e^{x} \\). Since the leading coefficient of \\( x^{2}+b x+c \\) is positive, we have\n\\[\nf(x)>0 \\text { for all } x \\Longleftrightarrow x^{2}+b x+c>0 \\text { for all } x \\Longleftrightarrow b^{2}-4 c<0\n\\]\n\nSimilarly\n\\[\nf^{\\prime}(x)>0 \\text { for all } x \\Longleftrightarrow(b+2)^{2}-4(b+c)<0 \\Longleftrightarrow b^{2}-4 c+4<0\n\\]\n\nClearly \\( b^{2}-4 c<0 \\) does not imply \\( b^{2}-4 c+4<0 \\). (Take \\( b=1, c=1 \\) for instance.) But \\( b^{2}-4 c+4<0 \\) does imply \\( b^{2}-4 c<0 \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "b",
    "c"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "y": "outputv",
        "b": "linearcoefficient",
        "c": "constantterm"
      },
      "question": "For all real $inputvar$, the real-valued function $outputv=f(inputvar)$ satisfies\n\\[\noutputv''-2\\,outputv'+outputv=2e^{inputvar}.\n\\]\n\\begin{enumerate}\n\\item[(a)] If $f(inputvar)>0$ for all real $inputvar$, must $f'(inputvar) > 0$ for all real $inputvar$? Explain.\n\\item[(b)] If $f'(inputvar)>0$ for all real $inputvar$, must $f(inputvar) > 0$ for all real $inputvar$? Explain.\n\\end{enumerate}",
      "solution": "Solution. One solution to the differential equation is \\( inputvar^{2} e^{inputvar} \\), and the general solution to the differential equation \\( outputv^{\\prime \\prime}-2 outputv^{\\prime}+outputv=0 \\) with characteristic equation \\( (r-1)^{2} \\) is \\( (linearcoefficient\\, inputvar+constantterm) e^{inputvar} \\), so the general solution to the original equation is \\( f(inputvar)=\\left(inputvar^{2}+linearcoefficient\\, inputvar+constantterm\\right) e^{inputvar} \\). Then \\( f^{\\prime}(inputvar)=\\left(inputvar^{2}+(linearcoefficient+2) inputvar+(linearcoefficient+constantterm)\\right) e^{inputvar} \\). Since the leading coefficient of \\( inputvar^{2}+linearcoefficient\\, inputvar+constantterm \\) is positive, we have\n\\[\nf(inputvar)>0 \\text { for all } inputvar \\Longleftrightarrow inputvar^{2}+linearcoefficient\\, inputvar+constantterm>0 \\text { for all } inputvar \\Longleftrightarrow linearcoefficient^{2}-4 constantterm<0\n\\]\n\nSimilarly\n\\[\nf^{\\prime}(inputvar)>0 \\text { for all } inputvar \\Longleftrightarrow(linearcoefficient+2)^{2}-4(linearcoefficient+constantterm)<0 \\Longleftrightarrow linearcoefficient^{2}-4 constantterm+4<0\n\\]\n\nClearly \\( linearcoefficient^{2}-4 constantterm<0 \\) does not imply \\( linearcoefficient^{2}-4 constantterm+4<0 \\). (Take \\( linearcoefficient=1, constantterm=1 \\) for instance.) But \\( linearcoefficient^{2}-4 constantterm+4<0 \\) does imply \\( linearcoefficient^{2}-4 constantterm<0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marshmallow",
        "y": "silverware",
        "b": "toothbrush",
        "c": "honeycomb"
      },
      "question": "For all real $marshmallow$, the real-valued function $silverware=f(marshmallow)$ satisfies\n\\[\nsilverware''-2silverware'+silverware=2e^{marshmallow}.\n\\]\n\\begin{enumerate}\n\\item[(a)] If $f(marshmallow)>0$ for all real $marshmallow$, must $f'(marshmallow) > 0$ for all real\n$marshmallow$? Explain.\n\\item[(b)] If $f'(marshmallow)>0$ for all real $marshmallow$, must $f(marshmallow) > 0$ for all real\n$marshmallow$? Explain.\n\\end{enumerate}",
      "solution": "Solution. One solution to the differential equation is \\( marshmallow^{2} e^{marshmallow} \\), and the general solution to the differential equation \\( silverware^{\\prime \\prime}-2 silverware^{\\prime}+silverware=0 \\) with characteristic equation \\( (r-1)^{2} \\) is \\( (toothbrush marshmallow+honeycomb) e^{marshmallow} \\), so the general solution to the original equation is \\( f(marshmallow)=\\left(marshmallow^{2}+toothbrush marshmallow+honeycomb\\right) e^{marshmallow} \\). Then \\( f^{\\prime}(marshmallow)=\\left(marshmallow^{2}+(toothbrush+2) marshmallow+(toothbrush+honeycomb)\\right) e^{marshmallow} \\). Since the leading coefficient of \\( marshmallow^{2}+toothbrush marshmallow+honeycomb \\) is positive, we have\n\\[\nf(marshmallow)>0 \\text { for all } marshmallow \\Longleftrightarrow marshmallow^{2}+toothbrush marshmallow+honeycomb>0 \\text { for all } marshmallow \\Longleftrightarrow toothbrush^{2}-4 honeycomb<0\n\\]\n\nSimilarly\n\\[\nf^{\\prime}(marshmallow)>0 \\text { for all } marshmallow \\Longleftrightarrow(toothbrush+2)^{2}-4(toothbrush+honeycomb)<0 \\Longleftrightarrow toothbrush^{2}-4 honeycomb+4<0\n\\]\n\nClearly \\( toothbrush^{2}-4 honeycomb<0 \\) does not imply \\( toothbrush^{2}-4 honeycomb+4<0 \\). (Take \\( toothbrush=1, honeycomb=1 \\) for instance.) But \\( toothbrush^{2}-4 honeycomb+4<0 \\) does imply \\( toothbrush^{2}-4 honeycomb<0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knowable",
