Equations And Inequalities Question 20
Question: Number of integral values of x satisfying the inequality $ {{( \frac{3}{4} )}^{6x+10-x^{2}}}<\frac{27}{64} $ is
Options:
A) 5
B) 6
C) 7
D) 8
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ {{( \frac{3}{4} )}^{6x+10-x^{2}}}<\frac{27}{64} $
$ \Rightarrow {{( \frac{3}{4} )}^{6x+10-x^{2}}}<{{( \frac{3}{4} )}^{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.
BETA
