Let \(f: R \rightarrow R\) be defined as \(f(x)=\left\{\begin{array}{ccc}\frac{a-b \cos 2 x}{x^{2}} & ; & x 1\end{array}\right.\) If \(f\) is continuous everywhere in \(\mathbf{R}\) and m is the number of points where \(f\) is NOT differential then \(\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}\) equals :

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Continuity and Differentiability

Let

f : R \to R

be defined as

f (x) = ⎩ ⎨ ⎧ \frac{a - b c o s 2 x}{x ^{2}} x^{2} + c x + 2 2 x + 1;;; x < 0 0 \leq x \leq 1 x > 1

f

is continuous everywhere in

R

and m is the number of points where

f

is NOT differential then

m + a + b + c

equals :

Answer: (D)

f (x) = \frac{a - b cos 2 x}{x ^{2}}

for

x < 0

;

f (x) = x^{2} + c x + 2

for

0 \leq x \leq 1

;

f (x) = 2 x + 1

for

x > 1

.

**Continuity at

x = 1

:**

f (1^{-}) = 1 + c + 2 = 3 + c

f (1^{+}) = 2 (1) + 1 = 3

. Equate:

c = 0

.

**Continuity at

x = 0

:**

f (0) = 2

. For

f (0^{-})

, expand

cos 2 x = 1 - 2 x^{2} + \frac{2}{3} x^{4} - \dots

f (x) = \frac{( a - b ) + 2 b x ^{2} - \frac{2 b}{3} x ^{4} + \dots}{x ^{2}}

For the limit to exist, the constant term must vanish:

a = b

. Then

f (0^{-}) = 2 b

. Setting

2 b = f (0) = 2

gives

b = 1

, hence

a = 1

.

Differentiability: At

x = 1

: LHD

= 2 x + c ∣_{x = 1} = 2

, RHD

= 2

— differentiable. At

x = 0

: with

a = b = 1

f (x) = \frac{1 - cos 2 x}{x ^{2}}

for

x < 0

, whose derivative from the left and the derivative of

x^{2} + 2

from the right both evaluate to

0

x = 0

— differentiable there too. So

m = 0

m + a + b + c = 0 + 1 + 1 + 0 = 2

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option