Revision: Mathematics >> Continuity and Differentiability CUET (UG) Continuity and Differentiability

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = |x − 1|, x ∈ R

f(x) = (x − 1), if x − 1 > 0

= −(x − 1), if x − 1 < 0

At x = 1

f(1) = 1 − 1 = 0

left-side limit:

`lim_(h -> 0^-) (f(1 - h) - f(1))/ -h`

= `lim_(h -> 0^-) (1 - (1 - h) - 0)/ (- h)`

= `lim_(h -> 0^-) (+ h)/(- h)`

= −1

Right-side limit:

= `lim_(h -> 0^+) (f(1 + h) - f(1))/h`

= `lim_(h -> 0^+) ((1 + h) - 1 - 0)/ h`

= `lim_(h -> 0^+) h/h`

= 1

Left-side limit and the right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

Any function will not be differentiable if the left-hand limit and the right-hand limit are not equal.

f(x) = [x], 0 < x < 3

(i) At x = 1

Left-side limit:

`lim_(h -> 0) ([1 - h] - [1])/-h`

= `lim_(h -> 0) (0 - 1)/-h`

= `lim_(h -> 0) 1/h`

= Infinite (∞)

Right-hand limit:

`lim_(h -> 0) ([1 + h] - [1])/h`

= `lim_(h -> 0) (1 - 1)/h`

= 0

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 1.

(ii) At x = 2

Left-side limit:

`lim_(h -> 0) (f(2 + h) - f(2))/h`

= `lim_(h -> 0) ([2 + h]-2)/h`

= `lim_(h -> 0) (2 -2)/h`

= 0

Right-hand limit:

`lim_(h -> 0) (f(2 - h) - f (2))/h`

= `lim_(h -> 0) ([2 - h] - [2])/-h`

= `lim_(h -> 0) (1 - 2)/-h`

= Infinite (∞)

Left-side limit and right-side limit are not equal.

Hence, f(x) is not differentiable at x = 2.

y = `|(f(x), g(x), h(x)),(l, m, n),(a, b, c)|`

`dy/dx= |(d/dx (f(x)), d/dx (g(x)), d/dx (h(x))), (l, m, n), (a, b, c)| + |(f(x), g(x), h(x)),(0, 0, 0),(a, b, c)| + |(f(x), g(x), h(x)),(l, m, n),(0, 0, 0)|`

`= |(f'(x), g'(x), h'(x)),(l, m, n),(a, b, c)|`

Given, (x – a)² + (y – b)² = c²  ...(1)

On differentiating with respect to x,

`=> 2 (x - a) + 2(y - b) dy/dx = 0`

`=> (x - a) + (y - b) dy/dx = 0`  ...(2)

Differentiating again with respect to x,

`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0

`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0

`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}`  ...(3)

Putting the value of (y – b) in (2),

`(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)`   ...(4)

Putting the values ​​of (x − a) and (y − b) from (3) and (4) in (1),

`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`

On multiplying by `((d^2y)/dx^2)^2`,

`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2 = c^2 ((d^2y)/dx)^2`

`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`

`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`

On taking the square root,

`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c  ...(a constant independent of a and b.)

`x sqrt(1 + y) + y sqrt(1 + x) = 0`

∴ `xsqrt(1 + y) = - y sqrt(1 + x) = 0`

On squaring both sides,

x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y²x

⇒ x² – y² – y²x + x²y = 0

⇒ (x – y)(x + y) + xy(x – y) = 0

⇒ (x – y)[x + y + xy] = 0

x – y = 0

As x ≠ y

x + y (1 + x) = 0

y (1 + x) = –x

∴ y = `-x/(1 - x)`

Differentiating w.r.t., x

∴ `dy/dx = ((1 + x)(1) - x (1))/(1 + x)^2`

= `-(1 + x - x)/(1 + x)^2`

= `-1/(1 + x)^2`

Given: x = `e^(x/y)`

Taking log on both the sides,

log x = `log e^(x/y)`

⇒ log x = `x/y log e`

⇒ log x = `x/y`  ...[∵ log e = 1]  ...(i)

Differentiating both sides w.r.t. x:

`d/dx log x = d/dx (x/y)`

⇒ `1/x = (y xx 1 - x xx dy/dx)/y^2`

⇒ `y^2 = xy - x^2 xx dy/dx`

⇒ `x^2 xx dy/dx = xy - y^2`

⇒ `dy/dx = (y(x - y))/x^2`

⇒ `dy/d = y/x xx ((x - y)/x)`

⇒ `dy/dx = 1/logx xx ((x - y)/x)   ...[∵ log x = x/y "from equation (i)"]`

`dy/dx = (x - y)/(xlogx)`

Hence proved.

cos y = x cos (a + y)

∴ x = `(cos y)/(cos (a + y))`

On differentiating with respect to y,

`cos (a + y) d/dy cos y - cos y d/dy`

`therefore dx/dy = (cos (a + y))/(cos^2 (a + y))`

`= (- sin y cos (a + y) + cos y sin (a + y))/(cos^2 (a + y))`

`= (sin (a + y) cos y - cos (a + y) sin y)/(cos^2 (a + y))`

`= (sin (a + y - y))/(cos^2 (a + y))`   ...[∵ sin (A − B) = sin A cos B − cos A sin B]

`= (sin a)/(cos^2  (a + y))`

`therefore dy/dx = (cos^2 (a + y))/(sin a)`

Given, y = 5 cos x – 3 sin x

Differentiating both sides with respect to x,

`dy/dx = 5 d/dx cos x - 3 d/dx sin x`

= 5 (−sin x) − 3 cos x

= −5 sin x − 3 cos x

Differentiating both sides again with respect to x,

`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`

= −5 cos x − 3 (−sin x)

= 3 sin x − 5 cos x

Hence, `(d^2 y)/dx^2 + y` = 0

(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)