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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main · Mathematics · Indefinite Integration

aligned & If (4 x^2+5 x^4+7)/((x^2+1)(x^4+x^2+1))=(A x+B)/(x^2+1) &...

 If 4x2+5x4+7(x2+1)(x4+x2+1)=Ax+Bx2+1+Cx3+Dx2+Ex+Fx4+x2+1, then B+2(D+F+E)−C⋅A=\begin{aligned} & \text { If } \frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1} \\ & +\frac{C x^3+D x^2+E x+F}{x^4+x^2+1}, \text { then } \\ & B+2(D+F+E)-C \cdot A=\end{aligned}

Options

  1. A0
  2. B3
  3. C1
  4. D-3

Correct answer

A. 0

Step-by-step solution

We have, 4x2+5x4+7(x2+1)(x4+x2+1)=Ax+Bx2+1+Cx3+Dx2+Ex+Fx4+x2+1⇒4x2+5x4+7=(Ax+B)(x4+x2+1)+(Cx3+Dx2+Ex+F)(x2+1)⇒5x4+4x2+7=Ax5+Ax3+Ax+Bx4+Bx2+B+Cx5+Cx3+Dx4+Dx2+Ex3+Ex+Fx2+F⇒5x4+4x2+7=(A+C)x5+(B+D)x4+(A+C+E)x3+(B+D+F)x2+(A+E)x+(B+F)\begin{aligned} & \frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1}+\frac{C x^3+D x^2+E x+F}{x^4+x^2+1} \\ & \Rightarrow 4 x^2+5 x^4+7=(A x+B)\left(x^4+x^2+1\right) \\ & +\left(C x^3+D x^2+E x+F\right)\left(x^2+1\right) \\ & \Rightarrow 5 x^4+4 x^2+7=A x^5+A x^3+A x+B x^4+B x^2 \\ & +B+C x^5+C x^3+D x^4+D x^2+E x^3+E x+F x^2+F \\ & \Rightarrow 5 x^4+4 x^2+7=(A+C) x^5+(B+D) x^4 \\ & +(A+C+E) x^3+(B+D+F) x^2+(A+E) x+(B+F) \\ \end{aligned} On comparing, we get A+C=0B+D=5\begin{aligned} & A+C=0 \\ & B+D=5 \end{aligned}A+C+E=0B+D+F=4A+E=0B+F=7\begin{array}{r} A+C+E=0 \\ B+D+F=4 \\ A+E=0 \\ B+F=7 \end{array}On solving, we get \begin{gathered}A=C=E=0, B=8 . D=-3, F=-1 \\ \therefore B+2(D+F+E)-C \cdotA=8+2(-3-1+0)-0 \times0 \\ =8-8=0 \end{gathered}
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