Test Level 1: Functions - 2 - CAT MCQ

F(2) = -2 and F1(-2) = -2
F(-2) = 0 and F1(2) = 0
So, according to this,
F(x) = F1(-x) or F1(x) = F(-x)

x³ + y³ = (x + y)³ - 3xy (x + y)
= (x + y) [(x + y)² - 3xy]
= h(x) [ {h(x)}² - 3g(x) ]

Given function is f(x) = a^3x
As point  lies on the given curve, it must satisfy the given equation.

a = 7

Here, y = f(x)

Which is defined for y ∈ R - {1}
Hence f(x) is invertible.
∴ One - one and onto

f(x) = x³ - 4x + p
f(0) = p
f(1) = 1 - 4 + p = p - 3
∵ f(0) and f(1) have opposite signs
∴ f(0) and f(1) < 0
 p(p - 3) < 0
 0 < p < 3

This shows that root of equation are real and equal. Thus, the graph touches the x-axis and is hence tangential to the x-axis.
Thus, answer option 3 is correct.

Given: f(x) = 3x - 5 and f(g(x)) = 2(x)
f(g(x)) = 3g(x) - 5 = 2x
3g(x) - 5 = 2x
g(x) = (2x + 5)/3

f(a, b) = a² + b³
g(a, b) = a + b
g(3, 4) = 7
f(3, g(3, 4)) = f(3, 7) = 3² + 7³ = 352

It is given that y = log₁₀x & y = 1/x
Draw their graphs approximately and check.

Here, we see the graph has only 1 intersection point.

 f (x) = x² - 12x + 27 = x² - 12x + 36 - 9
= (x - 6)² - 9.
Since (x - 6)² ≥ 0, the minimum value of the expression is - 9.

Let x be the radius and y be the volume of the balloon.
Then, y = 4/3πx³  dy/dx = 4πx²
 = 4π(7)² = 196π cm² 
Hence, the volume is increasing with respect to its radius at the rate of 196  π cm², when the radius is 7 cm.

Let the numbers be x and 12 - x.
∴ Sum of squares = x² + (12 - x)² = f(x), say.
∴ f` (x) = 2x + 2 (12 - x) (-1) = 4(x - 6)
∴ f`(x) = 0, only when x = 6.
The numbers are 6 and 6.

The sum of squares of two numbers is minimum when the numbers are equal.

sq(5, 10) = [R(5, 10)]

dr/dt = 3, we have A = πr² ⇒ dA/dt = 2 π r (dr/dt)
⇒ (dA/dt)_{r = 10} = 2π x 10 x 3 = 60 π cm²/s

Let z = 3y³ + 4y² - 6y + 7
Differentiating both sides w.r.t. y,

= 9y² + 8y - 6 + 0 = 9y² + 8y - 6

y – axis by definition. 

Origin by definition. 

The minimum value of the function would occur at the minimum value of (x² – 2x + 5) as this quadratic function has imaginary roots.
For
y = x² – 2x + 5
dy/dx = 2x – 2 = 0 ⇒ x = 1
⇒ x² – 2x + 5 = 4.