If a function fx defined byfx - aex - be-x- -1 - x - 1 cx2- 1 - x - 3 ax2 - 2cx 3 - x - 4 -be continuous for some a- b- c- and f-0 - f-2 - e- then the value of a is -

F(x ) is continuous at nbsp;x =1 ⇒ c=ae+be^ 1 at x= 3 ⇒ 9c = 9a +6c ⇒ c = 3a Now f'(0 ) + f'(2 ) =e ⇒ a b + 4c = e ⇒ a e (3a ae ) + 4·3a = e ⇒ a 3a ...

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Byju's Answer

Standard XII

Mathematics

One-One into Function

If a function...

Question

If a function $f (x)$ defined by
$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} a e^{x} + b e^{- x}, & - 1 \leq x < 1 c x^{2}, & 1 \leq x \leq 3 a x^{2} + 2 c x & 3 < x \leq 4 \end{matrix}$
be continuous for some $a, b, c \in R$ and $f^{'} (0) + f^{'} (2) = e,$ then the value of $a$ is :

\frac{1}{e^{2} - 3 e + 13}

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\frac{e}{e^{2} - 3 e - 13}

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\frac{e}{e^{2} + 3 e + 13}

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\frac{e}{e^{2} - 3 e + 13}

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Solution

The correct option is D $\frac{e}{e^{2} - 3 e + 13}$
$f (x)$ is continuous
at $x = 1 \Rightarrow c = a e + b e^{- 1}$
at $x = 3 \Rightarrow 9 c = 9 a + 6 c \Rightarrow c = 3 a$

Now $f^{'} (0) + f^{'} (2) = e$
$\Rightarrow a - b + 4 c = e$
$\Rightarrow a - e (3 a - a e) + 4 \cdot 3 a = e$
$\Rightarrow a - 3 a e + a e^{2} + 12 a = e$
$\Rightarrow 13 a - 3 a e + a e^{2} = e$
$\Rightarrow a = \frac{e}{13 - 3 e + e^{2}}$

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If a function f(x) defined by f(x)=⎧⎪⎨⎪⎩aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx3<x≤4 be continuous for some a,b,c∈R and f′(0)+f′(2)=e, then the value of a is :

The correct option is D ee2−3e+13f(x) is continuous at x=1⇒c=ae+be−1 at x=3⇒9c=9a+6c⇒c=3a Now f′(0)+f′(2)=e ⇒a−b+4c=e ⇒a−e(3a−ae)+4⋅3a=e ⇒a−3ae+ae2+12a=e ⇒13a−3ae+ae2=e ⇒a=e13−3e+e2

If a function $f (x)$ defined by
$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} a e^{x} + b e^{- x}, & - 1 \leq x < 1 c x^{2}, & 1 \leq x \leq 3 a x^{2} + 2 c x & 3 < x \leq 4 \end{matrix}$
be continuous for some $a, b, c \in R$ and $f^{'} (0) + f^{'} (2) = e,$ then the value of $a$ is :