J & K CET Engineering J and K - CET Engineering Solved Paper-2004

  • question_answer
    Let \[y=p+q\]where p varies directly as x and q varies inversely as\[{{x}^{2}}\]. If \[y=19\]when \[x=2\] or 3, then y in terms of x is

    A)       \[36x+\frac{5}{{{x}^{2}}}\]     

    B)  \[\frac{5}{x}\,36{{x}^{2}}\]

    C)  \[5x+\frac{36}{{{x}^{2}}}\]      

    D)  None of these

    Correct Answer: C

    Solution :

    Given that,  \[y=p+q\] and \[p\propto x\] and \[q\propto \frac{1}{{{x}^{2}}}\] Now, taking option [c] Let \[y=5x+\frac{36}{{{x}^{2}}}\] which is satisfied when \[x=2\] or 3 Hence, option [c] is correct.


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