Question

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is

विकल्प

• R

• R − {3}

• (0, ∞)

• none of these

Answer 1

(b)  

$$R-\left(3\right)$$

$$(\text{ LHD at x }=3)=\underset{x\to 3^{-}}{lim}\frac{f\left(x\right)-f\left(3\right)}{x-3}$$
$$\left(\text{ LHD at x = 3 }\right)=\underset{h\to 0}{lim}\frac{f(3-h)-f\left(3\right)}{3-h-3}$$
$$\left(\text{ LHD at x = 3 }\right)=\underset{h\to 0}{lim}\frac{f(3-h)-f\left(3\right)}{-h}$$
$$\left(\text{ LHD at x = 3}\right)=\underset{h\to 0}{lim}\frac{|3-h-3|\cos (3-h)-f\left(3\right)}{-h}$$
$$(\text{ LHD at x }=3)=\underset{h\to 0}{lim}\frac{h\cos (3-h)-0}{-h}=-\cos 3$$
$$(\text{ RHD at x }=3)=\underset{x\to 3^{+}}{lim}\frac{f\left(x\right)-f\left(3\right)}{x-3}$$
$$\left(\text{ RHD at x = 3 }\right)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{3+h-3}$$
$$(\text{ RHD at x }=3)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{h}$$
$$\left(\text{ RHD at x = 3 }\right)=\underset{h\to 0}{lim}\frac{|3+h-3|\cos (3+h)-f\left(3\right)}{h}$$
$$(\text{ RHD at x }=3)=\underset{h\to 0}{lim}\frac{h\cos (3+h)-0}{h}=\cos 3$$

So, f(x) is not differentiable at x = 3.
Also, f(x) is differentiable at all other points because both modulus and cosine functions are differentiable and the product of two differentiable function is differentiable.