If fx - a e-ax- x - 0 x3 - 4x - 2b- x - 0 -is differentiable at x - 0- then the value of -a- - -b2 - 1 is

At x = 0 L.H.D. = a^2 e^ ax |_x = 0 = a^2 R.H.D = nbsp; 3x^2 4 |_x = 0 = 4 L.H.D. = R.H.D. ⇒ a^2 = 4 ⇒ a^2 = 4 ⇒ |a| = 2 nbsp; Function is differen ...

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Standard XI

Mathematics

Differentiability in an Interval

If fx = a e-a...

Question

If $f (x) = {\begin{matrix} a e^{- a x}, & x \leq 0 x^{3} - 4 x + 2 b, & x > 0 \end{matrix}$

is differentiable at $x = 0$ , then the value of $| a | + \sqrt{b^{2} - 1}$ is

\sqrt{3}

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0

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None of the above

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Solution

The correct option is B $2$
At $x = 0$
L.H.D. $= - a^{2} e^{- a x} {∣ ∣}_{x = 0} = - a^{2}$
R.H.D $= 3 x^{2} - 4 {∣ ∣}_{x = 0} = - 4$
L.H.D. = R.H.D.
$\Rightarrow - a^{2} = - 4 \Rightarrow a^{2} = 4 \Rightarrow | a | = 2$

Function is differentiable at $x = 0$
$\Rightarrow$ Function is continuous at $x = 0$
L.H.L. $= lim x \to 0^{-} a e^{- a x}$
R.H.L. $= lim x \to 0^{+} x^{3} - 4 x + 2 b$
L.H.L. = R.H.L.
$\Rightarrow a = 2 b$
$\Rightarrow a^{2} = 4 b^{2}$
$\Rightarrow b^{2} = \frac{a^{2}}{4} = \frac{4}{4} = 1$
$∴ | a | + \sqrt{b^{2} - 1} = 2 + \sqrt{1 - 1} = 2$

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If f(x)={ae−ax,x≤0x3−4x+2b,x>0 is differentiable at x=0, then the value of |a|+√b2−1 is

If $f (x) = {\begin{matrix} a e^{- a x}, & x \leq 0 x^{3} - 4 x + 2 b, & x > 0 \end{matrix}$

is differentiable at $x = 0$ , then the value of $| a | + \sqrt{b^{2} - 1}$ is