Discuss the continuity & derivability of f x = x - 1 2 ; 0 ≤ x < 1 x x ; 1 ≤ x ≤ 2 , where x indicates the greatest integer x

We have, f x = x - 1 2 ; 0 ≤ x < 1 x x ; 1 ≤ x ≤ 2 ⇒ f x = 1 2 - x ; 0 ≤ x < 1 2 x - 1 2 ; 1 2 ≤ x < 1 x ; 1 ≤ x < 2 2 x ; x = 2 Now, at x = 1 2 ⇒ f 1 2 = 0 LHL = lim x → 1 - 2 f x = lim x → 1 - 2 1 2 - x = 0 RHL = lim x → 1 2 + f x = lim x → 1 2 + x - 1 2 = 0 Therefore, f x is continuous at x = 1 2 Similarly, at x = 1 , f 1 = 1 LHL = lim x → 1 - f x = lim x → 1 - x - 1 2 = 1 2 RHL = lim x → 1 + f x = lim x → 1 + x = 1 Since, LHL ≠ RHL , therefore f x is neither continuous nor differentiable at x = 1 Now, LHL = lim x → 2 - f x = lim x → 2 - x = 2 RHL = lim x → 2 + f x = lim x → 2 + 2 x = 4 Since, LHL ≠ RHL , therefore f x is neither continuous nor differentiable at x = 2 . Also, at x = 1 2 f ' x = - 1 ; 0 < x < 1 2 1 ; 1 2 < x < 1 f ' 1 2 - = - 1 and f ' 1 2 + = 1 Since, LHD ≠ RHD , therefore f x is not differentiable at x = 1 Ans. f x is continuous but not differentiable at x = 1 2 and neither continuous nor differentiable at x = 1 , 2 .

Check differentiability of f ( x ) = sgn x 2 / 3 + cos x 2 1 + x 2 + | x - 1 | 5 / 3 in [ - 2 , 2 ] , where [.] denotes GIF .

Here, f x = s g n x 2 / 3 + cos x 2 1 + x 2 + x - 1 5 / 3 Now, s g n x 2 / 3 = x 2 / 3 x 2 / 3 , x ≠ 0 0 , x = 0 = 1 , x ≠ 0 0 , x = 0 Similarly, cos x 2 1 + x 2 = 0 , x ≠ 0 1 , x = 0 and, x - 1 5 / 3 = 1 - x 5 / 3 , x ∈ [ - 2 , 1 ) x - 1 5 / 3 , x ∈ [ 1 , 2 ) Combining above cases f x = 1 + 1 - x 5 / 3 , x ∈ [ - 2 , 0 ) 2 , x = 0 1 + 1 - x 5 / 3 , x ∈ 0 , 1 1 + x - 1 5 / 3 , x ∈ [ 1 , 2 ) It can be seen that f x is continuous at x = 0 , 1 f ' x = 5 3 1 - x 2 / 3 x ∈ [ - 2 , 0 ) 5 3 1 - x 2 / 3 x ∈ ( 0 , 1 ) 5 3 x - 1 2 / 3 x ∈ [ 1 , 2 ) At, x = 0 LHD ⇒ f ' 0 - = 5 3 RHD ⇒ f ' 0 + = 5 3 At x = 1 LHD ⇒ f ' 1 - = 0 RHD ⇒ f ' 1 + = 0 i.e. f x is differentiable at x = 0 , 1 Hence, f x is differentiable in [ - 2 , 2 ]

Discuss the continuity and differentiability of h ( x ) = f ( x ) · g ( x ) in ( 0 , 3 ) , if f x = e x - e [ x ] + 1 {where [.] denote G . I . F . } and g x = | x - 1 | + | x - 2 | 2 | x - 1 | + | x - 2 | 3 ( | x - 1 | + | x - 2 | ) 2 , x ∈ ( 0 , 1 ) , x ∈ [ 1 , 2 ) , x ∈ [ 2 , 3 ) .

Since, we know that product of two continuous functions is also a continuous function. So we can check continuity of f x and g x separately. Also, in 0 , 3 , x can have values 0 or 1 or 2 . So, x + 1 ≠ 0 . e x is continuous for all R . Given f x = e x - e [ x ] + 1 For continuity at x = 1 , f 1 + = 0 , f 1 - = 0 and f 1 = 0 So f x is continuous at x = 1 Now for differentiability at x = 1 , f ' 1 + = e 2 , f ' 1 - = e Hence, f x is not differentiable at x = 1 . Given g x = | x - 1 | + | x - 2 | 2 | x - 1 | + | x - 2 | 3 ( | x - 1 | + | x - 2 | ) 2 , x ∈ ( 0 , 1 ) , x ∈ [ 1 , 2 ) , x ∈ [ 2 , 3 ) Now, for g x at x = 1 L . H . L . = lim x → 1 - x - 1 + x - 2 2 = lim x → 1 - - x + 1 - x + 2 2 = 3 - 2 2 = 1 2 R . H . L . = lim x → 1 + x - 1 + x - 2 = lim x → 1 + x - 1 - x + 2 = 1 Since, L . H . L . ≠ R . H . L . , therefore, g x is not continuous at x = 1 . Now, f 1 = 0 , therefore h 1 = 0 ⇒ h x is continuous at x = 1 . Also, g x is not differentiable at x = 1 as it is not continuous. Now, f 1 = 0 , therefore h 1 = 0 ⇒ h x is continuous at x = 1 . As we know that product of non differentiable function is a non differentiable function, hence, h x = f x · g x is not differentiable at x = 1 . For continuity at x = 2 , f 2 + = e 2 - e 3 , f 2 - = e 2 - e 2 So f x is discontinuous at x = 2 . Hence, f x is non differentiable at x = 2 . For g x , at x = 2 , R . H . L . = lim x → 2 + 3 x - 1 + x - 2 2 = lim x → 2 + 3 x - 1 + x - 2 2 = lim x → 2 + 3 2 x - 3 2 = 3 2 Also, g 2 = 1 ⇒ g x is not continuous. At x = 2 , f 2 = e 2 - e 3 and g x is not continuous. So, h x is not continuous at x = 2 . Hence, h x is also not differentiable at x = 2 .