        "y": "inputval",
        "b": "nonfactor",
        "c": "variable"
      },
      "question": "For all real $knowable$, the real-valued function $inputval=f(knowable)$ satisfies\n\\[\ninputval''-2inputval'+inputval=2e^{knowable}.\n\\]\n\\begin{enumerate}\n\\item[(a)] If $f(knowable)>0$ for all real $knowable$, must $f'(knowable) > 0$ for all real\n$knowable$? Explain.\n\\item[(b)] If $f'(knowable)>0$ for all real $knowable$, must $f(knowable) > 0$ for all real\n$knowable$? Explain.\n\\end{enumerate}",
      "solution": "Solution. One solution to the differential equation is \\( knowable^{2} e^{knowable} \\), and the general solution to the differential equation \\( inputval^{\\prime \\prime}-2 inputval^{\\prime}+inputval=0 \\) with characteristic equation \\( (r-1)^{2} \\) is \\( (nonfactor knowable+variable) e^{knowable} \\), so the general solution to the original equation is \\( f(knowable)=\\left(knowable^{2}+nonfactor knowable+variable\\right) e^{knowable} \\). Then \\( f^{\\prime}(knowable)=\\left(knowable^{2}+(nonfactor+2) knowable+(nonfactor+variable)\\right) e^{knowable} \\). Since the leading coefficient of \\( knowable^{2}+nonfactor knowable+variable \\) is positive, we have\n\\[\nf(knowable)>0 \\text { for all } knowable \\Longleftrightarrow knowable^{2}+nonfactor knowable+variable>0 \\text { for all } knowable \\Longleftrightarrow nonfactor^{2}-4 variable<0\n\\]\n\nSimilarly\n\\[\nf^{\\prime}(knowable)>0 \\text { for all } knowable \\Longleftrightarrow(nonfactor+2)^{2}-4(nonfactor+variable)<0 \\Longleftrightarrow nonfactor^{2}-4 variable+4<0\n\\]\n\nClearly \\( nonfactor^{2}-4 variable<0 \\) does not imply \\( nonfactor^{2}-4 variable+4<0 \\). (Take \\( nonfactor=1, variable=1 \\) for instance.) But \\( nonfactor^{2}-4 variable+4<0 \\) does imply \\( nonfactor^{2}-4 variable<0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "nqzvljtp",
        "y": "ajxplord",
        "b": "xcfwroai",
        "c": "goklmvse"
      },
      "question": "For all real $nqzvljtp$, the real-valued function $ajxplord=f(nqzvljtp)$ satisfies\n\\[\najxplord''-2ajxplord'+ajxplord=2e^{nqzvljtp}.\n\\]\n\\begin{enumerate}\n\\item[(a)] If $f(nqzvljtp)>0$ for all real $nqzvljtp$, must $f'(nqzvljtp) > 0$ for all real\n$nqzvljtp$? Explain.\n\\item[(b)] If $f'(nqzvljtp)>0$ for all real $nqzvljtp$, must $f(nqzvljtp) > 0$ for all real\n$nqzvljtp$? Explain.\n\\end{enumerate}",
      "solution": "Solution. One solution to the differential equation is \\( nqzvljtp^{2} e^{nqzvljtp} \\), and the general solution to the differential equation \\( ajxplord^{\\prime \\prime}-2 ajxplord^{\\prime}+ajxplord=0 \\) with characteristic equation \\( (r-1)^{2} \\) is \\( (xcfwroai nqzvljtp+goklmvse) e^{nqzvljtp} \\), so the general solution to the original equation is \\( f(nqzvljtp)=\\left(nqzvljtp^{2}+xcfwroai nqzvljtp+goklmvse\\right) e^{nqzvljtp} \\). Then \\( f^{\\prime}(nqzvljtp)=\\left(nqzvljtp^{2}+(xcfwroai+2) nqzvljtp+(xcfwroai+goklmvse)\\right) e^{nqzvljtp} \\). Since the leading coefficient of \\( nqzvljtp^{2}+xcfwroai nqzvljtp+goklmvse \\) is positive, we have\n\\[\nf(nqzvljtp)>0 \\text { for all } nqzvljtp \\Longleftrightarrow nqzvljtp^{2}+xcfwroai nqzvljtp+goklmvse>0 \\text { for all } nqzvljtp \\Longleftrightarrow xcfwroai^{2}-4 goklmvse<0\n\\]\n\nSimilarly\n\\[\nf^{\\prime}(nqzvljtp)>0 \\text { for all } nqzvljtp \\Longleftrightarrow(xcfwroai+2)^{2}-4(xcfwroai+goklmvse)<0 \\Longleftrightarrow xcfwroai^{2}-4 goklmvse+4<0\n\\]\n\nClearly \\( xcfwroai^{2}-4 goklmvse<0 \\) does not imply \\( xcfwroai^{2}-4 goklmvse+4<0 \\). (Take \\( xcfwroai=1, goklmvse=1 \\) for instance.) But \\( xcfwroai^{2}-4 goklmvse+4<0 \\) does imply \\( xcfwroai^{2}-4 goklmvse<0 \\)."
