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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-5

  • question_answer
    If polynomials \[2{{x}^{3}}+a{{x}^{2}}+3x-5\] and\[{{x}^{3}}+{{x}^{2}}-2x+a\]are divided by\[(x-2)\], then same remainders are obtained. Find the value of\[a\].

    A) \[-3\]                              

    B) \[3\]

    C) \[-4\]                

    D)        \[-9\]

    Correct Answer: A

    Solution :

    \[f(x)=2{{x}^{3}}+a{{x}^{2}}+3x-5\]     \[g(x)={{x}^{3}}+{{x}^{2}}-2x+a\] By Remainder Theorem, \[f(2)=(2\times {{2}^{3}}+a\times {{2}^{2}}+3\times 2-5)=17+4a\] Again,\[g(2)=({{2}^{3}}+{{2}^{2}}-2\times 2+a)=8+a\] \[\therefore \]      \[17+4a=8+a\] \[\Rightarrow \]   \[3a=-9\] \[\Rightarrow \]   \[a=-3\]


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