If f(x)=int_0^x e^{t^2}(t-2)(t-3) dt for all | vedclass

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(C) Given $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$.By the Fundamental Theorem of Calculus,$f^{\prime}(x)=e^{x^2}(x-2)(x-3)$.To find the critical points,se

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$(A, B, C, D)$

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(C) Given $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$.By the Fundamental Theorem of Calculus,$f^{\prime}(x)=e^{x^2}(x-2)(x-3)$.To find the critical points,set $f^{\prime}(x)=0$,which gives $x=2$ and $x=3$.Analyzing the sign of $f^{\prime}(x)$:For $x  0$ (increasing).For $2 For $x > 3$,$f^{\prime}(x) > 0$ (increasing).Thus,$f$ has a local maximum at $x=2$ and a local minimum at $x=3$. This confirms $(A)$,$(B)$,and $(D)$ are correct.Now,find $f^{\prime \prime}(x)$:$f^{\prime \prime}(x) = \frac{d}{dx} [e^{x^2}(x^2-5x+6)] = e^{x^2}(2x)(x^2-5x+6) + e^{x^2}(2x-5) = e^{x^2}(2x^3-10x^2+12x+2x-5) = e^{x^2}(2x^3-10x^2+14x-5)$.Let $g(x) = 2x^3-10x^2+14x-5$. Since $g(0) = -5$ and $g(1) = 2-10+14-5 = 1$,by the Intermediate Value Theorem,there exists $c \in (0, 1)$ such that $g(c) = 0$,implying $f^{\prime \prime}(c) = 0$. Thus,$(C)$ is also correct.Therefore,all options $(A), (B), (C), (D)$ are correct.

List-$I$	List-$II$
$(P)$ The minimum value of $n$ for which the function $f(x)=\left[\frac{10 x^3-45 x^2+60 x+35}{n}\right]$ is continuous on the interval $[1,2]$,is	$(1)$ $8$
$(Q)$ The minimum value of $n$ for which $g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right), x \in R$,is an increasing function on $R$,is	$(2)$ $9$
$(R)$ The smallest natural number $n$ which is greater than $5$,such that $x=3$ is a point of local minima of $h(x)=\left(x^2-9\right)^{n}\left(x^2+2 x+3\right)$,is	$(3)$ $5$
$(S)$ Number of $x_0 \in R$ such that $l(x)=\sum_{k=0}^4\left(\sin \|x-k\|+\cos \left\|x-k+\frac{1}{2}\right\|\right), x \in R$ is not differentiable at $x_0$,is	$(4)$ $6$
	$(5)$ $10$

If $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$ for all $x \in(0, \infty)$,then
$(A)$ $f$ has a local maximum at $x=2$
$(B)$ $f$ is decreasing on $(2,3)$
$(C)$ there exists some $c \in(0, \infty)$ such that $f^{\prime \prime}(c)=0$
$(D)$ $f$ has a local minimum at $x=3$

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If $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$ for all $x \in(0, \infty)$,then$(A)$ $f$ has a local maximum at $x=2$$(B)$ $f$ is decreasing on $(2,3)$$(C)$ there exists some $c \in(0, \infty)$ such that $f^{\prime \prime}(c)=0$$(D)$ $f$ has a local minimum at $x=3$