Prove that π e x - e + e π x - π + π π + e e x - π - e = 0 has one real root in e , π and other in π , π + e .

π e x - e + e π x - π + π π + e e x - π - e = 0 ⇒ π e x - π x - e - π + e π x - e x - e - π + π π + e e x - π x - e = 0 Let f x = π e x - π x - e - π + e π x - e x - e - π + π π + e e x - π x - e ⇒ f e = π e e - π - π = π e + 1 π - e > 0 , f π = e π π - e - e = - e π + 1 π - e < 0 and f π + e = π e e + π π e > 0 Now, we will use the result that if f x is a polynomial and a , b are two real numbers such that f a f b < 0 then, f x = 0 has an odd number of real roots between a and b . Clearly, f e f π < 0 and f π f e + π < 0 ∴ f x has one real root in e , π and other in π , π + e .

If | f p + q - f q | ≤ p q for all p and q ∈ Q & q ≠ 0 , show that ∑ i = 1 k f 2 k - f 2 i ≤ k ( k - 1 ) 2 .

Given, | f p + q - f q | ≤ p q , p and q ∈ Q , q ≠ 0 Now, f 2 k - f 2 i = f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . . . + f 2 i + 1 - f 2 i ⇒ f 2 k - f 2 i ≤ f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i . . . i ∵ a + b ≤ a + b Let us consider f 2 k - 1 + 2 k - 1 - f 2 k - 1 ≤ 2 k - 1 2 k - 1 { ∵ Given, | f p + q - f q | ≤ p q } ⇒ f 2 k - f 2 k - 1 ≤ 1 . ∵ x m · x n = x m + n Similarly, f 2 k - 1 - f 2 k - 2 ≤ 1 . . . . . . . . . . . . . . . . . . . . . . . f 2 i + 1 - f 2 i ≤ 1 So, f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i ≤ 1 + 1 + . . . . . k - i term ⇒ f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i ≤ k - i From equation i , f 2 k - f 2 i ≤ k - i ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ ∑ i = 1 k k - i ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ k - 1 + k - 2 + . . . . . + 1 ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ k k - 1 2 { ∵ Sum of first n natural number is n n + 1 2 } Hence, proved.

The function f : R → R satisfies x + f x = f f x for every x ∈ R . Find all solutions of the equation f f x = 0

Given, f : R → R ; x + f x = f f x ∀ x ∈ R We have to find solutions of the equation f f x = 0 So here f x = f f x - x Let f x 1 = f x 2 ∀ x 1 , x 2 ∈ R ⇒ f f x 1 - x 1 = f f x 2 - x 2 ⇒ f f x 2 - x 1 = f f x 2 - x 2 ⇒ x 1 = x 2 ⇒ f x is one-one function Now f f x = x + f x Put x = 0 f f 0 = f 0 ⇒ f 0 = 0 ( f x is a one-one function) So f f 0 = 0 If there is another x such that f f x = 0 = f f 0 Due to injectivity f x = f 0 and x = 0 So, the solution of f f x = 0 is only x = 0 .

If 2 f x = f x y + f x y ∀ x , y ∈ R + , f 1 = 0 and f ' ( 1 ) = 1 , find f ( x ) .

Given, 2 f x = f x y + f x y . . . i , f 1 = 0 and f ' 1 = 1 Let us assume x y = r 2 and x y = s 2 ⇒ x = r s Now, equation i will convert into 2 f r s = f r 2 + f s 2 Differentiating the both sides of above equation w.r.t., r , we get ⇒ 2 f ' r s s = f ' r 2 2 r + 0 ∵ d d x f g x = f ' g x g ' x ⇒ f ' r s = r f ' r 2 s Now, substitute r = 1 in the above equation, we get ⇒ f ' s = f ' 1 s . ⇒ f ' s = 1 s ∵ f ' 1 = 1 Integrating the above equation, we get ⇒ ∫ f ' s d s = ∫ 1 s d s ⇒ f s = ln s + C Put s = 1 in the above equation, we get ⇒ f 1 = ln 1 + C ⇒ C = 0 So, f s = ln s ⇒ f x = ln x .

