mathematics-class-12-unit-06-chapter-07-derivatives-l-7-12-s7sxkz4r3ey

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- Theorem: Suppose $f:(a, b) \rightarrow \mathbb{R}$ is a differential function.             
- (i) If $f^{\prime}(x)>0 \quad \forall x \in(a, b)$, then $f(x)$ is strictly increasing on $(a, b)$              
- (ii) If $f^{\prime}(x)<0 \quad \forall x \in(a, b)$, then $f(x)$ is strictly decreasing on $(a, b)$.

- Proof: (i) Let $a<x_1<x_2<b$.
- Then $f$ is cont. on        
- $\left[x_1, x_2\right] $ is diff. an       
- $\left(x_1, x_2\right)$.        
- $\therefore$ By the MVT, $\exists c \in\left(x_1, x_2\right)$ such that

- $\begin{aligned}& \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=f^{\prime}(c)>0 \\\\ \Rightarrow & f\left(x_2\right)-f\left(x_1\right)>0 \quad\left(\because x_2-x_1>0\right) . \\\\ \Rightarrow & f\left(x_2\right)>f\left(x_1\right) \text { whenever } x_2>x_1 . \end{aligned}$
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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-7 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- (ii) Similar to (i).

- Problem-1:

- Prove that $\sin x \leqslant x \quad \forall x \geqslant 0$.

- Solution:$ ~ $ Let $f(x)=x-\sin x$

- $\text { Then, } ~ f^{\prime}(x)=1-\cos x \geqslant 0 \quad(\because ~ -1 \leq \cos x \leqslant 1)$

- $\therefore f(x)$ is an increasing function. Also, $\quad f(0)=0$.

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<footer style = "font-size: 18px">Derivatives lecture-7 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore$ For $x \geqslant 0, f(x) \geqslant f(0)=0$           
- ie. $x-\sin x \geqslant 0$            
- ie. $\sin x \leq x \quad \forall x \geqslant 0$.

- Problem-2:

- Show that $x<\tan x \quad \forall x \in\left(0, \frac{\pi}{2}\right)$               
- Solution: $ ~ $ Put $f(x)=\tan x-x$              
- Then $f^{\prime}(x)=\sec ^2 x-1=\tan ^2 x>0$               
- $\therefore f(x)$ is strictly increasing $m\left(0, \frac{\pi}{2}\right)$           
- $\Rightarrow \quad f(x)>f(0)=0 \quad \forall x \in\left(0, \frac{\pi}{2}\right)$               
- ie. $\tan x>x \quad \forall x \in\left(0, \frac{\pi}{2}\right)$.           
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Intermediate Value Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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-  Prove that $|\sin x-\sin y| \leqslant|x-y|$ for all $x, y \in \mathbb{R}$.

- Solve: $\quad f(x)=\sin x$ is cont. and differentiable everywhere.           
- $\therefore$ By the MVT, given $x<y, \exists c \in(x, y)$ such that $f^{\prime}(c)=\frac{f(y)-f(x)}{y-x}$         
- ie. $\frac{\sin y-\sin x}{y-x}=f^{\prime}(c)=\cos c$       
- $\Rightarrow\left|\frac{\sin x-\sin y}{x-y}\right|=|\cos x| \leq 1 \text { if } x \neq y $              
- $\Rightarrow \quad|\sin x-\sin y| \leqslant|x-y| .$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Intermediate Value Theorem $\rightarrow$ Corollary </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Intermediate Value Theorem </primary>
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- Intermediate Value Theorem (IVT) Suppose $f:[a, b] \rightarrow \mathbb{R}$ be a continuous $f_x$. Let $y$ lie between $f(a)$ and $f(b) \cdot(f(a)<f(b))$ Then $\exists$ at least one $x \in(a, b)$ such that $f(x)=y$           
- The theorem. 'is intuitively clear but we will skip the formal proof.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Intermediate Value Theorem </span> $\rightarrow$ Corollary $\rightarrow$ Applications of IVT </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Corollary </primary>
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- Note that continuity is required for the IVT.

- Corollary: Suppose $f:[a, b] \rightarrow \mathbb{R}$ be continuous and assume that $f(a)$ and $f(b)$ are of opposite signs (ie. $f(a) f(b)<0$ ). Then $\exists$ at least one $x \in(a, b)$ s.t. $f(x)=0$.

