| Assertion [A]: Let |
| \[f:\text{ }R\to R,\text{ }f\left( x \right)={{x}^{3}}+{{x}^{2}}\text{+3x+ }sin\text{ }x,\text{ f(x)}\]is one-one. |
| Reason [R]: f(x) is decreasing function. |
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 :
Given \[f\left( x \right)={{x}^{3}}+{{x}^{2}}+3x+sinx\] \[\therefore \,\,\,\,\,f'\left( x \right)=3{{x}^{2}}+2x+3+cos\,x\] since \[-1\le \cos x\le 1\] Taking minimum value -1 \[\therefore \,\,f'\left( x \right)=3{{x}^{2}}+2x+3-1=3{{x}^{2}}+2x+2\] Here Discriment, \[D={{b}^{2}}-4ac={{\left( 2 \right)}^{2}}-4\times 3\times 2\] \[=4-24=-20<0~\] \[\Rightarrow f'\left( x \right)>0\] \[\therefore \,\,\,f\left( x \right)\]is one-one in its domain \[\therefore \] Assertion [A] is true Also \[f'\left( x \right)>0\Rightarrow f\left( x \right)\] is increasing function \[\Rightarrow \]Reason (R) is false \[\therefore \] Hence option [C] is the correct answer.
You need to login to perform this action.
You will be redirected in
3 sec
