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ICSE Class X Math 2016 Paper

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Question : 1 of 57
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SECTION - I
Using remainder theorem, find the value of kk if on dividing 2x3+3x2kx+52x^3+3x^2-kx+5 by x2x-2, leaves a remainder 7.
Solution:  
Let f(x)=2x3+3x2kx+5f(x)=2x^3+3x^2-kx+5
By remainder theorem, when f(x)f(x) is divided by (x2)(x-2),
x2=0,x=2x-2=0, \Rightarrow x=2
  the remainder    =f(2)\; \text{the remainder} \;\; = f(2)
  =2(2)3+3(2)2k(2)+5\; = 2(2)^3+3(2)^2-k(2)+5
  =16+122k+5\; = 16+12-2k+5
  =332k\; = 33-2k
According to question
332k  =733-2k \; = 7
2k  =262k \; = 26
k  =13k \; = 13
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