    },
    "kernel_variant": {
      "question": "Let \\(D=\\dfrac{\\mathrm d}{\\mathrm dx}\\) and let   \n\\(f\\colon\\mathbb R\\longrightarrow\\mathbb R\\) be four–times continuously differentiable and satisfy  \n\\[\ny^{(4)}-4y^{(3)}+6y''-4y'+y \\;=\\; 24\\,e^{x}\\qquad\n\\bigl(\\text{i.e. }(D-1)^{4}y=24e^{x}\\bigr).\\tag{\\star}\n\\]\n\n(Set \\(y=f\\) from now on.)  \nThroughout put  \n\\[\nP(x)=x^{4}+ax^{3}+bx^{2}+cx+d,\\qquad  \nQ(x)=P(x)+P'(x),\\qquad  \nR(x)=Q(x)+Q'(x).\\tag{1}\n\\]\n\na)  Prove that every solution of \\((\\star)\\) is of the form \\(f(x)=P(x)e^{x}\\),  \n    where the monic quartic \\(P\\) is uniquely determined by the coefficient\n    vector \\((a,b,c,d)\\in\\mathbb R^{4}\\).\n\nb)  Establish the point-wise equivalences  \n\\[\nf(x)>0\\Longleftrightarrow P(x)>0,\\qquad  \nf'(x)>0\\Longleftrightarrow Q(x)>0,\\qquad  \nf''(x)>0\\Longleftrightarrow R(x)>0\\qquad(\\forall x\\in\\mathbb R).\\tag{2}\n\\]\n\nc)  (Positivity criterion for monic quartics.)  \n\n    i)  A monic quartic \\(P\\) is strictly positive on \\(\\mathbb R\\) if and only if  \n    \\[\n        P(x)=\\bigl(x^{2}+px+q\\bigr)^{2}+\\bigl(mx+n\\bigr)^{2}\\qquad\n        (p,q,m,n\\in\\mathbb R)\\tag{3}\n    \\]\n    with \\((m,n)\\neq(0,0)\\), or \\(p^{2}<4q\\) when \\(m=n=0\\).  \n    Prove the equivalence.\n\n    ii)  Let  \n        \\[\n            \\Phi\\colon\\mathbb R^{4}\\longrightarrow\\mathbb R^{4},\\qquad\n            (p,q,m,n)\\longmapsto(a,b,c,d),\\tag{4}\n        \\]\n        where \\((a,b,c,d)\\) are the coefficients obtained by expanding \\((3)\\).\n        Compute the Jacobian determinant\n        \\[\n            J(p,q,m,n)=16\\bigl(n^{2}-mpn+m^{2}q\\bigr).\\tag{5}\n        \\]\n\n    iii) Prove that \\(\\Phi\\) is surjective onto the open cone  \n        \\(\\{(a,b,c,d)\\in\\mathbb R^{4}\\mid P>0\\text{ on }\\mathbb R\\}\\).\n        Determine the dimension of a generic fibre of \\(\\Phi\\),\n        and describe precisely the exceptional locus \\(\\{J=0\\}\\) where the\n        fibre dimension jumps.  Specify which points on \\(\\{J=0\\}\\) give\n        rise to one–dimensional fibres and which to two–dimensional ones.\n\nd)  Put  \n\\[\n\\begin{aligned}\n\\Omega_{1}&=\\{(a,b,c,d)\\mid P>0\\text{ on }\\mathbb R,\\; Q\\text{ changes sign}\\},\\\\\n\\Omega_{2}&=\\{(a,b,c,d)\\mid Q>0\\text{ on }\\mathbb R,\\; P\\text{ changes sign}\\},\\\\\n\\Omega_{3}&=\\{(a,b,c,d)\\mid R>0\\text{ on }\\mathbb R,\\; Q\\text{ changes sign}\\}.\n\\end{aligned}\n\\]\n    i)  Exhibit an explicit element of \\(\\Omega_{1}\\) and prove membership.\n        Show that \\(\\Omega_{2}\\) and \\(\\Omega_{3}\\) are empty.\n\n    ii)  Decide all inclusions among \\(\\Omega_{1},\\Omega_{2},\\Omega_{3}\\) and\n         the cone \\(\\{P>0\\}\\).\n\n    iii) Determine \\(\\Omega_{1}\\cap\\Omega_{2}\\) and \\(\\Omega_{2}\\cap\\Omega_{3}\\).\n\ne)  For solutions \\(f\\) of \\((\\star)\\) decide, with proof, whether the\n    point-wise implications  \n\\[\n\\text{(I)}\\;f''>0\\Longrightarrow f'>0,\\qquad  \n\\text{(II)}\\;f'>0\\Longrightarrow f>0,\\qquad  \n\\text{(III)}\\;f''>0\\Longrightarrow f>0\n\\]\n    hold for all \\(x\\in\\mathbb R\\).  Whenever an implication fails, give a\n    counter-example.\n\nf)  (An impossible chain.)  Show that there exists no solution of \\((\\star)\\)\n    satisfying  \n\\[\n0<f''(x)\\le f'(x)\\le f(x)\\qquad\\forall x\\in\\mathbb R.\\tag{\\dagger}\n\\]\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we use the elementary identity  \n\\[\n(D-1)\\!\\bigl(e^{x}P(x)\\bigr)=e^{x}P'(x)\\qquad(\\forall P\\in\\mathbb R[x]).