A possible positive value of 'a',for which f^ | vedclass

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(A) Given,$f(x) = \int_0^x [(a+1)(t+1)^2 - (a-1)(t^2+t+1)] dt$. By the Fundamental Theorem of Calculus,$f^{\prime}(x) = (a+1)(x+1)^2 - (a-1)(x^2+x+1)$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given,$f(x) = \int_0^x [(a+1)(t+1)^2 - (a-1)(t^2+t+1)] dt$. By the Fundamental Theorem of Calculus,$f^{\prime}(x) = (a+1)(x+1)^2 - (a-1)(x^2+x+1)$. Expanding the terms: $f^{\prime}(x) = (a+1)(x^2+2x+1) - (a-1)(x^2+x+1)$. $f^{\prime}(x) = (a+1)x^2 + 2(a+1)x + (a+1) - (a-1)x^2 - (a-1)x - (a-1)$. $f^{\prime}(x) = (a+1-a+1)x^2 + (2a+2-a+1)x + (a+1-a+1)$. $f^{\prime}(x) = 2x^2 + (a+3)x + 2 = 0$. For the quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$. Here,$A = 2$,$B = (a+3)$,and $C = 2$. $D = (a+3)^2 - 4(2)(2) = 0$. $(a+3)^2 - 16 = 0$. $(a+3)^2 = 16$. $a+3 = \pm 4$. Case $1$: $a+3 = 4 \Rightarrow a = 1$. Case $2$: $a+3 = -4 \Rightarrow a = -7$. Since the question asks for a positive value of '$a$',the answer is $1$.

$A$ possible positive value of '$a$',for which $f^{\prime}(x)=0$ has equal roots,is

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