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Question Number 158988 by cortano last updated on 11/Nov/21

Answered by Rasheed.Sindhi last updated on 11/Nov/21

$$\mathcal{T}hegraphoff\left(x\right)intersectsy=2x-1$$$$atx=1,2,3$$$$\therefore \mathcal{T}hepointsofintersections:$$$$(1,2\left(1\right)-1),(2,2\left(2\right)-1)\mathrm{\&}(3,2\left(3\right)-1)$$$$Or(1,1),(2,3)\mathrm{\&}(3,5)$$$$alsothepointsonf\left(x\right):$$$$(1,1):$$$$f\left(1\right)={\left(1\right)}^4+a{\left(1\right)}^3+b{\left(1\right)}^2+c\left(1\right)+d=1$$$$\Rightarrow a+b+c+d=0............A$$$$(2,3):$$$$f\left(2\right)={\left(2\right)}^4+a{\left(2\right)}^3+b{\left(2\right)}^2+c\left(2\right)+d=3$$$$\Rightarrow 8a+4b+2c+d=-13.....B$$$$(3,5):$$$$f\left(3\right)={\left(3\right)}^4+a{\left(3\right)}^3+b{\left(3\right)}^2+c\left(3\right)+d=5$$$$\Rightarrow 27a+9b+3c+d=-76........C$$$$f\left(0\right)=d$$$$f\left(4\right)=4^4+a{\left(4\right)}^3+b{\left(4\right)}^2+c\left(4\right)+d=e\left(say\right)$$$$=256+64a+16b+4c+d$$$$f\left(0\right)+f\left(4\right)=256+64a+16b+4c+2d$$$$=256+64a+16b+4c+2d$$$$=256+64a+16b+2(-13-8a-4b)$$$$=230+48a+8b$$$$f\left(0\right)+f\left(4\right)=230+8(6a+b)$$$$B-A:7a+3b+c=-13...........D$$$$C-B:19a+5b+c=-63........E$$$$E-D:12a+2b=-50\Rightarrow 6a+b=-25$$$$f\left(0\right)+f\left(4\right)=230+8(6a+b)$$$$=230+8(-25)=30$$

Commented by cortano last updated on 13/Nov/21

$$thankyousir$$

Answered by mr W last updated on 11/Nov/21

$$sayg\left(x\right)=f\left(x\right)-(2x-1)$$$$g\left(1\right)=g\left(2\right)=g\left(3\right)=0$$$$\Rightarrow g\left(x\right)=(x-1)(x-2)(x-3)(x+p)$$$$\Rightarrow f\left(x\right)-(2x-1)=(x-1)(x-2)(x-3)(x+p)$$$$\Rightarrow f\left(x\right)=(x-1)(x-2)(x-3)(x+p)+2x-1$$$$f\left(0\right)=(0-1)(0-2)(0-3)(0+p)+2\times 0-1$$$$\Rightarrow f\left(0\right)=-6p-1$$$$f\left(4\right)=(4-1)(4-2)(4-3)(4+p)+2\times 4-1$$$$\Rightarrow f\left(4\right)=31+6p$$$$f\left(0\right)+f\left(4\right)=-6p-1+31+6p=30$$

Commented by Rasheed.Sindhi last updated on 11/Nov/21

$$ANovelMethod!$$$$Wonderful\mathcal{S}ir!$$

Commented by mr W last updated on 12/Nov/21

$$thankssir!weseeagain:$$$$manyroadsleadtoRome!$$

Commented by Rasheed.Sindhi last updated on 12/Nov/21

$$Ofcoursesir!$$$$I^{\prime }malwaysinterestedin{}^{\prime }many$$$${roads}^{\prime }thatisvarietyofsolutions.$$$$Butafterallsomeroadsarebetter$$$$thanothers!$$

Commented by cortano last updated on 13/Nov/21

$$nice$$

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