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  • The point <img alt="\mathrm{\left(2 t^{2}+2 t+4, t^{2}+t+1\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%282%20t%5E%7B2%7D&plus;2%20t&plus;4%2C%20t%5E%7B

The point \mathrm{\left(2 t^{2}+2 t+4, t^{2}+t+1\right)} lies on the line \mathrm{x+2 y=1}, for

Option: 1

All real values of t


Option: 2

Some real values of t

 


Option: 3

\mathrm{t=\frac{-4 \pm \sqrt{7}}{8}}


Option: 4

for no value of t


Answers (1)

best_answer

The given point lies on the line if \left(2 t^{2}+2 t+4\right)+2\left(t^{2}+t+1\right)=1

\Rightarrow 4 \mathrm{t}^{2}+4 \mathrm{t}+5=0
Here discriminant =16-4.4 .5=-64<0

Hence for no real value of ’t’ it is possible.

 

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