मान लीजिए कि psi_1:[0, infty) rightarrow R , | vedclass

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Question

$f^{\prime}(x)=\left(|x|-x^2\right) e^{-x^2}+\left(|x|-x^2\right) e^{-x^2}, x \geq 0$

$f^{\prime}=2\left(x-x^2\right) e^{-x^2}$

hence option $(D

Vedclass · Accepted Answer

$C,D$

Vedclass · Answer

$f^{\prime}(x)=\left(|x|-x^2 ight) e^{-x^2}+\left(|x|-x^2 ight) e^{-x^2}, x \geq 0$ $f^{\prime}=2\left(x-x^2 ight) e^{-x^2}$ hence option $(D)$ is wrong $g^{\prime}(x)=x^{-x^2} 2 x$ $f^{\prime}(x)+g^{\prime}(x)=2 imes e^{-x^2}$ $f(x)+g(x)=-e^{-x^2}+c$ $f(x)+g(x)=-e^{-x^2}+1$ $F (\ell n 3)+ g (\sqrt{\ell n 3})=1-\frac{1}{3}=\frac{2}{3} ext { (option (A) is wrong) }$ $H ( x )=\psi_1( x )-1-\alpha x = e ^{- x }+ x -1-\alpha x , \quad x \geq 1 \& \alpha \in(1, x )$ $H(1)= e ^{-1}+1-1-\alpha<0$ $H^{\prime}(x)=-e^{-x}+1-\alpha>0 \quad \Rightarrow H(x) ext { is } \downarrow\Rightarrow ext { option (B) is wrong }$ $ ext { (C) } \psi_2(x)=2\left(\psi_1(\beta)-1 ight)$ $ ext { Applying L.M.V.T to } \Psi_2( x ) ext { in }[0, x ]$ $\psi_2^{\prime}(\beta)=\frac{\psi_2(x)-\psi_2(0)}{x}$ $2 \beta-2+2 e ^{-\beta}=\frac{\Psi_2( x )-0}{ x }$ $\Rightarrow \psi_2(x)=2 x\left(\psi_1(\beta-1) ight) ext { has one solution }$ $ ext { option }( C ) ext { is correct. }$ $ ext { (A) } \psi_1( x )= e ^{- x }+ x , \quad x \geq, 0$ $\psi_1^{\prime}( x )=1- e ^{- x }>0 \Rightarrow \psi_1( x ) ext { is } \uparrow$ $\psi_1( x ) \geq \psi_1(0) \quad \forall x \geq 0 \Rightarrow \psi_1( x ) \geq 1$ $ ext { (B) } \psi_2( x )= x ^2-2 x +2-2 e ^{- x } \quad x \geq 0$ $\psi_2^{\prime}( x )=2 x -2+2 e ^{- x }=2 \psi_1( x )-2 \geq 0 \quad \forall x \geq 0$ $\Rightarrow \psi_2( x ) ext { is } \uparrow \Rightarrow \psi_2( x ) \geq \psi_2(0)$ $\Rightarrow \psi_2( x ) \geq 0$ $ ext { (C) } f ( x )=2 \int_0^x\left( t - t ^2 ight) e ^{- t ^2} dt \quad \& \quad x \in\left(0, \frac{1}{2} ight)$ $=\int_0^{ x } 2 te ^{- t ^2} dt -\int_0^x 2 t ^2 e ^{- t ^2} dl$ $=-\left. e ^{- x ^2} ight|_0 ^{ x }-$ Let $H ( x )= f ( x )-1+ e ^{- x ^2}+\frac{2}{3} x ^3-\frac{2}{5} x ^5, \quad x \in\left(0, \frac{1}{2} ight)$ $H(0)=0$ $H^{\prime}(x)=2\left(x-x^2 ight) e^{-x^2}-2 x e^{-x^2}+2 x^2-2 x^4$ $=-2 x^2 e^{-x^2}+2 x^2-2 x^4$ $=2 x^2\left(1-x^2-e^{-x^2} ight)$ $\because e ^{-x} \geq 1-x \quad \forall x \geq 0$ $\Rightarrow H^{\prime}(x) \leq 0$ $\Rightarrow H(x) ext { is } \downarrow \Rightarrow 1-1(x)<0 \forall x \in\left(0, \frac{1}{2} ight)$ $ ext { Let } P(x)=g(x)-\frac{2}{3} x^3+\frac{2}{5} x^5-\frac{1}{7} x^7 x \in\left(0, \frac{1}{2} ight)$ $P^{\prime}(x)=2 x^2 e^{-x^2}-2 x^2+2 x^4-x^6$ $\quad=2 x^2\left(1-\frac{x^2}{1}+\frac{x^4}{2}-\frac{x^6}{3}+\ldots ight)-2 x^2+2 x^4-x^6$ $\quad=-\frac{x^8}{3}+\frac{x^{10}}{12} \ldots \ldots \ldots . .$ $\Rightarrow P^{\prime}(x) \leq 0$ $\Rightarrow P(x) ext { is } \downarrow$ $\Rightarrow P(x) \leq 0$ option $(D)$ is correct

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