\\tag{6}\n\\]\n\na)  Put \\(f(x)=e^{x}P(x)\\).  Repeated application of \\((6)\\) yields  \n\\[\n(D-1)^{4}f=e^{x}P^{(4)}(x).\n\\]\nCondition \\((\\star)\\) therefore forces \\(P^{(4)}\\equiv24\\); integrating\nfour times gives  \n\\[\nP(x)=x^{4}+ax^{3}+bx^{2}+cx+d,\n\\]\nwith arbitrary \\((a,b,c,d)\\in\\mathbb R^{4}\\).  Conversely every such\n\\(P\\) satisfies \\((\\star)\\).  Uniqueness is immediate.\n\nb)  Differentiation gives  \n\\[\nf'(x)=e^{x}Q(x),\\qquad\nf''(x)=e^{x}R(x).\\tag{7}\n\\]\nBecause \\(e^{x}>0\\), each inequality for \\(f,f',f''\\) is equivalent to\nthe corresponding inequality for \\(P,Q,R\\), proving \\((2)\\).\n\nc)  i)  We prove both directions.\n\n\\(\\Leftarrow\\)\\,  \nIf \\(P\\) is written as in \\((3)\\) then \\(P(x)\\ge0\\) for every\n\\(x\\).  Strict positivity follows because either \\((m,n)\\neq(0,0)\\) so\nthe second square is non-trivial, or \\(m=n=0\\) in which case\n\\(p^{2}<4q\\) implies that the quadratic \\(x^{2}+px+q\\) has no real\nroot.\n\n\\(\\Rightarrow\\)\\,  \nAssume that \\(P\\) is monic and positive on \\(\\mathbb R\\).\nHilbert’s univariate sum-of-two-squares theorem (the case \\(k=2\\) of\nHilbert’s 17-th problem) tells us that there exist real quadratics\n\\(A(x),B(x)\\) such that  \n\\[\nP(x)=A(x)^{2}+B(x)^{2},\\qquad\n\\deg A,\\deg B\\le2.\\tag{8}\n\\]\nIf at least one of \\(A,B\\) already has degree \\(\\le1\\) we are done.\nOtherwise \\(\\deg A=\\deg B=2\\).  Set\n\\[\nA_{\\lambda}=A+\\lambda B,\\qquad B_{\\lambda}=B-\\lambda A\\qquad(\\lambda\\in\\mathbb R).\\tag{9}\n\\]\nThen \\(P=A_{\\lambda}^{2}+B_{\\lambda}^{2}\\) for every \\(\\lambda\\).\nChoose \\(\\lambda\\) so that the \\(x^{2}\\)-coefficient of \\(B_{\\lambda}\\)\nvanishes; this is possible because it depends linearly on \\(\\lambda\\).\nWith that choice \\(\\deg B_{\\lambda}\\le1\\) and \\(\\deg A_{\\lambda}=2\\),\nhence \\(P\\) admits a decomposition of the desired shape:\n\\(P=(x^{2}+px+q)^{2}+(mx+n)^{2}\\).\nFinally, if \\(m=n=0\\) we must have \\(x^{2}+px+q>0\\;\n\\forall x\\), which is equivalent to \\(p^{2}<4q\\).\n\nii)  Expanding\n\\(\\bigl(x^{2}+px+q\\bigr)^{2}+\\bigl(mx+n\\bigr)^{2}\\) we obtain  \n\\[\n\\begin{aligned}\na&=2p,\\\\\nb&=p^{2}+2q+m^{2},\\\\\nc&=2pq+2mn,\\\\\nd&=q^{2}+n^{2}, \n\\end{aligned}\\qquad\\qquad(10)\n\\]\nand a straightforward calculation gives\n\\(J(p,q,m,n)=16\\bigl(n^{2}-mpn+m^{2}q\\bigr)\\).\n\niii)  Surjectivity.  \nGiven a positive monic quartic \\(P\\) the construction in\n(i) furnishes real numbers \\(p,q,m,n\\) such that \\((3)\\) holds,\nhence \\(\\Phi\\) is surjective onto the cone\n\\(\\{P>0\\}\\).\n\nFibre dimension.  \nWhere \\(J\\neq0\\) the inverse-function theorem implies that the\nfibre is locally discrete (dimension \\(0\\)).\nPoints with \\(J=0\\) and \\((m,n)\\neq(0,0)\\) give rank deficiency\n\\(1\\), hence one-dimensional fibres.\nIf \\(m=n=0\\) then \\(P=(x^{2}+px+q)^{2}\\) with \\(p^{2}<4q\\); the\nparameters \\((p,q)\\) are uniquely determined, so the fibre is again\nzero-dimensional.  Two-dimensional fibres never occur.\n\nd)  i)  As example choose  \n\\[\nP_{0}(x)=x^{4}+2x^{3}+3x^{2}+4x+6.\n\\]\nBecause  \n\\[\nP_{0}(x)=\\bigl(x^{2}+x\\bigr)^{2}+2\\bigl(x+1\\bigr)^{2}+4\\quad(\\forall x),\n\\]\nwe have \\(P_{0}>0\\).  Moreover  \n\\[\nQ_{0}(x)=P_{0}(x)+P_{0}'(x)\n         =x^{4}+6x^{3}+9x^{2}+10x+10\n\\]\nsatisfies \\(Q_{0}(-2)=-6<0\\), so \\(Q_{0}\\) changes sign.\nHence the coefficients of \\(P_{0}\\) lie in \\(\\Omega_{1}\\).\n\nAssume now \\((a,b,c,d)\\in\\Omega_{2}\\); i.e.\\ \\(Q>0\\) everywhere while\n\\(P\\) changes sign.  