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Question 94 of 148

Q.Find the intervals of convexity and concavity of the Gaussian curve y=e−x2y=e^{-x^2} and also find the points of inflection. OR Show that (Z,∗)(Z, *) is an infinite abelian group, where '*' is defined as a∗b=a+b+2a*b=a+b+2 and Z is the set of all integers.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★
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Main part: use the sign of y′′y'' for y=e−x2y=e^{-x^2} to find convex/concave intervals and inflection points. OR alternative: verify the group axioms for (Z,∗)(Z,*) with a∗b=a+b+2a*b=a+b+2.

Main part — convexity, concavity and inflection points of y=e−x2y=e^{-x^2}

  1. First derivative.

    y=e−x2⇒y′=−2x e−x2y=e^{-x^2}\Rightarrow y'=-2x\,e^{-x^2}

  2. Second derivative (product rule on −2x-2x and e−x2e^{-x^2}):

    y′′=−2e−x2+(−2x)(−2x e−x2)=−2e−x2+4x2e−x2=e−x2(4x2−2)=2e−x2(2x2−1)y''=-2e^{-x^2}+(-2x)\big(-2x\,e^{-x^2}\big)=-2e^{-x^2}+4x^2e^{-x^2}=e^{-x^2}(4x^2-2)=2e^{-x^2}(2x^2-1)

  3. Find where y′′=0y''=0. Since e−x2>0e^{-x^2}>0 for all xx, y′′=0  ⟺  2x2−1=0  ⟺  x2=12  ⟺  x=±12y''=0 \iff 2x^2-1=0 \iff x^2=\dfrac12 \iff x=\pm\dfrac{1}{\sqrt2}.

  4. Determine the sign of y′′y'' on each interval (sign follows the sign of 2x2−12x^2-1, since e−x2>0e^{-x^2}>0 always):

    • For ∣x∣>12|x|>\dfrac{1}{\sqrt2}: 2x2−1>0⇒y′′>02x^2-1>0 \Rightarrow y''>0 — curve is convex (concave up).

    • For ∣x∣<12|x|<\dfrac{1}{\sqrt2}: 2x2−1<0⇒y′′<02x^2-1<0 \Rightarrow y''<0 — curve is concave (concave down).

  5. Points of inflection. At x=±12x=\pm\dfrac{1}{\sqrt2}, y′′y'' changes sign, so these are genuine inflection points.

    y(±12)=e−1/2=1ey\left(\pm\dfrac{1}{\sqrt2}\right)=e^{-1/2}=\dfrac{1}{\sqrt e}

    Inflection points: (12,1e)\left(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt e}\right) and (−12,1e)\left(-\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt e}\right).

OR — alternative: (Z,∗)(Z,*) with a∗b=a+b+2a*b=a+b+2 is an infinite abelian group

  1. Closure. For any a,b∈Za,b\in Z, a∗b=a+b+2a*b=a+b+2 is a sum of integers plus 22, hence an integer. So a∗b∈Za*b\in Z — closure holds.

  2. Associativity. For any a,b,c∈Za,b,c\in Z:

    (a∗b)∗c=(a+b+2)∗c=(a+b+2)+c+2=a+b+c+4(a*b)*c=(a+b+2)*c=(a+b+2)+c+2=a+b+c+4

    a∗(b∗c)=a∗(b+c+2)=a+(b+c+2)+2=a+b+c+4a*(b*c)=a*(b+c+2)=a+(b+c+2)+2=a+b+c+4

    Both equal a+b+c+4a+b+c+4, so (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c) — associativity holds.

  3. Identity element. Seek e∈Ze\in Z with a∗e=aa*e=a for all aa: a+e+2=a⇒e=−2a+e+2=a\Rightarrow e=-2.

    Check: e∗a=−2+a+2=ae*a=-2+a+2=a, and a∗e=a+(−2)+2=aa*e=a+(-2)+2=a. So e=−2∈Ze=-2\in Z is a two-sided identity.

  4. Inverse of each element. For a∈Za\in Z, seek a′∈Za'\in Z with a∗a′=e=−2a*a'=e=-2: a+a′+2=−2⇒a′=−4−a=−(a+4)a+a'+2=-2\Rightarrow a'=-4-a=-(a+4).

    …

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