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Let $f (x) = \{\begin{array}{cc} a x, & x < 2 \\ a x^{2} + b x + 3, & x \geq 2 \end{array}$ If $f$ is differentiable for all $x$ then the value of $(a, b)$ is equal to

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(1, 2)

(\frac{3}{2}, \frac{9}{2})

(\frac{3}{4}, - \frac{9}{2})

(\frac{3}{4}, - \frac{9}{4})

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detailed solution

Correct option is D

$lim_{x \to 2 -} f (x) = lim_{x \to 2 -} a x = 2 a$ and $lim_{x \to 2 +} f (x) = lim_{x \to 2 +}$

$(a x^{2} + b x + 3) = 4 a + 2 b + 3$

Since $f$ is differentiable so continuous, hence

$2 a + 2 b + 3 = 0$

Also $lim_{h \to 0 +} \frac{f (2 + h) - f (2)}{h}$

$= lim_{h \to 0 +} \frac{a (2 + h)^{2} + b (2 + h) + 3 - (4 a + 2 b + 3)}{h}$

$= lim_{h \to 0 +} \frac{a h^{2} + 4 a h + b h}{h}$

$= 4 a + b$

$lim_{h \to 0 -} \frac{f (2 + h) - f (2)}{h}$

$= lim_{h \to 0 -} \frac{a (2 + h) - (4 a + 2 b + 3)}{h}$

$= a (using (i))$

So $a = 4 a + b \Rightarrow - 3 a = b$ . Putting this in (i),

we get $a = \frac{3}{4}, b = - \frac{9}{4}$

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Let f(x)=ax,x<2ax2+bx+3,x≥2 If f is differentiable for all x then the value of (a, b) is equal to