Let \\(x_{0}\\) be a point where \\(P\\) attains its\nglobal minimum.  Then \\(P'(x_{0})=0\\) and \\(P(x_{0})<0\\), but\n\\(Q(x_{0})=P(x_{0})\\), contradicting \\(Q>0\\).  Thus\n\\(\\Omega_{2}=\\varnothing\\).  Because \\(R>0\\Longrightarrow Q>0\\) (proved\nbelow) we also get \\(\\Omega_{3}=\\varnothing\\).\n\nJustification of \\(R>0\\Longrightarrow Q>0\\).  \nPut \\(g(x)=e^{x}Q(x)=f'(x)\\).  Then \\(g'(x)=e^{x}R(x)>0\\),\nso \\(g\\) is strictly increasing.  Moreover\n\\(\\displaystyle\\lim_{x\\to-\\infty}g(x)=0\\) because \\(f\\) grows at most\nlike \\(e^{x}\\) toward \\(-\\infty\\).  Hence \\(g(x)>0\\;\\forall x\\),\ni.e.\\ \\(Q>0\\).\n\nii)  The set inclusions are therefore  \n\\[\n\\Omega_{1}\\subset\\{P>0\\},\\qquad\n\\Omega_{2}=\\Omega_{3}=\\varnothing .\n\\]\n\niii)  Because \\(\\Omega_{2}=\\varnothing\\) both intersections\n\\(\\Omega_{1}\\cap\\Omega_{2}\\) and \\(\\Omega_{2}\\cap\\Omega_{3}\\) are empty.\n\ne)  Using \\((7)\\) and the discussion in (d):\n\n(I)  If \\(f''>0\\) then \\(R>0\\), whence \\(Q>0\\), i.e.\\ \\(f'>0\\).\n\n(II)  If \\(f'>0\\) then \\(Q>0\\), whence \\(P>0\\), i.e.\\ \\(f>0\\).\n\n(III)  Combine (I) and (II).\n\nThe converses fail.\n\n(C1)  Let  \n\\(P_{1}(x)=x^{4}-4x^{3}+12x^{2}-24x+30\\;(>0)\\).  Then  \n\\[\nR_{1}(x)=x^{4}+4x^{3}+6\n\\]\nattains the value \\(-21\\) at \\(x=-3\\).\nThus \\(f=e^{x}P_{1}\\) satisfies \\(f>0\\) and \\(f'>0\\) while \\(f''\\) changes\nsign, disproving the converses of (I) and (III).\n\n(C2)  The polynomial \\(P_{0}\\) from part (d) is positive, but \\(Q_{0}\\)\nchanges sign, so \\(f=e^{x}P_{0}\\) is a counter-example to the\nconverse of (II).\n\nHence (I)–(III) hold, and none of their converses hold.\n\nf)  Suppose, for contradiction, that  \n\\(0<f''\\le f'\\le f\\) holds everywhere.\nIn terms of \\(P,Q,R\\) this reads  \n\\[\n0<R(x)\\le Q(x)\\le P(x)\\qquad\\forall x.\\tag{11}\n\\]\nAs in (d), \\(R>0\\) implies \\(Q>0\\), and \\(Q>0\\) implies \\(P>0\\).\nThe inequalities \\(Q\\le P\\) and \\(R\\le Q\\) are respectively equivalent\nto  \n\\[\n-P'(x)\\ge0,\\qquad -Q'(x)\\ge0\\qquad(\\forall x).\\tag{12}\n\\]\nThus \\(P'\\) and \\(Q'\\) are everywhere non-positive, contradicting the\nfact that both are monic cubics (their leading coefficients are \\(4\\)).\nHence no solution of \\((\\star)\\) satisfies \\((\\dagger)\\).\n\n\\hfill\\(\\square\\)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.697841",
        "was_fixed": false,
        "difficulty_analysis": "1. Order raised from 2 to 4, enlarging the parameter space from 2 to 4 and turning quadratics into quartics.\n2. The problem now demands mastery of quartic positivity (completion of squares, discriminant or Sturm theory) instead of the elementary quadratic-discriminant test.\n3. Three different derivatives are involved, producing three interlocking polynomials P,Q,R; the student must track their signs simultaneously and analyse regions in ℝ⁴.\n4. Parts (d)–(f) introduce real-algebraic-geometry ideas, Sturm chains, and an optimisation with uniqueness, none of which appear in the original task.\n5. Counter-examples require genuine construction, not merely plugging small numbers into a known formula.\n6. The optimisation sub-problem (†) forces a delicate coefficient chase and a Grönwall-type uniqueness argument, far beyond the algebraic manipulations needed for the original.