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If the function $f:R\rightarrow R,$ defined by is differentiable, then the value of

$f'(-3)+f'(3)$ is equal to :

Option 1)
0

Option 2)
3

Option 3)
4

Option 4)
$\frac{15}{2}$

Answers (1)

best_answer

As we learnt in

Condition for differentiable -

A function f(x) is said to be differentiable at $x=x_{\circ }$ if $Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })$ both exist and are equal otherwise non differentiable

-

$f(x)=\left\{\begin{matrix} ax &x< 2 \\ ax^{2}-bx+3 & x\geqslant 2 \end{matrix}$

$f'(x)=\left\{\begin{matrix} a\ \, \, \, \, \,\;\: \: \: \: \: \: \: \: x< 2 \\ 2ax-b\, \, x\geqslant 2 \end{matrix}$

$\therefore a=4a-b$

$b=3a$ ----------------(i)

Now f(x) is continuous at x=2

$\therefore 2a=4a-2b+3$

$2b=2a+3$

$6a=2a+3$

$4a=3$

$\therefore a=\frac{3}{4}$

$b=\frac{9}{4}$

$\therefore f{}'(x)=\left\{\begin{matrix} \frac{3}{4} & x< 2\\ \frac{3x}{2}-\frac{9}{4} & x\geqslant 2 \end{matrix}\right.$

$f{}'(-3)+f{}'(3)=>\frac{3}{4}\times \frac{3}{2}\times 3-\frac{9}{4}$

$=\frac{3}{4}\times \frac{9}{2}-\frac{9}{4}$

$=>\frac{3+18-9}{4}=\frac{12}{4}=3$

Option 1)

0

This option is incorrect

Option 2)

3

This option is correct

Option 3)

4

This option is incorrect

Option 4)

$\frac{15}{2}$

This option is incorrect

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