Abhishek Singh
Last Activity: 3 Years ago
The above equation is quadratic in (lnx +1)/x and hence will give solution of the form (lnx+1)/x = k1,k2.
Now we can check the number of solution the equation (lnx+1)/x = K has for general K. Consider f(x) = (ln(x)+1)/x
→ f(x) is increases in (0,1) and decreases on (1,infinity). Max value = f(1) = 1
→ Use this to sketch the graph of f(x) (easy to do if you know graph sketching )
You can clearly see if k1 Solution & 0
==> Acc to question The equation has 3 roots ==> the roots k1,k2 are such that 1 is negative and other is positive and less than1
==> Putting these condition on solution of quadratic will give the range of a
Since roots exist D>0 ==> D = (a+1)^2 – 4(a-1) = (a-1)^2 +4 which is always >0
Since product of roots (k1.k2) a-1 a
Now finding the roots
taking the +ve sign we will get the +ve root ==> Now +ve root must be
Hence the two conditions we get is a>-1/2 and a a lies in (-1/2,1) ==> OPTION A