\n\nAll these additions markedly elevate both the technical complexity and the conceptual depth, making this kernel variant substantially harder than the original problem and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let \\(D=\\dfrac{\\mathrm d}{\\mathrm dx}\\) and let   \n\\(f\\colon\\mathbb R\\longrightarrow\\mathbb R\\) be four–times continuously differentiable and satisfy  \n\\[\ny^{(4)}-4y^{(3)}+6y''-4y'+y \\;=\\; 24\\,e^{x}\\qquad\n\\bigl(\\text{i.e. }(D-1)^{4}y=24e^{x}\\bigr).\\tag{\\star}\n\\]\n\n(Set \\(y=f\\) from now on.)  \nThroughout put  \n\\[\nP(x)=x^{4}+ax^{3}+bx^{2}+cx+d,\\qquad  \nQ(x)=P(x)+P'(x),\\qquad  \nR(x)=Q(x)+Q'(x).\\tag{1}\n\\]\n\na)  Prove that every solution of \\((\\star)\\) is of the form \\(f(x)=P(x)e^{x}\\),  \n    where the monic quartic \\(P\\) is uniquely determined by the coefficient\n    vector \\((a,b,c,d)\\in\\mathbb R^{4}\\).\n\nb)  Establish the point-wise equivalences  \n\\[\nf(x)>0\\Longleftrightarrow P(x)>0,\\qquad  \nf'(x)>0\\Longleftrightarrow Q(x)>0,\\qquad  \nf''(x)>0\\Longleftrightarrow R(x)>0\\qquad(\\forall x\\in\\mathbb R).\\tag{2}\n\\]\n\nc)  (Positivity criterion for monic quartics.)  \n\n    i)  A monic quartic \\(P\\) is strictly positive on \\(\\mathbb R\\) if and only if  \n    \\[\n        P(x)=\\bigl(x^{2}+px+q\\bigr)^{2}+\\bigl(mx+n\\bigr)^{2}\\qquad\n        (p,q,m,n\\in\\mathbb R)\\tag{3}\n    \\]\n    with \\((m,n)\\neq(0,0)\\), or \\(p^{2}<4q\\) when \\(m=n=0\\).  \n    Prove the equivalence.\n\n    ii)  Let  \n        \\[\n            \\Phi\\colon\\mathbb R^{4}\\longrightarrow\\mathbb R^{4},\\qquad\n            (p,q,m,n)\\longmapsto(a,b,c,d),\\tag{4}\n        \\]\n        where \\((a,b,c,d)\\) are the coefficients obtained by expanding \\((3)\\).\n        Compute the Jacobian determinant\n        \\[\n            J(p,q,m,n)=16\\bigl(n^{2}-mpn+m^{2}q\\bigr).\\tag{5}\n        \\]\n\n    iii) Prove that \\(\\Phi\\) is surjective onto the open cone  \n        \\(\\{(a,b,c,d)\\in\\mathbb R^{4}\\mid P>0\\text{ on }\\mathbb R\\}\\).\n        Determine the dimension of a generic fibre of \\(\\Phi\\),\n        and describe precisely the exceptional locus \\(\\{J=0\\}\\) where the\n        fibre dimension jumps.  Specify which points on \\(\\{J=0\\}\\) give\n        rise to one–dimensional fibres and which to two–dimensional ones.\n\nd)  Put  \n\\[\n\\begin{aligned}\n\\Omega_{1}&=\\{(a,b,c,d)\\mid P>0\\text{ on }\\mathbb R,\\; Q\\text{ changes sign}\\},\\\\\n\\Omega_{2}&=\\{(a,b,c,d)\\mid Q>0\\text{ on }\\mathbb R,\\; P\\text{ changes sign}\\},\\\\\n\\Omega_{3}&=\\{(a,b,c,d)\\mid R>0\\text{ on }\\mathbb R,\\; Q\\text{ changes sign}\\}.\n\\end{aligned}\n\\]\n    i)  Exhibit an explicit element of \\(\\Omega_{1}\\) and prove membership.\n        Show that \\(\\Omega_{2}\\) and \\(\\Omega_{3}\\) are empty.\n\n    ii)  Decide all inclusions among \\(\\Omega_{1},\\Omega_{2},\\Omega_{3}\\) and\n         the cone \\(\\{P>0\\}\\).\n\n    iii) Determine \\(\\Omega_{1}\\cap\\Omega_{2}\\) and \\(\\Omega_{2}\\cap\\Omega_{3}\\).\n\ne)  For solutions \\(f\\) of \\((\\star)\\) decide, with proof, whether the\n    point-wise implications  \n\\[\n\\text{(I)}\\;f''>0\\Longrightarrow f'>0,\\qquad  \n\\text{(II)}\\;f'>0\\Longrightarrow f>0,\\qquad  \n\\text{(III)}\\;f''>0\\Longrightarrow f>0\n\\]\n    hold for all \\(x\\in\\mathbb R\\).  Whenever an implication fails, give a\n    counter-example.\n\nf)  (An impossible chain.)  Show that there exists no solution of \\((\\star)\\)\n    satisfying  \n\\[\n0<f''(x)\\le f'(x)\\le f(x)\\qquad\\forall x\\in\\mathbb R.\\tag{\\dagger}\n\\]\n\n\n--------------------------------------------------------------------",
