Class 11th

27. Given, f (x)= $\frac{x^2+2x+1}{x^2-8x+12}$

The given function is valid if denominator is not zero.

So, if x² – 8x+12=0.

⇒ x² – 2x – 6x+12=0

⇒ x (x – 2) –6 (x – 2)=0

⇒ (x – 2) (x – 6)=0

⇒ x=2 and x=6.

So,  f (x) will be valid for all real number x except x=2,6.

∴ Domain of f (x)=R – {2,6}

61. The given Eqⁿ of the line is $\frac{x}{4}+\frac{y}{6}$ = 1 ______ (1)

so, Slope of line = -$\frac{6}{4}=-\frac{3}{2}.$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{-1}{(-3/2)}=\frac{2}{3}.$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then,  $\frac{0}{4}+\frac{y}{6}=1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4}+\frac{y}{6}=1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 0

26. Given, f(x)=x².

$\frac{f(1.1)-f(1)}{1.1-1}=\frac{{(1.1)}^2-1^2}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

60. The given eqn of lines are

x - 7y + 5 = 0 ______ (1) ⇒ x = 7y - 5

and 3x + y = 0 _________ (2)

Solution (1) and (2) we get,

3 [7y – 5] + y = 0 .

⇒ 21y - 15 + y = 0

⇒ 22y = 15

25. Given, f(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 3\hfill \\ 3x,\hfill & 3\le x\le 10\hfill \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f  has one and only one image.

 ? f(x) is a function.

Given, g(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 2\hfill \\ 3x,\hfill & 2\le x\le 10\hfill \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the element 2 of the domain has more than one image i.e., 4 and 6.  

? g(x) is not a function.

24. (i) f(x)=2 – 3x, x $\in$ R, x>0.

Given, x>0

3x>3 * 0

3x>0

(–1) * 3x<(1) * 0.

–3x<0

2 – 3x<0+2

2 – 3x<2

i.e., f(x) < 2

Hene, range of f(x) = (– ?, 2)

(ii) Given, f(x) = x²+2, x is a real number.

Since, x is a real number,

x² ≥ 0 (x²=0 for x=0)

x²+2 ≥ 0+2

x²+2 ≥ 2

f(x) ≥ 2

?Range of f(x) = [2, ?) 

(iii) Given, f(x) = x, x is a real number.

As, f(x) = x, the range of f(x) is also real.

i.e., Range of f(x) = R.

Kindly go through the solution

58. Let (0, y) be the point on y-axis which is at a distance 4 unit from the line $\frac{x}{3}+\frac{y}{4}=1$

Then, the line $\frac{x}{3}+\frac{y}{4}=1$

4x + 3y = 12.

4x + 3y - 12 = 0

57. Let a and b be the x & y intercept. Then,

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

Given, a + b = 1. ______ (2) b = 1 -a _____ (3)

and ab = -6 _____ (4)

Putting eqn (B) in (iii) we get a

a (1- a) = - 6

a - a² = - 6

a² - a - 6 = 0

a² + 2a - 3a - 6 = 0

a (a + 2) - 3 (a + 2) = 0

(a + 2) (a -3) = 0

(a + 2) (a -3) = 0.

a = 3 or a = -2.

When a = 3, b = 1- a = 1 - 3 = - 2

When a = - 2, b = 1 - (-2) = 1 + 2 = 3

So, (a, b) = (3, -2) and (-2, 3)

Hence, eqⁿ (1) becomes,

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

2x – 3y = 6 and 2y - 3x = 6

Gives the read eqⁿ of lines