Functions

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & f(x)=a \sin \left(x+\frac{\pi}{4}\right)+b \cos x+c \\ & f(x)=\frac{a \sin x}{\sqrt{2}}+\left(\frac{a}{\sqrt{2}}+b\right) \cos x+c=r \sin (x+\theta)+c \\ & \text { where } r=\sqrt{\left(\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}+b\right)^2} \text { and } \\ & \theta=\tan ^{-1}\left(\frac{a+\sqrt{2} b}{a}\right) \end{aligned} \end{aligned} $

For $f$ to be one-one, we must have

$ \begin{aligned} & -\frac{\pi}{2} \leq x+\theta \leq \frac{\pi}{2} \\ & \Rightarrow-\frac{\pi}{2}-\theta \leq x \leq \frac{\pi}{2}-\theta \Rightarrow x \in\left[-\frac{\pi}{2}-\theta, \frac{\pi}{2}-\theta\right] \end{aligned} $

Hence, $X=\left[\frac{-\pi}{2}-\theta, \frac{\pi}{2}-\theta\right]$, where

$ \theta=\tan ^{-1}\left(\frac{a+\sqrt{2} b}{a}\right) $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } f^{-1}(9)=x \\ & \Rightarrow \quad f(x)=9 \\ & \Rightarrow \quad x^2+5 x+9=9 \\ & \Rightarrow \quad x^2+5 x=0 \\ & \Rightarrow \quad x(x+5)=0 \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & x & =0,-5 \text { which are in } R . \\ \therefore & & f^{-1}(9) & =\{0,-5\} \end{aligned} $

Given that $ f(x) = [x] $ represents the greatest integer less than or equal to $ x $, and $ g(x) = |x| $, let's solve the expression $ f\left(g\left(\frac{8}{5}\right)\right) - g\left(f\left(-\frac{8}{5}\right)\right) $.

First, calculate the individual components:

Calculate $ f\left(-\frac{8}{5}\right) $:

$ f\left(-\frac{8}{5}\right) = \left[-\frac{8}{5}\right] = -2 $

Calculate $ g\left(\frac{8}{5}\right) $:

$ g\left(\frac{8}{5}\right) = \left|\frac{8}{5}\right| = \frac{8}{5} $

Now, substitute these values into the expression:

Evaluate $ f\left(g\left(\frac{8}{5}\right)\right) $:

$ f\left(g\left(\frac{8}{5}\right)\right) = f\left(\frac{8}{5}\right) = \left[\frac{8}{5}\right] = 1 $

Evaluate $ g\left(f\left(-\frac{8}{5}\right)\right) $:

$ g\left(f\left(-\frac{8}{5}\right)\right) = g(-2) = |-2| = 2 $

Finally, calculate the expression:

$ f\left(g\left(\frac{8}{5}\right)\right) - g\left(f\left(-\frac{8}{5}\right)\right) = 1 - 2 = -1 $

Hence, the value of the expression is $-1$.

We have, $f(x)=x^2-2 x+1$ and

$\begin{array}{ll} & f \circ g(x)=x^2+2 x+1 \\ \therefore & f(g(x))=x^2+2 x+1 \\ \Rightarrow & (g(x))^2-2\left(g(x)+1=x^2+2 x+1\right. \\ \Rightarrow & (g(x))^2-2\left(g(x)-x^2-2 x=0\right. \\ \Rightarrow & g(x)=\frac{-(-2) \pm \sqrt{(-2)^2-4\left(-x^2-2 x\right)(1)}}{2(1)} \\ \Rightarrow & g(x)=\frac{2 \pm \sqrt{4+4 x^2+8 x}}{2} \\ \Rightarrow & g(x)=\frac{2 \pm 2 \sqrt{x^2+2 x+1}}{2} \\ \Rightarrow & g(x)=1 \pm(x+1) \\ \Rightarrow & g(x)=x+2-x \end{array}$

We have, ${1 \over 2}gof(x) = 2{x^2} - 5x + 2$

$ \Rightarrow g(f(x)) = 4{x^2} - 10x + 4$

$ \Rightarrow {(f(x))^2} + f(x) - 2 = 4{x^2} - 10x + 4$

$ \Rightarrow {(f(x))^2} + f(x) - (4{x^2} - 10x + 6) = 0$

$ \Rightarrow f(x) = {{ - 1\, \pm \,\sqrt {1 + 4(4{x^2} - 10x + 6)} } \over 2}$

$ \Rightarrow f(x) = {{ - 1\, \pm \,\sqrt {16{x^2} - 40x + 25} } \over 2}$

$ \Rightarrow f(x) = {{ - 1\, \pm \,(4x - 5)} \over 2} = 2x - 3, - 2x + 2$

Hence, $f(x) = 2x - 3$

Given, f(x) = 4x $-$ x²

f(a + 1) $-$ f(a $-$ 1)

= [4(a + 1) $-$ (a + 1)²] $-$ [4(a $-$ 1) $-$ (a $-$ 1)²]

= [4a + 4 $-$ a² $-$ 1 $-$ 2a] $-$ [4a $-$ 4 $-$ a² $-$ 1 + 2a]

$\Rightarrow$ 8 $-$ 4a = 0

$\therefore$ a = 2

We have, $y = x{(x - 1)^2}$

$\therefore$ ${{dy} \over {dx}} = {(x - 1)^2}\,.\,1 + 2x(x - 1) = 3{x^2} - 4x + 1 = (x - 1)(3x - 1)$

For maximum or minimum, ${{dy} \over {dx}} = 0$

$ \Rightarrow (x - 1)(3x - 1) = 0 \Rightarrow x = {1 \over 3},1$

Now, ${{{d^2}y} \over {d{x^2}}} = 6x - 4 \Rightarrow {\left[ {{{{d^2}y} \over {d{x^2}}}} \right]_{x = {1 \over 3}}} = - 2 < 0$

$\therefore$ y is maximum when $x = {1 \over 3}$ and maximum value is

${[y]_{x = {1 \over 3}}} = {4 \over {27}}$

Also, ${\left[ {{{{d^2}y} \over {d{x^2}}}} \right]_{x = 1}} = 2 > 0$

$\therefore$ y is minimum when x = 1.

