Class 11th

Kindly go through the solution

Let A have the co-ordinate (x, o)

By laws of reflection

∠PAB = ∠ QAB = θ

And ∠ CAQ + θ = 90°

As normal is ⊥ to surface (x-axis)

⇒ ∠CAQ = 90° - θ

and ∠CAP = 90° + θ

73. The given eqⁿ of limes are.

9x + 6y – 7 = 0 ______ (1)

3x + 2y + b = 0 ______ (2)

Let P (x₀, y₀) be a point equidistant from (1) and (2) so

9x₀ + 6y - 7 = ± 3 (3x₀ + 2y₀ + 6)

When, 9x₀ + 6y₀ – 7 = 3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ - 7 = 9x₀ + 6y₀ + 18

⇒ - 7 = 18 which in not true

So, 9x₀ + 6y₀ - 7 = -3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ -7 = -9x₀ -6y₀ -18

⇒ 18x₀ + 12y₀ + 11= 0.

Hence, the required eqn of line through (x₀, y₀) & equidistant  from parallel line 9x + 6y - 7 = 0

and 3x + 2y + 6 = 0 is 18x + 12y + 11 = 0.

72. The given eqⁿ of the lines are.

x + y ? 5 = 0 _______ (1)

3x ? 2y + 7 = 0 ______ (2)

Given, sum of perpendicular distance of P (x, y)  from the two lines is always 10 .

The above eqⁿ can be expressed as a linear combination Ax + By + C = 0 where A, B & C are constants representing a straight line

P (x, y) mover on a line.

71. 

36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

f(11)=11 [? prime factor of 11 = 11]

f(12)=3 [? prime factor of 12 = 2, 3]

f(13)=13 [? prime factor of 13 = 13]

?Range of f=set of all image of f(x) = {3,5,11,13}.

35. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 * 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b $\in$ z

So, ab=(–1) * (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1, –2).

?The element 1 has two image i.e., 2 and –2.

Hence, f is not a function.

34. Given,

A={1,2,3,4}

B={1,5,9,11,15,16}

f={(1,5),(2,9),(3,1),(4,5),(2,11)}.

(i) As every element of f is an element of A * B

We can clearly say that f $\subset$ A * B.

?f is a relation from A to B.

(ii) As the element 2 of the domain has two image i.e., 9 and 11. f is not a function.

69. 

The given eqⁿ of the lines are.

4x + 7y + 5 = 0______ (1)

2x - y = 0 ______ (2)

Solving (1) and (2) we get,

4 x + 7 (2 x)+5 = 0

4x +14 x + 5= 0

x = $\frac{-5}{18}$

and y = 2x = $2\left(\frac{-5}{18}\right)=\frac{-5}{9}$