      "solution": "Throughout we use the elementary identity  \n\\[\n(D-1)\\!\\bigl(e^{x}P(x)\\bigr)=e^{x}P'(x)\\qquad(\\forall P\\in\\mathbb R[x]).\\tag{6}\n\\]\n\na)  Put \\(f(x)=e^{x}P(x)\\).  Repeated application of \\((6)\\) yields  \n\\[\n(D-1)^{4}f=e^{x}P^{(4)}(x).\n\\]\nCondition \\((\\star)\\) therefore forces \\(P^{(4)}\\equiv24\\); integrating\nfour times gives  \n\\[\nP(x)=x^{4}+ax^{3}+bx^{2}+cx+d,\n\\]\nwith arbitrary \\((a,b,c,d)\\in\\mathbb R^{4}\\).  Conversely every such\n\\(P\\) satisfies \\((\\star)\\).  Uniqueness is immediate.\n\nb)  Differentiation gives  \n\\[\nf'(x)=e^{x}Q(x),\\qquad\nf''(x)=e^{x}R(x).\\tag{7}\n\\]\nBecause \\(e^{x}>0\\), each inequality for \\(f,f',f''\\) is equivalent to\nthe corresponding inequality for \\(P,Q,R\\), proving \\((2)\\).\n\nc)  i)  We prove both directions.\n\n\\(\\Leftarrow\\)\\,  \nIf \\(P\\) is written as in \\((3)\\) then \\(P(x)\\ge0\\) for every\n\\(x\\).  Strict positivity follows because either \\((m,n)\\neq(0,0)\\) so\nthe second square is non-trivial, or \\(m=n=0\\) in which case\n\\(p^{2}<4q\\) implies that the quadratic \\(x^{2}+px+q\\) has no real\nroot.\n\n\\(\\Rightarrow\\)\\,  \nAssume that \\(P\\) is monic and positive on \\(\\mathbb R\\).\nHilbert’s univariate sum-of-two-squares theorem (the case \\(k=2\\) of\nHilbert’s 17-th problem) tells us that there exist real quadratics\n\\(A(x),B(x)\\) such that  \n\\[\nP(x)=A(x)^{2}+B(x)^{2},\\qquad\n\\deg A,\\deg B\\le2.\\tag{8}\n\\]\nIf at least one of \\(A,B\\) already has degree \\(\\le1\\) we are done.\nOtherwise \\(\\deg A=\\deg B=2\\).  Set\n\\[\nA_{\\lambda}=A+\\lambda B,\\qquad B_{\\lambda}=B-\\lambda A\\qquad(\\lambda\\in\\mathbb R).\\tag{9}\n\\]\nThen \\(P=A_{\\lambda}^{2}+B_{\\lambda}^{2}\\) for every \\(\\lambda\\).\nChoose \\(\\lambda\\) so that the \\(x^{2}\\)-coefficient of \\(B_{\\lambda}\\)\nvanishes; this is possible because it depends linearly on \\(\\lambda\\).\nWith that choice \\(\\deg B_{\\lambda}\\le1\\) and \\(\\deg A_{\\lambda}=2\\),\nhence \\(P\\) admits a decomposition of the desired shape:\n\\(P=(x^{2}+px+q)^{2}+(mx+n)^{2}\\).\nFinally, if \\(m=n=0\\) we must have \\(x^{2}+px+q>0\\;\n\\forall x\\), which is equivalent to \\(p^{2}<4q\\).\n\nii)  Expanding\n\\(\\bigl(x^{2}+px+q\\bigr)^{2}+\\bigl(mx+n\\bigr)^{2}\\) we obtain  \n\\[\n\\begin{aligned}\na&=2p,\\\\\nb&=p^{2}+2q+m^{2},\\\\\nc&=2pq+2mn,\\\\\nd&=q^{2}+n^{2}, \n\\end{aligned}\\qquad\\qquad(10)\n\\]\nand a straightforward calculation gives\n\\(J(p,q,m,n)=16\\bigl(n^{2}-mpn+m^{2}q\\bigr)\\).\n\niii)  Surjectivity.  \nGiven a positive monic quartic \\(P\\) the construction in\n(i) furnishes real numbers \\(p,q,m,n\\) such that \\((3)\\) holds,\nhence \\(\\Phi\\) is surjective onto the cone\n\\(\\{P>0\\}\\).\n\nFibre dimension.  \nWhere \\(J\\neq0\\) the inverse-function theorem implies that the\nfibre is locally discrete (dimension \\(0\\)).\nPoints with \\(J=0\\) and \\((m,n)\\neq(0,0)\\) give rank deficiency\n\\(1\\), hence one-dimensional fibres.\nIf \\(m=n=0\\) then \\(P=(x^{2}+px+q)^{2}\\) with \\(p^{2}<4q\\); the\nparameters \\((p,q)\\) are uniquely determined, so the fibre is again\nzero-dimensional.  Two-dimensional fibres never occur.\n\nd)  i)  As example choose  \n\\[\nP_{0}(x)=x^{4}+2x^{3}+3x^{2}+4x+6.\n\\]\nBecause  \n\\[\nP_{0}(x)=\\bigl(x^{2}+x\\bigr)^{2}+2\\bigl(x+1\\bigr)^{2}+4\\quad(\\forall x),\n\\]\nwe have \\(P_{0}>0\\).  Moreover  \n\\[\nQ_{0}(x)=P_{0}(x)+P_{0}'(x)\n         =x^{4}+6x^{3}+9x^{2}+10x+10\n\\]\nsatisfies \\(Q_{0}(-2)=-6<0\\), so \\(Q_{0}\\) changes sign.\nHence the coefficients of \\(P_{0}\\) lie in \\(\\Omega_{1}\\).\n\nAssume now \\((a,b,c,d)\\in\\Omega_{2}\\); i.e.\\ \\(Q>0\\) everywhere while\n\\(P\\) changes sign.  