Substitute 0 for y in the equation 4y² = 4x² $-$ x³,

0 = x²(4 $-$ x) $\Rightarrow$ x = 0, x = 4

It means curve makes the loop symmetric about X-axis between 0 and 4.

Area $ = 2\int_0^4 {ydx = {2 \over 2}\int_0^4 {\sqrt {4{x^2} - {x^3}} dx = \int_0^4 {x\sqrt {4 - x} \,dx} } } $

Let $4 - x = t \Rightarrow - dx = dt$

Area $ = - \int_4^0 {(4 - t)\sqrt t \,dt} $

$ = \int_0^4 {(4\sqrt t - t\sqrt t )dt = \left[ {4 \times {2 \over 3}{t^{3/2}} - {2 \over 5}{t^{5/2}}} \right]_0^4} $

$ = {8 \over 3} \times 8 - {2 \over 5} \times 32 = {{64} \over 3} - {{64} \over 5} = 64 \times {2 \over {15}} = {{128} \over {15}}$

We have, $2f(xy) = {(f(x))^x} + {(f(y))^x}$, $x,y \in R$ and $f(1) = a$.

Putting x = 1, y = 2, we get

$2f(2) = {(f(1))^2} + f(2) \Rightarrow f(2) = {a^2}$

Putting x = 1, y = 3, we get

$2f(3) = {(f(1))^3} + f(3) \Rightarrow f(3) = {a^3}$, and so on.

$\therefore$ $f(n) = {a^n}$

$\sum\limits_{k = 1}^n {f(i) = a + {a^2} + .. + {a^n} = {{a({a^n} - 1)} \over {a - 1}}} $.

The given inequality is equivalent to

$1 < |x - 3| + |4 - x|$

i.e., $1 < |x - 3| + |x - 4|$ ...... (i)

For x < 3, we have

$3 - x + 4 - x = 7 - 2x > 1$

i.e., x < 3, which is true in the domain.

For $3 \le x < 4$, Eq. (i) gives

$x - 3 + 4 - x = 1 < 1$ ..... (ii)

For 4 $\le$ x, we have

$2x - 7 > 1 \Rightarrow x > 4$ ...... (iii)

From Eqs. (i) and (iii), we have the solution of the inequality as x < 3 or x > 4.

i.e., $x \in ] - \infty ,3) \cup (4,\infty $ [which is equal to R $-$ [3, 4].

Here, ${{2x - 1} \over {{x^3} + 4{x^2} + 3x}} \in R$ only when ${x^3} + 4{x^2} + 3x \ne 0$

$ \Rightarrow x({x^2} + 4x + 3) = 0$

$ \Rightarrow x(x + 1)(x + 3) = 0$

$ \Rightarrow x = 0, - 1, - 3$

$\therefore$ $\left\{ {x \in R:{{2x - 1} \over {{x^3} + 4{x^2} + 3x}} \in R} \right\}$

$ = R - \{ 0, - 1, - 3\} $

Given, ${{|x - 2|\, - 1} \over {|x - 2|\, - 2}} \le 0$

Let $|x - 2|\, = k$

Then, given equation,

${{k - 1} \over {k - 2}} \le 0 \Rightarrow {{(k - 1)(k - 2)} \over {{{(k - 2)}^2}}} \le 0$

$ \Rightarrow (k - 1)(k - 2) \le 0 \Rightarrow 1 \le k \le 2$

$ \Rightarrow 1 \le |x - 2|\, \le 2$

Case I When $1 \le |x - 2|$

$ \Rightarrow |x - 2| \ge 1$

$ \Rightarrow x - 2 \ge 1$ or $x - 2 \le - 1$

$ \Rightarrow x \ge 3$ and $x \le 1$ ..... (i)

Case II When $|x - 2| \le 2$

$ \Rightarrow - 2 \le x - 2 \le 2$

$ \Rightarrow - 2 + 2 \le x \le 2 + 2$

$ \Rightarrow 0 \le x \le 4$ ..... (ii)

From Eqs. (i) and (ii), $x \in [0,1] \cup [3,4]$.

We have, $f(x) = {x \over {\sqrt {1 + {x^2}} }}$

$ \Rightarrow f(f(x)) = {{f(x)} \over {\sqrt {1 + {{(f(x))}^2}} }} = {{{x \over {\sqrt {1 + {x^2}} }}} \over {\sqrt {1 + {{{x^2}} \over {1 + {x^2}}}} }} = {x \over {\sqrt {1 + 2{x^2}} }}$

Similarly, $f(f(f(x))) = {x \over {\sqrt {1 + 3{x^2}} }}$

$\underbrace {fofofo.....of(x)}_{x\,times} = {x \over {\sqrt {1 + n{x^2}} }} = {x \over {\sqrt {1 + \left( {\sum\limits_{r = 1}^n 1 } \right){x^2}} }}$