Question

If the function g\left( x \right)=\left\\{ \begin{aligned} & k\sqrt{x+1}\text{ }0\le x\le 3 \\\ & mx+2\text{ }3< x\le 5 \\\ \end{aligned} \right. is differentiable, then the value of k+m is:
(a) 2
(b) $\dfrac{16}{5}$
(c) $\dfrac{10}{3}$
(d) 4

Answer 1

We know that a function is differentiable at a point if and only if it is continuous as well as differentiable at that point. For f(x) to be continuous at x=a, the value of the function at x=a must be equal to $\displaystyle \lim_{x \to {{a}^{-}}}f(x)\text{ and }\displaystyle \lim_{x \to {{a}^{+}}}f(x)$ and for being differentiable the left hand derivative and right hand derivative must be equal. So, form two equations in k and m and solve the equation to get the values of k and m.

Complete step-by-step answer:
We know that a function is differentiable at a point if and only if it is continuous as well as differentiable at that point. For f(x) to be continuous at x=a, the value of the function at x=a must be equal to $\displaystyle \lim_{x \to {{a}^{-}}}f(x)\text{ and }\displaystyle \lim_{x \to {{a}^{+}}}f(x)$ .
Therefore, checking the continuity of the function g\left( x \right)=\left\\{ \begin{aligned} & k\sqrt{x+1}\text{ }0\le x\le 3 \\\ & mx+2\text{ }3< x\le 5 \\\ \end{aligned} \right. . at x=3.
First, finding the right-hand limit of the given function.
$\displaystyle \lim_{x \to {{3}^{+}}}g\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}(mx+2)$
Now, if we put the limit, we get:
$\displaystyle \lim_{x \to {{3}^{+}}}g\left( x \right)=3m+2$
Next, finding the left hand limit.
$\displaystyle \lim_{x \to {{3}^{-}}}g\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\left( k\sqrt{x+1} \right)$
Now, if we put the limit, we get:
$\displaystyle \lim_{x \to {{3}^{-}}}g\left( x \right)=k\sqrt{3+1}=2k$
Now, we know that the function g(x) is differentiable in its domain, so it is differentiable at 3 as well. So, LHL=RHL.
$\displaystyle \lim_{x \to {{3}^{-}}}g(x)=\displaystyle \lim_{x \to {{3}^{+}}}g(x)$
$\Rightarrow 2k=3m+2...............(i)$
We also know that for being differentiable the left hand derivative and right hand derivative must be equal.
$\dfrac{d\left( mx+2 \right)}{dx}=\dfrac{d\left( k\sqrt{x+1} \right)}{dx}$ at x=3.
The derivative of x is 1 and that of $\sqrt{x+1}=\dfrac{1}{2\sqrt{x+1}}$ .
$m=\dfrac{k}{2\sqrt{x+1}}$
Putting x=3, we get
$m=\dfrac{k}{2\sqrt{3+1}}$
$\Rightarrow m=\dfrac{k}{4}...........(i)$
If we substitute m in equation (i), we get
$2k=3m+2$
$\Rightarrow 2k=\dfrac{3k}{4}+2$
$\Rightarrow \dfrac{5k}{4}=2$
$\Rightarrow k=\dfrac{8}{5}$
If we substitute k in in equation (ii), we get
$m=\dfrac{k}{4}=\dfrac{8}{5\times 4}=\dfrac{2}{5}$
So, the value of k+m is:
$k+m=\dfrac{8}{5}+\dfrac{2}{5}=\dfrac{10}{5}=2$

So, the correct answer is “Option a”.

Note: If you have noticed we have not completely used the condition that for f(x) to be continuous at x=a, the value of the function at x=a must be equal to $\displaystyle \lim_{x \to {{a}^{-}}}f(x)\text{ and }\displaystyle \lim_{x \to {{a}^{+}}}f(x)$. We have only used LHL=RHL but not used LHL=RHL=exact value, because LHL and the exact value would be the same for the above case, i.e., equal to 2k. However, in some of the questions it might be helpful in getting an additional equation if needed.