A) \[9\]
B) \[10\]
C) \[11\]
D) \[12\]
Correct Answer: C
Solution :
[c] We have curve \[x={{t}^{2}}-3t+1,\] \[y=2{{t}^{2}}+3t-4\] For point \[M(-1,\,\,10)\] \[{{t}^{2}}-3t+1=-1\] \[\Rightarrow \,\,\,{{t}^{2}}-3t+2=0\] \[\Rightarrow \,\,\,(t-1)\,\,(t-2)=0\] \[\therefore \,\,\,\,t=1,\,2\] And from \[2{{t}^{2}}+3t-4=10\] \[\Rightarrow \,\,2{{t}^{2}}+3t-14=0\] \[\Rightarrow \,\,(2t+7)\,(t-2)=0\] \[\therefore \,\,\,\,t=\frac{-7}{2},2\] Hence, to the given point \[M\,(-1,\,10),\] there corresponds the value\[t=2\]. \[\therefore \] The slope of tangent at \[M(-1,10)={{\left. \frac{dy}{dx} \right|}_{t=2}}\] \[={{\left. \frac{dy/dt}{dx/dt} \right|}_{t=2}}\] \[={{\left. \frac{4t+3}{2t-3} \right|}_{t=2}}=11\]
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