If Δ 1 = x s i n θ c o s θ - s i n θ - x 1 c o s θ 1 x and Δ 2 = x s i n 2 θ c o s 2 θ - s i n 2 θ - x 1 c o s 2 θ 1 x , x ≠ 0 ; then for all θ ∈ 0 , π 2 :

Δ 1 + Δ 2 = - 2 ( x 3 + x - 1 )

Δ 1 - Δ 2 = x ( c o s 2 θ - c o s 4 θ )

If A = 1 2 4 2 then show that | 2 A | = 4 | A | .

Given A = 1 2 4 2 ⇒ A = 1 × 2 - 4 × 2 = - 6 ⇒ 4 A = 4 - 6 = - 24 Now 2 A = 2 1 2 4 2 = 2 4 8 4 2 A = 2 × 4 - 8 × 4 = - 24 Hence proved.

Show that the lines x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 are coplanar.

Consider any two points A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) and consider two vectors a → and b → having direction ratios as ( a 1 , b 1 , c 1 ) and ( a 2 , b 2 , c 2 ) respectively. Then the condition for them to be coplanar is A B → . ( a → × b → ) = 0 The two lines given in the question are x + 3 - 3 = y - 1 1 = z - 5 5 and x + 1 - 1 = y - 2 2 = z - 5 5 . These are in the form of x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 . So, on comparing all the three lines, we get The points for the first line, x 1 = - 3 , y 1 = 1 , z 1 = 5 , a 1 = - 3 , b 1 = 1 , c 1 = 5 and Similarly, for the second line, x 2 = - 1 , y 2 = 2 , z 2 = 5 , a 2 = - 1 , b 2 = 2 , c 2 = 5 . Now the direction ratios of the vector A B → = ( - 1 - - 3 ) , ( 2 - 1 ) , ( 5 - 5 ) = 2 , 1 , 0 . Now applying the condition for coplanarity, we get x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 2 1 0 - 3 1 5 - 1 2 5 = 2 5 - 10 - 1 - 15 + 5 + 0 = 2 × - 5 - 1 × - 10 = 2 × - 5 - 1 × - 10 = - 10 + 10 = 0 Since the condition is satisfied, the two lines are coplanar.

Find the order and the degree of the differential equation: x 2 d 2 y d x 2 = 1 + d y d x 2 4 .

Given differential equation is x 2 d 2 y d x 2 = 1 + d y d x 2 4 ⇒ x 2 d 2 y d x 2 1 = 1 + d y d x 2 4 We know that, The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. Hence, Order = 2 and Degree = 1 .

Find g o f and f o g when f : R → R and g : R → R is defined by f ( x ) = x 2 + 2 x + 3 and g ( x ) = 3 x − 4 .

Given f ( x ) = x 2 + 2 x + 3 and g ( x ) = 3 x - 4 Therefore, f o g x = f g x = f 3 x - 4 = 3 x - 4 2 + 2 3 x - 4 + 3 = 9 x 2 - 24 x + 16 + 6 x - 8 + 3 = 9 x 2 - 18 x + 11 And g ∘ f x = g f x = g x 2 + 2 x + 3 = 3 x 2 + 2 x + 3 - 4 = 3 x 2 + 6 x + 9 - 4 = 3 x 2 + 6 x + 5

If g : R → R be two functions defined as f x = | x | + x and g x = | x | - x , ∀ x ∈ R . Then find f o g and g o f . Hence find f o g ( – 3 ) , f o g ( 5 ) and g o f ( – 2 ) .

Given : f ( x ) = | x | + x and g ( x ) = | x | - x , ∀ x ∈ R To find: f o g , and g o f We know that f o g = f ( g ( x ) ) = | g ( x ) | + g ( x ) = | | x | − x | + ( | x | − x ) . . . . i [Since g x = x - x ] Now, we know that x = x if x ≥ 0 - x if x < 0 Thus, for x ≥ 0 f o g x = x - x + x - x = 0 [from i ] For x < 0 , f o g x = - x - x + - x - x = - 4 x [From i ] Therefore, f o g = f g x = 0 x ≥ 0 - 4 x x < 0 . . . . . . i i Similarly, g o f = g ( f ( x ) ) = | f ( x ) | − f ( x ) = | | x | + x | − ( | x | + x ) [Since f x = x + x ] g f x = x + x - x + x = 0 for x ≥ 0 - x + x - - x + x = 0 for x < 0 Therefore, g ( f ( x ) ) = g o f = 0 ∀ x Now, f o g ( − 3 ) = ( 4 ) ( − 3 ) = − 12 (Since from i i , f o g = 4 x f o r x < 0 ) f o g ( 5 ) = 0 (since, f o g = 0 f o r x ≥ 0 ) g o f ( − 2 ) = 0 (since, g o f = 0 ∀ x )

Let f : R → R and g : R → R be defined by f ( x ) = x + 1 and g ( x ) = x − 1 . Show that f o g = g o f = I R .

Given, f : R → R and g : R → R ⇒ f o g : R → R and g o f : R → R (Also, we know that I R : R → R ) So, the domains of all f o g , g o f and I R are the same. ( f o g ) ( x ) = f ( g ( x ) ) = f ( x − 1 ) = x − 1 + 1 = x = I R ( x ) . . . ( 1 ) ( g o f ) ( x ) = g ( f ( x ) ) = g ( x + 1 ) = x + 1 − 1 = x = I R ( x ) . . . ( 2 ) From ( 1 ) and ( 2 ) , ( f o g ) ( x ) = ( g o f ) ( x ) = I R ( x ) , ∀ x ∈ R Hence , f o g = g o f = I R