Find the point of discontinuity of y = f u , where f u = 3 2 u 2 + 5 u - 3 and u = 1 x + 2 .

Given, f u = 3 2 u 2 + 5 u - 3 and u = 1 x + 2 Now, factorize the denominator of f u , we get f u = 3 2 u 2 + 6 u - u - 3 ⇒ f u = 3 2 u - 1 u + 3 ⇒ u ≠ 1 2 & u ≠ - 3 ⇒ 1 x + 2 ≠ 1 2 ⇒ x ≠ 0 . . . i ( Using u = 1 x + 2 ) ⇒ 1 x + 2 ≠ - 3 ⇒ x ≠ - 7 3 . . . i i (Using u = 1 x + 2 ) Also, x + 2 ≠ 0 ⇒ x ≠ - 2 . . . i i i Now, if u is continuous at x = a , and f is continuous at x = u a , then the composite function f u is also continuous at x = a . So, from equations i , i i & i i i , the points of discontinuity are x = 0 , - 2 , - 7 3 .

Find the point of discontinuity of y = f u , where f u = 3 2 u 2 + 5 u - 3 and u = 1 x + 2 .

Given, f u = 3 2 u 2 + 5 u - 3 and u = 1 x + 2 Now, factorize the denominator of f u , we get f u = 3 2 u 2 + 6 u - u - 3 ⇒ f u = 3 2 u - 1 u + 3 ⇒ u ≠ 1 2 & u ≠ - 3 ⇒ 1 x + 2 ≠ 1 2 ⇒ x ≠ 0 . . . i ( Using u = 1 x + 2 ) ⇒ 1 x + 2 ≠ - 3 ⇒ x ≠ - 7 3 . . . i i (Using u = 1 x + 2 ) Also, x + 2 ≠ 0 ⇒ x ≠ - 2 . . . i i i Now, if u is continuous at x = a , and f is continuous at x = u a , then the composite function f u is also continuous at x = a . So, from equations i , i i & i i i , the points of discontinuity are x = 0 , - 2 , - 7 3 .

Show that the function f x = x 3 4 - sin π x + 3 takes the value 7 3 within the interval - 2 , 2 .

The given function is f x = x 3 4 - sin π x + 3 Since, x 3 4 , sin π x , 3 all are continuous function separately. So, f x will be a continuous function. Now, f - 2 = - 2 3 4 - sin - 2 π + 3 = - 2 + 3 = 1 Also, f 2 = 2 3 4 - sin 2 π + 3 = 2 + 3 = 5 Using intermediate value theorem, f x in the domain - 2 , 2 will take all values between f - 2 and f 2 . So, it takes all values between 1 and 5 . Hence, f x takes 7 3 within the interval - 2 , 2 .

Examine the differentiability of f x = 1 - e - x 2 at x = 0 .

Given, f x = 1 - e - x 2 at x = 0 Here, f x = 0 Now, LHD = f ' 0 - = lim h → 0 f 0 - h - f 0 - h = lim h → 0 1 - e - h 2 - h = lim h → 0 - 1 e h 2 e h 2 - 1 h 2 = lim h → 0 - 1 e h 2 × lim h → 0 e h 2 - 1 h 2 = - 1 , lim x → 0 e x - 1 x = 1 Similarly, RHD = f ' 0 + = lim h → 0 f 0 + h - f 0 h = lim h → 0 1 - e - h 2 h = lim h → 0 1 e h 2 e h 2 - 1 h 2 = 1 Since, LHD ≠ RHD f x is not differentiable at x = 0

If f x = a x 2 - b if x < 1 - 1 x if x ≥ 1 is derivable at x = 1 . Find the values of a and b .

Given, f x = a x 2 - b if x < 1 - 1 x if x ≥ 1 is derivable at x = 1 We observe that f x = 1 x if x ≤ - 1 a x 2 - b if - 1 < x < 1 - 1 x if x ≥ 1 Therefore, at x = 1 LHL = lim x → 1 - f x = lim x → 1 - a x 2 - b = a - b RHL = lim x → 1 + f x = lim x → 1 + - 1 x = - 1 f 1 = - 1 Since, f x is differentiable at x = 1 Therefore, it is also continuous at x = 1 So, we say that lim x → 1 - f x = lim x → 1 + f x = f 1 Therefore, a - b = - 1 . . . ( i ) Now differentiating above function, f ' x = - 1 x 2 if x ≤ - 1 2 a x if - 1 < x < 1 1 x 2 if x ≥ 1 As it is differentiable at x = 1 Therefore, LHD at x → 1 - = 2 a RHD at x → 1 + = 1 As, LHD = RHD Therefore, 2 a = 1 , a = 1 2 Now, b = a + 1 , b = 3 2 using i

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 33

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Find the point of discontinuity of $y = f (u)$ , where $f (u) = \frac{3}{2 u^{2} + 5 u - 3}$ and $u = \frac{1}{x + 2} .$

Important Questions on Continuity and Differentiability

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 34

Show that the function

f (x) = \frac{x^{3}}{4} - \sin π x + 3

takes the value

\frac{7}{3}

within the interval

[- 2, 2]

.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 35

If $g (x) = (|x - 1| + |4 x - 11|) [x^{2} - 2 x - 2],$ then find the number of point of discontinuity of $g (x)$ in $(\frac{1}{2}, \frac{5}{2})$

{where $[.]$ denotes GIF}

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 36

Test the continuity & differentiability of the function defined as under at $x = 1 & x = 2 .$

$f (x) = \{\begin{matrix} x & ; & x < 1 \\ 2 - x & ; & 1 \leq x \leq 2 \\ - 2 + 3 x - x^{2} & ; & x > 2 \end{matrix}\}$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 38

Prove that

f (x) = |x| \cos (x)

is not differentiable at

x = 0 .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-1>Q 39

Show that the function $f (x) = \{\begin{cases} x^{m} \sin (\frac{1}{x}) \\ 0 \end{cases} \begin{matrix} ; & x > 0 \\ ; & x = 0 \end{matrix}$ is,

$(i)$ Differentiable at $x = 0,$ if $m > 1$
$(ii)$ Continuous but not differentiable at $x = 0,$ if $0 < m \leq 1$
$(iii)$ Neither continuous nor differentiable, if $m \leq 0 .$