Differentiability

Given,
$ \left\{ \begin{matrix} \frac{2x-3}{2x-3}, & if & x\ge \frac{3}{2} \\ \frac{-(2x-3)}{2x-3}, & if & x

Let

and

Now,

Given,
Put,

Given,
On differentiating w. r. t. x, we get

Again, differentiating, we get

The correct answer is A:
Let ..(i)
Also, given

..(ii)
and
on differentiating E (i), w. r. t. x, we get
..(iii)

..(iv)
Again, differentiating E (iii) 0, we get
But (given)



On putting in E (iv), we get


On putting and
in E (i), we get

Given,
LHL=




RHL


Using L- Hospital rule,



Since, is continuous at

We have Taking log on both sides, we get


On differentiating w. r. t. x, we get

Let Then,

Let
and

and
Then,

Answer (b)

Answer (d) the chord joining the points and

We have,



Differentiating w.r.t. , we get

} \\[2ex]
-(x-10) & \text{if }
\end{cases}\)
Continuity at



Also,
Since,
is continuous at
is continuous everywhere on real numbers.
Differentiability at :




Since,
is not differentiable at .

Let . We need to differentiate this function using the chain rule. The derivative of is: In our case, .
The derivative of is: By applying the quotient rule and simplifying, the result is: Conclusion. Thus, the derivative is , which corresponds to option (b).

The function is not differentiable at because the left-hand and right-hand derivatives do not match. The left-hand derivative is (approaching from the negative side). The right-hand derivative is (approaching from the positive side). Since the derivatives do not match, the derivative at does not exist. Conclusion.
Thus, the correct answer is (d) None of these.