| Assertion [A]: The function \[f\left( x \right)~={{x}^{2}}+2x+1\] is strictly increasing on (1, 2). |
| Reason [R]: The least value of x is 1 in interval (1, 2) |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: C
Solution :
\[f\left( x \right)={{x}^{2}}+2x+1\] \[f'(x)=2x+2=2(x+1)\] Now \[x\in \left( 1,\text{ }2 \right)\Rightarrow 1<x<2\Rightarrow 2<2x<4\] \[\Rightarrow 2+2<2x+2<4+2\Rightarrow 4<f'\left( x \right)<6\] \[\Rightarrow f'\left( x \right)>0\] \[\Rightarrow \,f\left( x \right)\]is strictly increasing in interval (1, 2) \[\Rightarrow \]Assertion [A] is true In interval \[\left( 1,2 \right),\text{ }1<x<2\,\Rightarrow \text{ }x>1\] \[\therefore \] Reason R is false Hence option [C] is the correct answer.
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