f : ( 0 , 4 ] → R is a differentiable function. Then prove that for some a , b ∈ 0 , 4 , f 2 4 - f 2 0 = 8 f ' a · f b

Given: f is a differentiable function in ( 0 , 4 ] and as we know every differentiable function is a continuous function ⇒ f is also a continuous function in ( 0 , 4 ] So, using Langrange's Mean value theorem we can say that there exists at least one real number a in ( 0 , 4 ] such that f ' a = f 4 - f 0 4 - 0 ⇒ f ' a = f 4 - f 0 4 . . . i Also, we know that is a function f is continuous in ( 0 , 4 ] , then there exists a point b such that f 0 < f b < f 4 Also f 0 < f 0 + f 4 2 < f 4 ⇒ f b = f 0 + f 4 2 . . . ii By equations ( i ) & ( ii ) , f ' a f b = f 4 - f 0 4 f 0 + f 4 2 ⇒ f ' a f b = f 2 4 - f 2 0 8 ⇒ 8 f ' a f b = f 2 4 - f 2 0 Hence proved.

Let F x = f x 2 + f ' x 2 , F 0 = 7 , where f ( x ) is thrice differentiable function such that f x ≤ 1 ∀ x ∈ − 1 , 1 , then prove the followings. (i) there is at least one point in each of the intervals - 1 , 0 and 0 , 1 when f ' x ≤ 2 (ii) there is at least one point in each of the intervals - 1 , 0 and 0 , 1 where F ( x ) ≤ 5 (iii) there exist at least one maximum of F ( x ) in - 1 , 1 (iv) for some c ∈ − 1 , 1 , F c ≥ 7 , F ' c = 0 & F " c ≤ 0

∙ Given, F x = f x 2 + f ' x 2 ∙ F 0 = 7 ∙ ∀ x ∈ - 1 , 1 , f x ≤ 1 Since, f x is thrice differentiability. So, f x is also continuous in - 1 , 1 . By L M V T , there exists an α ∈ a , b such that f ' α = f b - f a b - a ∙ Let b = 1 , a = 1 ⇒ α ∈ 0 , 1 f ' α = f 1 - f 0 1 - 0 ⇒ f ' α ≤ f 1 + - f 0 ⇒ f ' α ≤ f 1 + f 0 ⇒ f ' α ≤ 2 , as f x ≤ 1 ∙ Consider, a = - 1 , b = 0 f ' α = f 0 - f - 1 0 - - 1 = f 0 - f - 1 ⇒ f ' x ≤ f 0 + - f - 1 ≤ 2 ∙ i There exists at least one point in each of - 1 , 0 & 0 , 1 such that f ' x ≤ 2 ∙ i i F x = f x 2 + f ' x 2 There exists an x ∈ - 1 , 0 & 0 , 1 such that f ' x ≤ 2 & also for x ∈ - 1 , 1 , f x ≤ 1 ⇒ There exists an x ∈ - 1 , 0 & 0 , 1 such that F x ≤ 1 2 + 2 2 ⇒ F x ≤ 5 ∙ i i i From i i , there exists at least one point in each of - 1 , 0 & 0 , 1 such that F x ≤ 5 . ∙ Also, F x is continuous & differentiable. ∙ F 0 = 7 ∙ Possible graph of F x are ⇒ F x has at least one local maxima in - 1 , 1 . ∙ i v From i i i , there exists a point of maxima in - 1 , 1 say x = c . ∙ F e ≥ 7 , as x = c is a point of local maxima ∙ F ' c = 0 ∙ F " c ≤ 0 Hence proved.

If m - 1 a 1 2 ≥ 2 m a 2 < 0 , then prove that x m + a 1 x m - 1 + a 2 x m - 2 + … . . + a m - x + a 0 = 0 has at least one non real root a 1 , a 21 , … , a m ∈ R .

