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The sum of the squares of the roots of $x^2-(a-2)x-a-1=0$ becomes minimum when $a$ is

  1. $0$
  2. $1$
  3. $2$
  4. $5$
in Numerical Ability by Boss (17.5k points)
recategorized by | 25 views

1 Answer

+1 vote

Let $\alpha,  \beta$ are the roots of the given equation.

Then $\alpha + \beta = a-2$, $\alpha. \beta= 1-a$

$(\alpha + \beta)^2= \alpha^2 + \beta^2 + 2 \alpha \beta$

$(a-2)^2= \alpha^2 + \beta^2 + 2 - 2a$

$\alpha^2 + \beta^2= a^2 -2a +2= x$(let)

We have to minimize $x$, differentiating $x$ wrt to $a$

$\frac{dx}{da} = 2a-2 = 0, a=1$

$\frac{d^2 x} {d a^2} = 2 > 0$(minima)

So for $a=1$, $x$ will be minimum.

Hence option B) is correct

by Boss (16.3k points)
edited by
0
Here in the second last line  for getting local minima $\dfrac{\mathrm{d}^{2}x }{\mathrm{d} a^{2}}=2>0$

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