⨍ x=ex2+x,x>0ax+b,x≤0 is differentiable at x=0 then

fx=ex2+x ; x>0ax+b ; x≤0 is differentiable at x=0 ⇒f is continuous at x=0⇒limx→0-fx=limx→0+fx ∴limx→0+ex2+x=limx→0-ax+b ⇒1=bf1x=ex22x+1 ;x>0a ;x<0 f10+=f10-⇒e00+1=a⇒a=1 ⇒a=1, b=1

⨍ x=ex2+x,x>0ax+b,x≤0 is differentiable at x=0 then

fx=ex2+x ; x>0ax+b ; x≤0 is differentiable at x=0 ⇒f is continuous at x=0⇒limx→0-fx=limx→0+fx ∴limx→0+ex2+x=limx→0-ax+b ⇒1=bf1x=ex22x+1 ;x>0a ;x<0 f10+=f10-⇒e00+1=a⇒a=1 ⇒a=1, b=1

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$⨍ (x) = \{\begin{matrix} e^{x^{2}} + x, & x > 0 \\ ax + b, & x \leq 0 \end{matrix}$ is differentiable at $x = 0$ then

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$a = 1, b = - 1$

$a = - 1, b = - 1$

$a = 1, b = 1$

$a = - 1, b = 1$

answer is C.

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Detailed Solution

$f (x) = \{\begin{cases} e^{x^{2}} + x; x > 0 \\ ax + b; x \leq 0 is differentiable at x = 0 \end{cases} \Rightarrow f is continuous at x = 0$ $\Rightarrow \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) ∴ \lim_{x \to 0^{+}} e^{x^{2}} + x = \lim_{x \to 0^{-}} ax + b \Rightarrow 1 = b$

$f^{1} (x) = \{\begin{cases} e^{x^{2}} (2 x) + 1; x > 0 \\ a; x < 0 \end{cases} f^{1} (0^{+}) = f^{1} (0^{-}) \Rightarrow e^{0} (0) + 1 = a \Rightarrow a = 1 \Rightarrow a = 1, b = 1$