Maths

55. We have (k - 3) x - (4 - k²) y + k² - 7 y + 6 = 0.

(i) When the line is parall to x-axis, all x coefficient = 0. then,

(k - 3)x - (4 -k²)y + k² - 7y + 6 = 0 x.x - a x y       where a = constant

Equating the co-efficient,

K – 3 = 0

=> k = 3

(ii) When the line is parallel to y-axis all y co-efficient = 0 then

- (4 -k)² = 0

=> – 4 + x² = 0

k² = 4

k = ± 2.

(iii) When the line pares through origin, (0, 0) need satisfy the given eqn then,

k² - 7k + 6 = 0

k² - k – 6k + 6 = 0

k (k- 1) - 6 (k - 1) = 0

(k = 1) (k - 6) = 0

k = 1 and k = 6

It is given that $f:R-\left\{-\frac{4}{3}\right\}\to R$ is defined as $f\left(x\right)=\frac{4x}{3x-4}$

Let y be an arbitrary element of Range f.

Then, there exists $x\in R-\left\{-\frac{4}{3}\right\}$ such that $y=f\left(x\right)$

$\Rightarrow y=\frac{4x}{3x+4}$

$\Rightarrow 3xy+4y=4x$ let us define g: Range $f\to R-\left\{-\frac{4}{3}\right\}as,g\left(y\right)=\frac{4y}{4-3y}$

$\begin{array}{l}\Rightarrow x\left(4-3y\right)=4y\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(\frac{4x}{3x+4}\right)\\ \Rightarrow x=\frac{4y}{4-3y}\\ =\frac{4\left(4\frac{x}{3x+4}\right)}{4-3\left(\frac{4x}{3x+4}\right)}=\frac{16x}{12x+16-12x}=\frac{16x}{16}=x\\ And,\left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{4y}{4-3y}\right)\\ =\frac{4\left(\frac{4y}{4-3y}\right)}{3\left(\frac{4y}{4-3y}\right)+4}=\frac{16y}{12y+16-12y}=\frac{16y}{16}=y\\ \therefore gof=I_{R-\left(-\frac{4}{3}\right)}and,fog=I\_"Range,f"\end{array}$

Thus, g is the inverse of f i.e., $f^{-1}=g.$

Hence, the inverse of f is the map g: Range $f\to R-\left\{-\frac{4}{3}\right\}$ which is given by

$g\left(y\right)=\frac{4y}{4-3y}$

The correct answer is B.

$f:R\to R$ is given as $f\left(x\right)={\left(3-x^{\wedge }3\right)}^{\wedge }\left(1/3\right)$

$\begin{array}{l}f\left(x\right)={\left(3-x^3\right)}^{\frac{1}{3}}\\ \therefore fof\left(x\right)=f\left(f\left(x\right)\right)=f\left({\left(3-x^3\right)}^{\frac{1}{3}}\right)={\left[3-{\left({\left(3-x^3\right)}^{\frac{1}{3}}\right)}^3\right]}^{\frac{1}{3}}\\ ={\left[3-\left(3-x^3\right)\right]}^{\frac{1}{3}}={\left(x^3\right)}^{\frac{1}{3}}=x\\ \therefore fof\left(x\right)=x\end{array}$

The correct answer is C.

Let f: X → Y be an invertible function. 

Then, there exists a function g: Y → X such that gof = IX and fog = IY. 

Here, f⁻¹ = g. 

Now, gof = IX and fog = IY

⇒ f⁻¹ of = IX and fof⁻¹ = IY

Hence, f⁻¹ : Y → X is invertible and f is the inverse of f⁻¹ 

i.e., (f−1)⁻¹  = f.

$f\left(1\right)=a,f\left(2\right)b,and,f\left(3\right)=c$

If we define $g:\left\{a,b,c\right\}\to \left\{1,2,3\right\}as,g\left(a\right)=1,g\left(b\right)=2,g\left(c\right)=3,$ then we have:

  $\begin{array}{l}\left(fog\right)\left(a\right)=f\left(g\left(a\right)\right)=f\left(1\right)=a\\ \left(fog\right)\left(b\right)=f\left(g\left(b\right)\right)=f\left(2\right)=b\\ \left(fog\right)\left(c\right)=f\left(g\left(c\right)\right)=f\left(3\right)=c\\ And\\ \left(gof\right)\left(1\right)=g\left(f\left(1\right)\right)=f\left(a\right)=1\\ \left(gof\right)\left(2\right)=g\left(f\left(2\right)\right)=f\left(b\right)=2\\ \left(gof\right)\left(3\right)=g\left(f\left(3\right)\right)=f\left(c\right)=3\\ \therefore gof=I_{X}and,fog=I_Y\\ Where,X=\left\{1,2,3\right\},and,Y=\left\{a,b,c\right\}.\end{array}$

Thus, the inverse of f exists and $f^{-1}=g.$

$\therefore f^{-1}:\left\{a,b,c\right\}\to \left\{1,2,3\right\}$ is given by,

$\begin{array}{l}{f}^{-1}\left(a\right)=1,f^{-1}\left(b\right)=2,f^{-1}\left(c\right)=3\\ \end{array}$

Let us now find the inverse of $f^{-1}$ i.e., find the inverse of g.

