When A = A^-1 and B is a matrix, then

rank (AB) = rank (B)

f(x) = x³ – 3x + [a]

Let [a] = t (where t will be an integer)

f(x) = x³ – 3x + t ……….(i)

⇒ f ’(x) = 3x² – 3

⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = -1

so f(x) = 0 will have three distinct and real solution when f (1). f(-1) < 0 ……………. (ii)

Now,

f(1) = (1)³ -3(1) + t = t – 2

f(–1) = (–1)³ – 3 (–1) + t = t + 2

From equation (ii)

(t –2) (t + 2) < 0

⇒ t ∈ (-2, 2)

Now t = [a]

Hence [a] ∈ (-2, 2)

⇒ a ∈ [-1, 2)

n^th term of 1^st series =n^th term of 2^nd series

⇒ 3 + (n − 1) 7 = 63 + (n − 1) 2

⇒ (n − 1)5 = 60

⇒ n -1=12

⇒ n = 13

The total number of words that can be formed is 10⁵ and number of these words in which no letters are repeated is ¹⁰P₅.

Hence the required number

= 10⁵ − ¹⁰P⁵

= 100000 − 10 × 9 × 8 × 7 × 6

= 69760

Tangent at (1, − 2) to x² + y² = 5 is

x − 2y − 5 = 0 ... (i).

Centre and radius of

x² + y² − 8x + 6y + 20 = 0 are

C (4, − 3) and radius r = √5.

Perpendicular distance from

C (4, − 3) to (i) is radius.

∴ (i) is also a tangent to the second circle.

Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)
h-4/1 = k=3/-2 = - [1.4 -2. (-3)-5]

∴ (h, k) = (3, −1)

f (4) = x² - 3x

= (4)² - 3(4)

= 16 - 12 = 4

From the first function we can draw the following output chart:

x y

- 1 0

0 1

1 4

2 9

As g(x) is the reflection of f(x) about line y = x, it must satisfy the opposite of the graph, i.e.

x y

0 - 1

1 0

4 1

9 2, which is satisfied only by option 4.

f(0) = cos 0 = 1 and f(2π) = cos (2π) = 1

So, f is not one to one and cos(x) lie between - 1 and 1.

Therefore, range is not equal to its codomain.

Hence, it is not onto.

We can write as 1+ 1/x² + x + 1. After solving, we get third option is correct.

sec (cosec^-1x) = cosec (sec^-1x) = |x|/√x² -1.

So options (2) = (3) = (4); Hence option (1) is different from the other three.