A) 22
B) 11
C) 0
D) 20
E) none of these
Correct Answer: B
Solution :
We have \[h(x)={{\{f(x)\}}^{2}}+{{\{g(x)\}}^{2}}\] \[\Rightarrow \]\[h(x)=2f(x)f(x)+2g(x)g(x)\] Now,\[f(x)=g(x)\]and\[f(x)=-f(x)\] \[\Rightarrow \]\[f(x)=g(x)\]and\[f(x)=-f(x)\] \[\Rightarrow \]\[-f(x)=g(x)\] Thus,\[f(x)=g(x)\]and\[g(x)=-f(x)\] \[\therefore \] \[h(x)=-2g(x)g(x)+2g(x)g(x)\] \[=0,\forall x\] \[\Rightarrow \]\[h(x)=\]constant for all\[x\] But \[h(5)=11\] Hence\[h(x)=11\]for all\[x\].You need to login to perform this action.
You will be redirected in
3 sec