A) 5
B) 6
C) 7
D) 8
Correct Answer: C
Solution :
[c] \[{{\left( \frac{3}{4} \right)}^{6x+10-{{x}^{2}}}}<\frac{27}{64}\] \[\Rightarrow {{\left( \frac{3}{4} \right)}^{6x+10-{{x}^{2}}}}<{{\left( \frac{3}{4} \right)}^{3}}\] \[\Rightarrow 6x+10-{{x}^{2}}>3\] (as base (3/4)<1) \[\therefore \,\,\,{{x}^{2}}-6x-7<0\,\,\,\,\,\,\,\therefore \,\,\,(x+1)(x-7)<0\] Thus, integral values of x are 0, 1, 2, 3, 4, 5 and 6.
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