Given f x = x m + a 1 x m - 1 + a 2 x m - 2 + … . . + a m - x + a 0 . Let f x = 0 has m real roots, say x 1 , x 2 , x 3 … . . , x m . Also, f x is continuous and differentiable for all real values of x . f x 1 = f x 2 = f x 3 = … . . = f x m = 0 Then, by Rolle's theorem f ' x = 0 has m - 1 real roots, f '' x = 0 has m - 2 real roots. So, we can say that f m - 2 x = 0 has 2 real roots. Here, f m - 2 x = m ! 2 x 2 + a 1 m - 1 ! x + a 2 m - 2 ! Further simplifying, f m - 2 x = m - 2 ! 2 m m - 1 x 2 + 2 a 1 m - 1 x + 2 a 2 For real roots, its discriminant should be greater than or equal to zero. ⇒ m - 2 ! 2 2 2 a 1 2 m - 1 2 - 4 m m - 1 2 a 2 ≥ 0 ⇒ a 1 2 m - 1 - 2 m a 2 ≥ 0 But as per question the above inequality doesn't hold good. That means our assumption is wrong. Hence, f x = 0 has at least one non-real root.

f x = ( 2013 ) x 2012 - ( 2012 ) x 2011 - 2014 x + 1007 , then show that for x ∈ 0 , 1007 1 / 2011 , f x = 0 has at least one real root.

According to Rolle's theorem, if a function y = f x i is continuous in the interval x ∈ a , b i i is differentiable in the interval x ∈ a , b i i i f a = f b Then, f ' c = 0 for at least one c ∈ a , b . Let, g ' x = f x = ( 2013 ) x 2012 - ( 2012 ) x 2011 - 2014 x + 1007 ⇒ d g x = ( 2013 ) x 2012 - ( 2012 ) x 2011 - 2014 x + 1007 d x ⇒ ∫ d g x = ∫ ( 2013 ) x 2012 - ( 2012 ) x 2011 - 2014 x + 1007 d x ⇒ g x = ( 2013 ) x 2013 2013 - ( 2012 ) x 2012 2012 - 2014 x 2 2 + 1007 x + K ⇒ g x = x 2013 - x 2012 - 1007 x 2 + 1007 x + K Here, g x is a polynomial, so it is continuous and differentiable for all x ∈ R . And g 1 = g 0 = K According to Rolle's theorem, there exist at least one x ∈ 0 , 1 where g ' x = f x = 0 . Hence, f x = 0 has at least one root in x ∈ 0 , 1 and also in x ∈ 0 , 1007 1 / 2011 .

A function f is differentiable in the interval 0 ≤ x ≤ 5 such that f ( 0 ) = 4 & f ( 5 ) = - 1 . If g x = f x x + 1 , then prove that there exists some c ∈ ( 0 , 5 ) such that g ' c = - 5 6 .

According to Lagrange's Mean Value Theorem, if a function F x is i Continuous in the interval x ∈ a , b . i i Differentiable in the interval x ∈ a , b Then, there exists at least one c ∈ a , b which satisfy F ' c = F b - F a b - a . Given, a function f is differentiable in the interval 0 ≤ x ≤ 5 , so it must be continuous in the interval 0 ≤ x ≤ 5 . Also given, f ( 0 ) = 4 & f ( 5 ) = - 1 . If g x = f x x + 1 , then g x is also continuous in x ∈ 0 , 5 and differentiable in 0 , 5 According to Lagrange's Mean Value Theorem, there exist at least one c ∈ 0 , 5 for which g ' c = g 5 - g 0 5 - 0 . ⇒ g ' c = f 5 5 + 1 - f 0 0 + 1 5 ⇒ g ' c = - 1 5 + 1 - 4 0 + 1 5 ⇒ g ' c = - 5 6 .

Let f x and g x be differentiable functions having no common zeros so that f x g ' x ≠ f ' x g x Prove that between any two zeros of f x , there exist atleast one zero of g x .

Let a & b be two consecutive zero of f x . ⇒ f a = f b = 0 Lets us assume that g x has no zero. Now, consider h x = f x g x . Since, g x has no zero so differentiability of f x & g x implies that h x will also be differentiable in a , b . h a = h b = 0 , as f a = f b = 0 & g x ≠ 0 . So, h x satisfies all conditions of Rolle's Theorem. ⇒ There exists at least one x ∈ a , b for which h ' x = 0 ⇒ f ' x g x - f x g ' x g x 2 = 0 , x ∈ a , b ⇒ f ' x g x = f x g ' x , x ∈ a , b which is contradicting the condition given in the question. Hence our assumption is wrong and we can say that between any two zeros of f x there exist at least one zero of g x .

If ϕ ( x ) is a differentiable function ∀ x ∈ R and a ∈ R + such that ϕ 0 = ϕ 2 a , ϕ a = ϕ 3 a and ϕ 0 ≠ ϕ a , then show that there is at least one root of equation ϕ ' x + a = ϕ ' x in 0 , 2 a .