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Intermediate Value Theorem $\rightarrow$ <span style = "color:blue"> Corollary </span> $\rightarrow$ Applications of IVT $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Applications of IVT </primary>
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- Pf: since o lies between $f(a)$ \& $f(b)$, by the IVT $\exists x \in(a, b)$ s.t. $f(x)=0$.

- **Applications of IVT:**

- Theorem: Every polynomial of odd degree must have at least one zero.           
- (ie. if $p(x)=a_0+a_1 x+\cdots+a_n x^n$ where $n$ is odd & $ a_n \neq 0$ then $\exists$ at least one $c \in \mathbb{R}$ s.t. $\quad p(c)=0)$

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<footer style = "font-size: 18px">Intermediate Value Theorem $\rightarrow$ Corollary $\rightarrow$ <span style = "color:blue"> Applications of IVT </span> $\rightarrow$ Remark $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Remark </primary>
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- Remark: The result is not true for even degree polynomials, e.g. $p(x)=x^2+1$ has no real zeros.

- Pf of the them: $p(x)=a_0+a_1 x+\cdots+a_n x^n$
- Since $n$ is an odd integer, $\lim _{x \rightarrow+\infty} x^n=+\infty$ and     
- $\lim _{x \rightarrow-\infty} x^n=-\infty \text {. }$

- $\therefore p(x)$ must be negative at some $x$ and positive at some other $x$.

- $\therefore$ \&y th cor. to IVT, $p(x)$ must be zero at same point $x$.

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<footer style = "font-size: 18px">Corollary $\rightarrow$ Applications of IVT $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Problem-4 $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Show that $x^5+4 x-1=0$ has exactly one solution in $\mathbb{R}$.

- Solve:$ ~ $ Let $f(x)=x^5+4 x-1$

- Since $p(x)$ is an odd degree polynomial, $p(x)=0$ has at least one solution in $\mathbb{R}$.

- Suppose there are two solutions $x_1<x_2$.

- Then $p\left(x_1\right)=0=p\left(x_2\right)$.

- By the Rolle's theorem, $\exists c \in\left(x_1, x_2\right)$ s.t. $p^{\prime}(c)=0 \quad \therefore p(x)=0$ has          
- But, $p^{\prime}(x)=5 x^4+4 \geqslant 4 \quad \therefore p(x)=0$ has

- $\therefore$ We get a contradiction.

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<footer style = "font-size: 18px">Applications of IVT $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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-  What is the solution. to $p(x)=0$ ?
 
-  Solution :
 
-  We can use the IVT to find approximate solution as follows.
 
- $\begin{aligned}& p(x)=x^5+4 x-1 \\\\& p(0)=-1<0, p(1)=4>0\end{aligned} $

- By the IVT, we know $p(c)=0$ for same $c \in(0,1)$ Find the value at the mid point $\frac{1}{2}$.

- $p\left(\frac{1}{2}\right)=\frac{1}{32}+2-1=1+\frac{1}{32}>0$

- Since $p(0)<0, p\left(\frac{1}{2}\right)>0$, the zero must lie in $\left(0, \frac{1}{2}\right)$.

- $p\left(\frac{1}{4}\right)=\frac{1}{4^5}+1-1=\frac{1}{4^5}>0 \text {, so, ........... }\left(0, \frac{1}{4}\right)$

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<footer style = "font-size: 18px">Remark $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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-  Prove that $\cos x=x$ for some $x \in\left(0, \frac{\pi}{2}\right)$.        
 
- Solution:$ ~ $ Let $f(x)=\cos x-x$
- $f(x)$ is continuous everywhere          
- Also, $f(0)=\cos 0-0=1>0$         
- $f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}-\frac{\pi}{2}=0-\frac{\pi}{2}=-\frac{\pi}{2}<0$

- By the IVT, $\exists x \in\left(0, \frac{\pi}{2}\right)$ s.t. $f(x)=0$
- $\text { ie. } \cos x=x \text {. }$

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Sol $\quad g(1)=-g(0)$

- Thus by the IVT, $\exists x \in[0,1]$ s.t. $g(x)=0$

- i.e $f(x+1)-f(x)=0$           
- i.e. $f(x+1)=f(x)$       
- Hence, done.

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<footer style = "font-size: 18px"> Problem-6 $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue">  Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>