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a possible positive value of ' $a$ ', for which $f^{\prime}$ ' $(x)=0$ has equal roots, is
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The correct answer is:
$1$
Given,
$f(x)=\int_0^x\left[(a+1)(t+1)^2-(a-1)\left(t^2+t+1\right)\right] d t$
Now, $f^{\prime}(x)=(a+1)(x+1)^2-(a-1)\left(x^2+x+1\right)=0$
$\Rightarrow 2 x^2+(3+a) x+2 a=0$
for equal roots,
$(3+a)^2-4 \times 4 a=0 \Rightarrow a=1$
$f(x)=\int_0^x\left[(a+1)(t+1)^2-(a-1)\left(t^2+t+1\right)\right] d t$
Now, $f^{\prime}(x)=(a+1)(x+1)^2-(a-1)\left(x^2+x+1\right)=0$
$\Rightarrow 2 x^2+(3+a) x+2 a=0$
for equal roots,
$(3+a)^2-4 \times 4 a=0 \Rightarrow a=1$
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