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SSC Sample Paper Mock Test-20 SSC CGL Tear-II Paper-1

  • question_answer
    The remainder, when \[f(x)\]is divide by \[(x-5)\] is where \[f(x)={{x}^{5}}-4{{x}^{4}}+7{{x}^{3}}-11x-13\]

    A) 289      

    B)  432    

    C)  1432   

    D)  1289

    Correct Answer: C

    Solution :

    Here, \[f\,\,(x)={{x}^{5}}-4{{x}^{4}}+7{{x}^{3}}-11x-13\] Rule 1 Now, \[{{a}_{0}}=1,\]\[{{a}_{1}}=-4,\]\[{{a}_{2}}=7,\]\[{{a}_{3}}=0,\]\[{{a}_{4}}=-11,\]\[{{a}_{5}}=-13\] Rule 2 5 Rule 3
    1 \[-\,\,4\] 7 0 \[-\,\,11\] \[-\,\,13\]
    5 5 60 300 1445
                \[{{b}_{0}}=1,\]\[{{b}_{1}}=1,\]\[{{b}_{2}}=12,\]\[{{b}_{3}}=60,\]\[{{b}_{4}}=289,\]\[R=1432\] The quotient \[Q={{b}_{0}}{{x}^{4}}+{{b}_{1}}{{x}^{3}}+{{b}_{2}}{{x}^{2}}+{{b}_{3}}x+{{b}_{4}}\] So,       \[Q={{x}^{4}}+{{x}^{3}}+12{{x}^{2}}+60x+289\]   and remainder R = 1432


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