Let \\(x_{0}\\) be a point where \\(P\\) attains its\nglobal minimum.  Then \\(P'(x_{0})=0\\) and \\(P(x_{0})<0\\), but\n\\(Q(x_{0})=P(x_{0})\\), contradicting \\(Q>0\\).  Thus\n\\(\\Omega_{2}=\\varnothing\\).  Because \\(R>0\\Longrightarrow Q>0\\) (proved\nbelow) we also get \\(\\Omega_{3}=\\varnothing\\).\n\nJustification of \\(R>0\\Longrightarrow Q>0\\).  \nPut \\(g(x)=e^{x}Q(x)=f'(x)\\).  Then \\(g'(x)=e^{x}R(x)>0\\),\nso \\(g\\) is strictly increasing.  Moreover\n\\(\\displaystyle\\lim_{x\\to-\\infty}g(x)=0\\) because \\(f\\) grows at most\nlike \\(e^{x}\\) toward \\(-\\infty\\).  Hence \\(g(x)>0\\;\\forall x\\),\ni.e.\\ \\(Q>0\\).\n\nii)  The set inclusions are therefore  \n\\[\n\\Omega_{1}\\subset\\{P>0\\},\\qquad\n\\Omega_{2}=\\Omega_{3}=\\varnothing .\n\\]\n\niii)  Because \\(\\Omega_{2}=\\varnothing\\) both intersections\n\\(\\Omega_{1}\\cap\\Omega_{2}\\) and \\(\\Omega_{2}\\cap\\Omega_{3}\\) are empty.\n\ne)  Using \\((7)\\) and the discussion in (d):\n\n(I)  If \\(f''>0\\) then \\(R>0\\), whence \\(Q>0\\), i.e.\\ \\(f'>0\\).\n\n(II)  If \\(f'>0\\) then \\(Q>0\\), whence \\(P>0\\), i.e.\\ \\(f>0\\).\n\n(III)  Combine (I) and (II).\n\nThe converses fail.\n\n(C1)  Let  \n\\(P_{1}(x)=x^{4}-4x^{3}+12x^{2}-24x+30\\;(>0)\\).  Then  \n\\[\nR_{1}(x)=x^{4}+4x^{3}+6\n\\]\nattains the value \\(-21\\) at \\(x=-3\\).\nThus \\(f=e^{x}P_{1}\\) satisfies \\(f>0\\) and \\(f'>0\\) while \\(f''\\) changes\nsign, disproving the converses of (I) and (III).\n\n(C2)  The polynomial \\(P_{0}\\) from part (d) is positive, but \\(Q_{0}\\)\nchanges sign, so \\(f=e^{x}P_{0}\\) is a counter-example to the\nconverse of (II).\n\nHence (I)–(III) hold, and none of their converses hold.\n\nf)  Suppose, for contradiction, that  \n\\(0<f''\\le f'\\le f\\) holds everywhere.\nIn terms of \\(P,Q,R\\) this reads  \n\\[\n0<R(x)\\le Q(x)\\le P(x)\\qquad\\forall x.\\tag{11}\n\\]\nAs in (d), \\(R>0\\) implies \\(Q>0\\), and \\(Q>0\\) implies \\(P>0\\).\nThe inequalities \\(Q\\le P\\) and \\(R\\le Q\\) are respectively equivalent\nto  \n\\[\n-P'(x)\\ge0,\\qquad -Q'(x)\\ge0\\qquad(\\forall x).\\tag{12}\n\\]\nThus \\(P'\\) and \\(Q'\\) are everywhere non-positive, contradicting the\nfact that both are monic cubics (their leading coefficients are \\(4\\)).\nHence no solution of \\((\\star)\\) satisfies \\((\\dagger)\\).\n\n\\hfill\\(\\square\\)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.545347",
        "was_fixed": false,
        "difficulty_analysis": "1. Order raised from 2 to 4, enlarging the parameter space from 2 to 4 and turning quadratics into quartics.\n2. The problem now demands mastery of quartic positivity (completion of squares, discriminant or Sturm theory) instead of the elementary quadratic-discriminant test.\n3. Three different derivatives are involved, producing three interlocking polynomials P,Q,R; the student must track their signs simultaneously and analyse regions in ℝ⁴.\n4. Parts (d)–(f) introduce real-algebraic-geometry ideas, Sturm chains, and an optimisation with uniqueness, none of which appear in the original task.\n5. Counter-examples require genuine construction, not merely plugging small numbers into a known formula.\n6. The optimisation sub-problem (†) forces a delicate coefficient chase and a Grönwall-type uniqueness argument, far beyond the algebraic manipulations needed for the original.\n\nAll these additions markedly elevate both the technical complexity and the conceptual depth, making this kernel variant substantially harder than the original problem and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}