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Q.(a) A wire of length 36 m is to be cut into two pieces. One of the pieces is to be made a square and the other, a circle. What would be the lengths of the two pieces, so that the combined area of the square and the circle is minimum ?

(OR)
(b) Find : ∫x3x4+3x2+2 dx\displaystyle\int \dfrac{x^3}{x^4 + 3x^2 + 2} \, dx
CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★
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  1. Minimising the combined area gives the square's piece =144π+4=\dfrac{144}{\pi+4} m and the circle's piece =36ππ+4=\dfrac{36\pi}{\pi+4} m;
  2. with t=x2t=x^2 the integral becomes log⁡∣x2+2∣−12log⁡∣x2+1∣+C\log|x^2+2|-\tfrac12\log|x^2+1|+C.

  1. Square side =x4=\dfrac{x}{4}, circle radius =36−x2π=\dfrac{36-x}{2\pi}; minimise total area via dAdx=0\dfrac{dA}{dx}=0, d2Adx2>0\dfrac{d^2A}{dx^2}>0.
  2. Substitution t=x2t=x^2 with partial fractions.

Part (a): Wire of length 36 m

  1. Let the piece bent into a square have length xx m; the other piece has length (36−x)(36-x) m for the circle.
  2. Square: side =x4=\dfrac{x}{4}, area =(x4)2=x216.=\left(\dfrac{x}{4}\right)^2=\dfrac{x^2}{16}. Circle: circumference =36−x⇒r=36−x2π=36-x\Rightarrow r=\dfrac{36-x}{2\pi}, area =πr2=(36−x)24π.=\pi r^2=\dfrac{(36-x)^2}{4\pi}.
  3. Combined area: A(x)=x216+(36−x)24π.A(x)=\dfrac{x^2}{16}+\dfrac{(36-x)^2}{4\pi}.
  4. dAdx=x8−(36−x)2π.\dfrac{dA}{dx}=\dfrac{x}{8}-\dfrac{(36-x)}{2\pi}. Set =0=0: x8=36−x2π⇒πx=4(36−x)⇒x(π+4)=144.\dfrac{x}{8}=\dfrac{36-x}{2\pi}\Rightarrow \pi x=4(36-x)\Rightarrow x(\pi+4)=144.
  5. So x=144π+4x=\dfrac{144}{\pi+4} m. The other piece =36−144π+4=36(π+4)−144π+4=36ππ+4=36-\dfrac{144}{\pi+4}=\dfrac{36(\pi+4)-144}{\pi+4}=\dfrac{36\pi}{\pi+4} m.
  6. d2Adx2=18+12π>0\dfrac{d^2A}{dx^2}=\dfrac{1}{8}+\dfrac{1}{2\pi}>0, so this gives a minimum.

Part (b): Evaluate ∫x3x4+3x2+2 dx\displaystyle\int\frac{x^3}{x^4+3x^2+2}\,dx …

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