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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The slope of the tangent to the curve\[x=3{{t}^{2}}+1,\,y={{t}^{3}}-1\] at \[x=1\] is :

    A)  \[\frac{1}{2}\]                                  

    B)  0

    C)  -2                                          

    D)  \[\infty \]

    Correct Answer: B

    Solution :

    Here we have\[x=3{{t}^{2}}+1\]                               ... (i) \[\therefore \]  \[\frac{dx}{dt}=6t\]                 \[y={{t}^{3}}-1\]                                               ?. (ii)                 \[\frac{dy}{dt}=3{{t}^{2}}\] \[\therefore \]  \[\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=3{{t}^{2}}\times \frac{1}{64}\]                 \[=\frac{t}{2}\] But from(i) when \[x=1\] we have                 \[1=3{{t}^{2}}+1\]                 \[3{{t}^{2}}=0\] \[\Rightarrow \]               \[t=0\] \[\therefore \] When \[x=1\] then \[t=0\] \[\therefore \]  \[\frac{dy}{dx}=0\]


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