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If f(x)=(ax+lnx+1)(x+lnx+1) where a belongs to real number has at least 3 intersection point with g(x)=x², find a

If f(x)=(ax+lnx+1)(x+lnx+1) where a belongs to real number has at least 3 intersection point with g(x)=x²,  find a

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Grade:12th pass

1 Answers

Abhishek Singh
105 Points
2 years ago
(ax+lnx+1)(x+lnx+1)-x^2 = 0 \\ \Rightarrow (lnx+1)^2+ (a+1)x(lnx+1)+ax^2-x^2=0 \\ \Rightarrow (lnx+1)^2+ (a+1)x(lnx+1)+(a-1)x^2=0 \\
 
 
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) = \frac{lnx+1}{x} \\ \Rightarrow f`(x) = \frac{x(\frac{1}{x})-(lnx+1).1}{x^2} = -\frac{lnx}{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  
 
k = \frac{-(a+1) \pm \sqrt{(a+1)^2-4(a-1)}}{2}
taking the +ve sign we will get the +ve root ==> Now +ve root must be
\Rightarrow \frac{-(a+1) + \sqrt{(a+1)^2-4(a-1)}}{2} < 1 \\
\Rightarrow \sqrt{(a+1)^2-4(a-1)} < a+3 \\ \Rightarrow (a+1)^2-4(a-1) < (a+3)^2\\ \Rightarrow (a+3)^2-(a+1)^2 +4(a-1) >0 \\ \Rightarrow 2(2a+4) +4(a-1) >0 \\ \Rightarrow 8a+4 > 0 \\ \Rightarrow a >-1/2
 
Hence the two conditions we get is a>-1/2 and a a lies in (-1/2,1) ==> OPTION A
 
 

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