Given equation is ϕ ' x + a - ϕ ' x = 0 … 1 Let f x = ∫ ϕ ' x + a - ϕ ' x d x , we get f x = ϕ x + a - ϕ x + k Now, f 0 = ϕ a - ϕ 0 + k and f 2 a = ϕ 3 a - ϕ 2 a + k Put ϕ 3 a = ϕ a and ϕ 2 a = ϕ 0 f 2 a = ϕ ( a ) - ϕ ( 0 ) + k Therefore, f 0 = f ( 2 a ) By Rolles theorem, we conclude that f ' c = 0 for at least one c ∈ 0 , 2 a . Therefore ϕ ' c + a - ϕ ' c = 0 for at least one c ∈ 0 , 2 a . Hence, equation ϕ ' x + a = ϕ ' x has at least one root in 0 , 2 a .

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

MEDIUM

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IMPORTANT

Earn 100

Let $f (x)$ is defined only for $x \in (0, 5)$ and defined as $f^{2} (x) = 1 \forall x \in (0, 5)$ . Function $f (x)$ is continuous for all $x \in (0, 5) - {1, 2, 3, 4}$ (at $x = 1, 2, 3, 4 f (x)$ may or may not be continuous). Find the number of possible functions $f (x)$ if it is discontinuous at
$(i)$ One integral point in $(0, 5)$
$(ii)$ Two integral points in $(0, 5)$
$(iii)$ Three integral points in $(0, 5)$
$(iv)$ Four integral points in $(0, 5)$

Important Questions on Continuity and Differentiability

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 27

f : (0, 4] \to R

is a differentiable function. Then prove that for some

a, b \in (0, 4), f^{2} (4) - f^{2} (0) = 8 f^{'} (a) \cdot f (b)

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 28

Let $F (x) = {(f (x))}^{2} + {(f^{'} (x))}^{2}, F (0) = 7$ , where $f (x)$ is thrice differentiable function such that $|f (x)| \leq 1 \forall x \in [- 1, 1]$ , then prove the followings.

(i) there is at least one point in each of the intervals $(- 1, 0)$ and $(0, 1)$ when $|f^{'} (x)| \leq 2$

(ii) there is at least one point in each of the intervals $(- 1, 0)$ and $(0, 1)$ where $F (x) \leq 5$

(iii) there exist at least one maximum of $F (x)$ in $(- 1, 1)$

(iv) for some $c \in (- 1, 1), F (c) \geq 7, F^{'} (c) = 0 & F^{"} (c) \leq 0$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 29

(m - 1) a_{1}^{2} \geq 2 m a_{2} < 0

, then prove that

x^{m} + a_{1} x^{m - 1} + a_{2} x^{m - 2} + \dots . . + a_{m - x} + a_{0} = 0

has at least one non real root

(a_{1}, a_{21}, \dots, a_{m} \in R)

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 30

f (x) = (2013) x^{2012} - (2012) x^{2011} - 2014 x + 1007

, then show that for

x \in [0, 1007^{1 / 2011}], f (x) = 0

has at least one real root.

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 31

A function

f

is differentiable in the interval

0 \leq x \leq 5

such that

f (0) = 4 & f (5) = - 1 .

g (x) = \frac{f (x)}{x + 1},

then prove that there exists some

c \in (0, 5)

such that

g^{'} (c) = - \frac{5}{6} .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 32

Let

f (x)

and

g (x)

be differentiable functions having no common zeros so that

f (x) g^{'} (x) \neq f^{'} (x) g (x)

Prove that between any two zeros of

f (x)

, there exist atleast one zero of

g (x)

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 33

f

is continuous in

[a, b]

and differentiable in

(a, b)

(where

a > 0

) such that

\frac{f (a)}{a} = \frac{f (b)}{b} .

Prove that there exist

x_{0} \in (a, b)

such that

f^{'} (x_{0}) = \frac{f (x_{0})}{x_{0}} .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 34

ϕ (x)

is a differentiable function

\forall x \in R

and

a \in R^{+}

such that

ϕ (0) = ϕ (2 a), ϕ (a) = ϕ (3 a)

and

ϕ (0) \neq ϕ (a)

, then show that there is at least one root of equation

ϕ^{'} (x + a) = ϕ^{'} (x)

(0, 2 a)

Important Questions on Continuity and Differentiability

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