Let function f x satisfying the relation f x + y + f x - y = 2 f x · f y , then prove that it is even a function.

f x + y + f x - y = 2 f x · f y For y = - x f x + - x + f x - - x = 2 f x · f - x f x - x + f x + x = 2 f x · f - x f 0 + f 2 x = 2 f x · f - x … 1 For y = x f x + x + f x - x = 2 f x · f x f 2 x + f 0 = 2 f x · f x … ( 2 ) From 1 and 2 2 f x · f - x = 2 f x · f x ⇒ f x = f - x or f x = 0 . If f x = f - x then f x is an even function. If f x = 0 then f x is an even function. ⇒ f x is an even function.

Let f ( x ) = x + 1 2 [ x ] , when - 2 ≤ x ≤ 2 , where [ . ] represents greatest integer function. Then

f ( x ) is discontinuous at x = 0 .

f ( x ) is continuous at x = 2 .

f ( x ) is continuous at x = 1 .

f ( x ) is continuous at x = - 1 .

The function f ( x ) is defined by f ( x ) = log ( 4 x - 3 ) x 2 - 2 x + 5 , if 3 4 < x < 1 or x > 1 4 , if x = 1

is discontinuous since neither f ( 1 - ) nor f ( 1 + ) exists.

is discontinuous at x = 1 since f ( 1 + ) does not exist though f ( 1 - ) exists.

is discontinuous at x = 1 since f ( 1 - ) does not exist though f ( 1 + ) exists.

Let f : [ a , b ] → [ 1 , ∞ ) be a continuous function and let g : R → R be defined as g x = 0 i f x < 0 ∫ a x f t d t i f a ≤ x ≤ b ∫ a b f t d t i f x > b . Then :

g x is continuous but not differentiable at b

g x is continuous and differentiable at either a or b but not both

For every twice differentiable function f : ℝ → - 2 , 2 with f 0 2 + f ′ 0 2 = 85 , which of the following statement(s) is (are) TRUE?

There exists α ∈ - 4 , 4 such that f α + f ′ ′ α = 0 a n d f ′ α ≠ 0

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 42

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Draw a graph of the function, $y = [x] + |1 - x|, - 1 \leq x \leq 3$ . Determine the points, if any, where this function is not differentiable, where $[\cdot]$ denotes the greatest integer function.

Important Questions on Continuity and Differentiability

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 43

Discuss the continuity & derivability of

f (x) = \{\begin{matrix} |x - \frac{1}{2}|; & 0 \leq x < 1 \\ x [x]; & 1 \leq x \leq 2 \end{matrix}

, where

[x]

indicates the greatest integer

x

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 44

Discuss continuity and differentiability of $y = f (x)$ in $[- 2, 5]$ , where [ .] denotes $GIF$ {.} denotes $FPF$

$f (x) = \{\begin{matrix} [x] & , & x \in [- 2, 0] \\ {x} & , & x \in (0, 2) \\ \frac{x^{2}}{4} & , & x \in [2, 3] \\ \frac{1}{\log_{4} (x - 3)} & , & x \in [3, 5] \end{matrix}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 45

Check differentiability of

f (x) = sgn (x^{2 / 3}) + [\cos (\frac{x^{2}}{1 + x^{2}})] + | x - 1 |^{5 / 3}

in

[- 2, 2]

, where [.] denotes

GIF

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 46

Discuss the continuity and differentiability of $h (x) = f (x) \cdot g (x)$ in $(0, 3)$ , if $f (x) = \frac{e^{x} - e}{[x] + 1}$ {where [.] denote $G . I . F .$ } and $g (x) = \{\begin{matrix} \frac{| x - 1 | + | x - 2 |}{2} \\ | x - 1 | + | x - 2 | \\ \frac{3 (| x - 1 | + | x - 2 |)}{2} \end{matrix} \begin{matrix} , & x \in (0, 1) \\ , & x \in [1, 2) \\ , & x \in [2, 3) \end{matrix}$ .