Inverse Trigonometric Functions

  $f\left(x\right)=e^{x^3-3x+1}$

$f\text{'}\left(x\right)=e^{x^3-3x+1}$  (3x² − 3)

$e^{x^3-3x+1}$ = ⋅ 3(x −1)(x +1)

For x ∈ (−∞, −1], f '(x) ≥ 0

∴     f(x) is increasing function

∴     a = e^–∞ = 0 = f (−∞)

b = e⁻¹⁺³⁺¹ = e³ = f (−1)

∴     P(4, e³ + 2)

$d=\frac{\left(e^3+2\right)\left(e^{-3}\right)}{\sqrt[]{1+e^{-6}}}=\frac{1+2e^{-3}}{\sqrt[]{1+e^{-6}}}=\frac{1+2e^{-3}}{\sqrt[]{1+e^{-6}}}=\frac{e^3+2}{\sqrt[]{e^6+1}}$

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

 =   $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$        

Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC*AL$

$\Delta =\frac{1}{2}*2\sqrt[]{2hr-a^2}*h$       

then   $x=\sqrt[]{2*\frac{3r}{2}*r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$      

So . $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$  

$g\left(2\right)=\underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{x^2-x-2}{2x^2-x-6}=\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)+3\left(x-2\right)}$           

$Nowfog=f\left(g\left(x\right)\right)=si{n}^{-1}g\left(x\right)=si{n}^{-1}\left(\frac{x^2-x-2}{2x^2-x-6}\right)$

$\frac{3x^2-2x-8}{2x^2-x-6}\ge 0\&\frac{-x^2+4}{2x^2-x-6}\le 0$          

On solving we get   $x\in \left(-\infty ,-2\right)\cup \left[\frac{-4}{3},\infty \right)$

As x = 2 also lies in domain since g(2) $=\underset{x\to 2}{lim}g\left(x\right)$  

$f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

$g\left(f\left(x\right)\right)=x\Rightarrow g\text{'}\left(f\left(x\right)\right).f\text{'}\left(x\right)=1$

$f\left(x\right)=1\Rightarrow x=0$

$g\text{'}\left(1\right).f\text{'}\left(0\right)=1$

$f\text{'}\left(x\right)=2x+e^{-x}$

$\begin{array}{l}f\text{'}\left(0\right)=1\\ g\text{'}\left(1\right)=1\end{array}$

Kindly consider the following Image 

$f\left(g\left(x\right)\right)=x\forall x\in R$

$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)$

For y = g (x)

x = y³ + y – 5

$\frac{dx}{dy}=3y^2+1\Rightarrow g\text{'}\left(63\right)=\frac{1}{3\left(16\right)+1}=\frac{1}{49}$

$g\text{'}\left(63\right)=\left(\frac{dy}{dx}\right)$

$\left(\begin{array}{l}x=63\\ y=4\end{array}\right)$

T? = cot? ¹ (2²? ¹ + 1/2? ) = cot? ¹ (1 + 2²? ¹)/2? ) = tan? ¹ (2? / (1 + 2²? ¹)
T? = tan? ¹ (2? ¹ - 2? ) / (1 + 2? ¹ . 2? ) = tan? ¹ (2? ¹) – tan? ¹ (2? )
S∞ = π/2 - tan? ¹ (2) = cot? ¹ (2) = tan? ¹ (1/2)

Solve sin? ¹ (3x/5) + sin? ¹ (4x/5) = sin? ¹x.
Using the formula sin? ¹a + sin? ¹b = sin? ¹ (a√ (1-b²) + b√ (1-a²):
sin? ¹ ( (3x/5)√ (1 - (4x/5)²) + (4x/5)√ (1 - (3x/5)²) ) = sin? ¹x
(3x/5) * √ (1 - 16x²/25) + (4x/5) * √ (1 - 9x²/25) = x
x * [ (3/5) * √ (25-16x²)/5 + (4/5) * √ (25-9x²)/5 - 1 ] = 0
So, x = 0 is one solution.
For the other part:
3√ (25-16x²) + 4√ (25-9x²) = 25
Let's check integer solutions. If x = 1:
3√ (9) + 4√ (16) = 33 + 44 = 9 + 16 = 25. So x = 1 is a solution.
If x = -1:
3√ (9) + 4√ (16) = 25. So x = -1 is a solution.
The solutions are x = 0, 1, -1.

Given f (x) = ax^2 + bx + c.
f (-1) = a - b + c = 2
f' (x) = 2ax + b, so f' (-1) = -2a + b = 1
f' (x) = 2a, so f' (-1) = 2a = 1/2

From 2a = 1/2, we get a = 1/4.
Substituting a into -2a + b = 1: -2 (1/4) + b = 1 => -1/2 + b = 1 => b = 3/2.
Substituting a and b into a - b + c = 2: 1/4 - 3/2 + c = 2 => -5/4 + c = 2 => c = 13/4.

So, f (x) = (1/4)x^2 + (3/2)x + 13/4 = (1/4) (x^2 + 6x + 13).
We need to find f (1):
f (1) = (1/4) (1^2 + 6 (1) + 13) = (1/4) (1 + 6 + 13) = (1/4) (20) = 5.