JEE Advanced Level Test: Relations and Functions- 4 - Airforce X Y / Indian Navy SSR MCQ

f (x) = |x- 1| is not diff at x = 1 critical point is at x = 1

Min. value = f(1)
Max. Value = f(-1)

Let y = (1/x)^x
Taking ln both sides, we get ln y =x ln 1/x = x ln (x)^-1 
= -x ln x {Since,ln x^a = a ln x  }
Now, differentiating both sides with respect to x we get,
1/y(dy/dx) = -1-ln x
⇒  (dy/dx) =(1/x)^x [-1-ln x} = 0 {Condition for y to be maximum}

⇒ln x = -1  {Since, (1/x)^x = 0}

⇒ x = e^-1  {Converting logarithmic equation to exponential equation}

⇒ So,  y(max) = (e)^1/e

Given 4x + 2πr = 2

For minimum value of area A



x = 2r

f '(x) = 20x⁴ -100x³ + 120x² f '(x) = 0
x = 2, 3 one maximum and one minimum

f (x) = x⁴ - 62x² + ax + 9
f '(x) = 4x³ -124x + a
f '(1) = 0 at x =1 4 -124 + a = 0
x = 120

f (x) = (PA)² = (x - a)² +2x - 2x²
f '(x) = 0x = (1-a)
 

Let Z=(x-3)²+y² = x²-6x+9+x⁴
 Z is minimum dz/dz = 2x-6+4x³ = 0
∴ x = 1 ∴ y = 1, (1,1)

z+x=k  z+zsinθ = k   z = k/1+sinθ

 

Let f (x) = ax+ b be the required liner function.
If a > 0 then f is increa sin g, f(-1) = 0,f(1) = 2 ⇒ -a + b = 0,a +b = 2 ⇒ a = -1,b = 1
If a < 0 then f decrea sin g, f(-1) = 2,f (1) ⇒ = 0 -a + b = 2,a+b = 0 = ⇒ a = -1,b = 1
∴ The required linear function are f(x) = x+1,f(x) = -x+1. 

(cos²x + sin⁴x) - (sin²x + cos⁴x) = (cos²x - sin²x) - (cos⁴x - sin⁴x)
= cos²x - cos²x = 0 

Givenf (x) = f(x-2)+f(x-3)
∴ f(3) = f(1)+f(0) = 1 + 0 = 1,
f(4) = f(2)+f(1) = 2 + 1 = 3
f(5) = f(3)+f(2) = 1+ 2 = 3,f(6). f (4) + f(3) = 3+1 = 4,
f(7) = f(5)+f(4) = 3+3 = 6,f(8) = f(6)+f(5) = 4+3 =7,
f(9) = f(7)+f(6) = 6+4 = 10.

cos 9x + cos (-10) x = cos 9x + cos10x

f (0) = cos0 + cos0 = 2,f (π/4) = cos9 π/4 + cos5π/2 = cos π/ 4 =
f (π/2) = cos9π/2 + cos5π = 0 -1 = -l,f (π) = cos9π + cos10π = - l +1 = 0.

f (x) f (1/x) = f (x) + f (1/x) ⇒ f(x) = xⁿ +1. 
f (2) = 33 ⇒ 2ⁿ +1= 33 ⇒ 2ⁿ = 32 = 2⁵ ⇒ n = 5 ⇒ f (x) = x⁵ +1.

f(x+y) + f(x) + f(y) ⇒ f(x) = a^x.f(3)
= 125 ⇒ a³ = 125 ⇒ a = 5.
∴ f (x) = 5^x. 

f (x) = x² -4x +5 = (x -2)² +1
x ∈ [2,∞) ⇒ x > 2 ⇒ x - 2 > 0 ⇒ (x - 2)² +1 > 1 ⇒ B = [1,∞). 

 is a constant function ⇒ bc ad 0 ⇒ ad bc.

Correct Answer :- c

Explanation : f(n-1)/2 ; n is odd        -n/2 ; n is even

f(1) = 0                f(3) = 1

f(2) = -1               f(4) = -2

f(5) = 2                f(6) = -3

f(n) + f(n+1) = -1    n is odd

if = 0     n is even

Range of f(n) = Z

f(n) is one-one, therefore f(n1) = f(n2)

(n₁-1)/2 = -n₂/2

=> 1 < (n1 + n2)/2 is not equal to 1/2

Therefore f(n) is onto.

Graph of y = f (x) is given in the figure.
∴ f is bijective.

Domain = {1, 3, 2}

∴ f is an even function.

If x > 0 then f x = 0. If x < 0 then f (x) = tanh x. If f (x) = -1then
ex - e^|x| = - e^x - e^|x| ⇒e^x = 0 ⇒ x ∉ R. 
Range does not contain -1.
∴ Range = (-1, 0].

log (x - 3) (5 - x) exists ⇒ (x - 3) (5 - x) > 0
⇒ (x -3)(x -5) < 0 ⇒ 3< x < 5
∴ Domain = (3, 5).

log₁₀ (x³ - x)is defmed ⇒ x³ - x > 0 ⇒ x(x² - 1) > 0 ⇒ (x +1)x(x - 1) > 0
x < -1 ⇒ (x+1)x(x-1) < 0; -1 < x < 0 ⇒ (x+1)x(x-1) > 0
0 < x < 1 ⇒ (x+1)x(x-1) < 0; x > 1 ⇒ (x+1)x(x-1) > 0.
∴ Domain = (-1,0)∪(1, ∞)