If we define $h:\left\{1,2,3\right\}\to \left\{a,b,c\right\}as$

$h\left(1\right)=a,h\left(2\right)=b,h\left(3\right)=c$ , then we have

$\begin{array}{l}\left(goh\right)\left(1\right)=g\left(h\left(1\right)\right)=g\left(a\right)=1\\ \left(goh\right)\left(2\right)=g\left(h\left(2\right)\right)=g\left(b\right)=2\\ \left(goh\right)\left(3\right)=g\left(h\left(3\right)\right)=g\left(c\right)=3\\ And\\ \left(hog\right)\left(a\right)=h\left(g\left(a\right)\right)=h\left(1\right)=a\\ \left(hog\right)\left(b\right)=h\left(g\left(b\right)\right)=h\left(2\right)=b\\ \left(hog\right)\left(c\right)=h\left(g\left(c\right)\right)=h\left(3\right)=c\\ \therefore goh=I_{X}and,hog=I_Y\\ Where,X=\left\{1,2,3\right\},and,Y=\left\{a,b,c\right\}.\end{array}$

Thus, the inverse of g exists and $g^{-1}=h\Rightarrow {\left(f^{-1}\right)}^{-1}=h.$

It can be noted that h=f.

Hence, ${\left(f^{-1}\right)}^{-1}=f.$

54. The equation of line whose intercept on axes are a and b is given by,

$\frac{x}{a}+\frac{y}{b}=1.$

Multiplying both sides by ab we get,

$\frac{abx}{a}+\frac{aby}{b}=ab$

$\Rightarrow bx+ay-ab=0.$

Let $f:X\to Y$ be an invertible function.

Also, suppose f has two inverses (say g₁ and g₂ ).

Then, for all y ∈ Y, we have:

$\begin{array}{l}fo{g}_1\left(y\right)=I_y\left(y\right)=fo{g}_2\left(y\right)\\ \Rightarrow f\left(g_1\left(y\right)\right)=f\left(g_2\left(y\right)\right)\end{array}$  [f is invertible => f is one-one]

$\Rightarrow g_1=g_2$  [g is one-one]

Hence, f has a unique inverse.

$f:R_{+}\to [4,\infty )$ is given as $f\left(x\right)=x^2+4$ .

One-one:

$\begin{array}{l}Let,f\left(x\right)=f\left(y\right).\\ \Rightarrow x^2+4=y^2+4\\ \Rightarrow x^2=y^2\\ \Rightarrow x=y\left[as,x=y\in R\right]\end{array}$

$\therefore f$ is a one-one function.

53. Let P be the point on the BC dropped from vertex A.

Slope of BC$=\mathrm{2\; -}$
$\frac{\mathrm{(-1)}}{1\; -4}$

$=\frac{2+1}{-3}$

$=\frac{3}{-3}$

= $-$ 1.

As A P $\perp$ BC,

Slope of AP= $\frac{-1}{\mathrm{slope\; of\; B}C}=\frac{-1}{-1}=1.$

Using slope-point form the equation of AP is,

$1=\frac{y-3}{x-2}$

$\Rightarrow$ x $-$ 2 = y $-$ 3

$\Rightarrow$ x – y – 2 + 3 = 0 $\Rightarrow$ x – y + 1 = 0

The equation of line segment through B(4, -1) and C(1, 2) is.

$y-(-1)=\frac{2-(-1)}{1-4}(x-4).$

$\Rightarrow y+1=\frac{2+1}{-3}(x-4)$

$\Rightarrow (y+1)=\frac{3}{-3}(x-4).$

$\Rightarrow y+1=-(x-4)$

$\Rightarrow xy+1=-x+4$

$\Rightarrow -x+y+1-4=0$

$\Rightarrow$ $x+y-3=0$

So, A=1, B=1 and C= $-$ 3.

Hence, length of AP=length of $\perp$ distance of A(2,3) from BC.