If f x × f y = f x y ∀ x , y ∈ R , y ≠ 0 , then prove that f x · f 1 x = 1 .

Given, f x × f y = f x y . . . i Let us substitute x = y = 1 in the above equation, we get f f 1 = f 1 . . . i i Now substitute x = 1 , y = f 1 in the equation i , we get ⇒ f f f 1 = f 1 f 1 ⇒ f f f 1 = 1 ⇒ f 1 = 1 { ∵ from equation i i , f f 1 = f 1 } . Let us substitute x = 1 in the equation i , we get ⇒ f f y = f 1 y = 1 y . . . i i i Now, put y = f x in the above equation, we get ⇒ f f f x = 1 f x From equation i i i , f f x = 1 x So, f 1 x = 1 f x ⇒ f 1 x · f x = 1 .

Find the period of f ( x ) satisfying the condition: i f x + p = 1 + 1 - 3 f x + 3 f 2 x - f 3 x 1 3 , p > 0 i i f ( x - 1 ) + f ( x + 3 ) = f ( x + 1 ) + f ( x + 5 ) .

i f x + p = 1 + 1 - 3 f ( x ) + 3 f 2 ( x ) - f 3 ( x ) 1 3 , p > 0 ⇒ f x + p = 1 + 1 - f x 3 1 3 ∵ a - b 3 = a 3 - b 3 - 3 a b a - b ⇒ f ( x + p ) = 1 + 1 - f x ∵ x m n = x m n ⇒ f ( x + p ) + f x = 2 . . . i Replacing x by x + p in the above equation, we get ⇒ f x + 2 p + f x + p = 2 . . . i i Subtract equation i from equation i i , we get ⇒ f x + 2 p - f x = 0 ⇒ f x + 2 p = f x ∵ Period of f x is 2 p . i i f ( x - 1 ) + f ( x + 3 ) = f ( x + 1 ) + f ( x + 5 ) . . . i Replacing x by x + 2 in the above equation, we get ⇒ f ( x + 1 ) + f ( x + 5 ) = f ( x + 3 ) + f ( x + 7 ) . . . i i Subtract equation i i from equation i , we get ⇒ f x - 1 = f x + 7 Replacing x by x + 1 in the above equation, we get ⇒ f x = f x + 8 ∵ Period of f x is 8 .

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.

Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4

Attempt the free practice questions on Chapter 26: Continuity and Differentiability, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4 with Hints & Solutions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 1

Prove that $\frac{π^{e}}{x - e} + \frac{e^{π}}{x - π} + \frac{π^{π} + e^{e}}{x - π - e} = 0$ has one real root in $(e, π)$ and other in $(π, π + e)$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 21

If $| f (p + q) - f (q) | \leq \frac{p}{q}$ for all $p$ and $q \in Q & q \neq 0,$ show that $\sum_{i = 1}^{k} |f (2^{k}) - f (2^{i})| \leq \frac{k (k - 1)}{2}$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 22

The function $f : R \to R$ satisfies $x + f (x) = f (f (x))$ for every $x \in R .$ Find all solutions of the equation $f (f (x)) = 0$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 23

If $2 f (x) = f (x y) + f (\frac{x}{y}) \forall x, y \in R^{+}, f (1) = 0$ and $f^{'} (1) = 1,$ find $f (x)$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 24

If $f (x \times f (y)) = \frac{f (x)}{y} \forall x, y \in R, y \neq 0,$ then prove that $f (x) \cdot f (\frac{1}{x}) = 1$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-4>Q 25

Find the period of $f (x)$ satisfying the condition:

$(i)$ $f (x + p) = 1 + {\{1 - 3 f (x) + 3 f^{2} (x) - f^{3} (x)\}}^{\frac{1}{3}}, p > 0$

$(i i)$ $f (x - 1) + f (x + 3) = f (x + 1) + f (x + 5)$ .

MEDIUM

JEE Main/Advance

IMPORTANT

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

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 $(0, 5)$

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

Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4

Questions from Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 4: Exercise-4 with Hints & Solutions

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