Manipal Entrance Test 2015 Math Paper

We have,
9x2−18|x|+5=0
⇒9|x|2−18|x|+5=0
⇒(3|x|−1)(3|x|−5)=0
⇒|x|=

1

3

or |x|=

5

3

⇒x=±

1

3

or x=±

5

3

Now, loge{(x+1)(x+2)} is defined, if
(x+1)(x+2)>0
⇒x<−2 or x>−1
Clearly, x=−

5

3

does not satisfy this condition. But all other values of x satisfy the above conditions.
Hence, the number of solutions lying in the domain of definition of the given function is 3.