Applications Of Derivatives Question 427
Question: The value of a so that the sum of the squares of the roots of the equation $ x^{2}-(a-2)x-a+1=0 $ assume the least value, is
[RPET 2000; AIEEE 2005]
Options:
A) 2
B) 1
C) 3
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ \alpha ,\ \beta $ be the roots of the equation $ x^{2}-(a-2)x-a+1=0, $ then $ \alpha +\beta =a-2,\alpha \beta =-a+1 $
$ \therefore z={{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta $
$ ={{(a-2)}^{2}}+2(a-1)=a^{2}-2a+2 $
$ \frac{dz}{da}=2a-2=0\Rightarrow a=1 $
$ \frac{d^{2}z}{da^{2}}=2>0, $ so z has minima at $ a=1 $
So $ {{\alpha }^{2}}+{{\beta }^{2}} $ has least value for $ a=1 $ .
This is because we have only one stationary value at which we have minima.
Hence $ a=1 $ .
BETA
