Continuity and Differentiability MCQs

Ans: (d) 43

Explanation: The given function is f(x) = 5x + 8, and its value at the point x = 7 is f(7) = 43.

Let us find the limit of the function at the point x = 7.

lim_x→7f(x)=f(7)

lim_x→7(5x+8)

=5(7)+8

=35+8

=43

Ans: (a) \(lim_{x→b}f(x)\)=f(b).

Explanation: A function is said to be continuous at point x = b if it satisfies the following:

if \(lim_{x→b}f(x)\)=f(b).

Ans: (b) Mass

Explanation: The continuity equation measures the rate at the mass of a substance that will leave the system to the mass of the substance that will enter the system. The process will take place at a steady rate.

Ans: (b) f′(c) = 0

Explanation: When a function becomes continuous and differentiable at such that f(a) = f(b) then in such case c value between (a,b) will be represented as f′(c) = 0.

Ans: (a) 1

Explanation: The function needs to meet the following requirements in order to use Rolle's theorem:

f(a) = f(b).

f(x) = x³ - 3x in the interval [0,√3]

Being a polynomial function, f'(x) = 3x² - 3 is differentiable everywhere.

f(√3) = (√3)³ - 3(√3) = 0 and f(0) = 0³ - 3(0) = 0.

Rolle's theorem states that f'(c) = 0 for at least one value c in the open interval (0, √3).

Now let's determine f'(x):

f'(x) = 3x²-3

Now, let's find c such that f'(c) = 0

⇒ 3c²-3=0

⇒ 3c²=3

⇒ c²=1

⇒ c= ±1

Since c must lie in the open interval (0, √3), c = 1.

So, the value of c in Rolle's theorem for the function f(x) = x³ - 3x in the interval [0, √3] is c = 1

Ans: (a) 68

Explanation: The given function is f(x) = 10x + 8, and its value at the point x = 6 is f(6) = 68.

Let us find out the value of the given equation by putting the value of x in the given function.

lim_x→6f(x) = 10x + 8

lim_x→6f(6) = (10 x 6+8)

= 60+8

= 68

Ans: (c) 6

Explanation: The given function is \(f(x) =\frac{(x^2 - 2x) }{ (x-2)}\), and its value at the point x = 6 is f(6) = 6.

To find out the continuity of the given function at the point x=6

lim_x→6f(x) = (x² - 2x) / (x-2)

lim_x→6f(6) = (6² - 2 × 6) / (6-2)

= (36 - 12)/4

= 24/4

= 6

Ans: (a) \(lim_{x→b}f(x)\)=(f(b+h)−f(b))/h

Explanation: A function is said to be differentiable at point x = b if it satisfies the following:

if \(lim_{x→b}f′(b)\)=(f(b+h)−f(b))/h

Ans: (d) \(\frac{4t}{3}\)

Explanation: x=t³

⇒\(\frac{dx}{dt}\) = 3t²

y=t⁴

⇒dy/dt = 4t³

⇒dy/dx = 4t³/3t² 

= 4t/3

Hence, d²y/dx² = 4t/3

Answer: (c) 12a + 4

Explanation: Given that, y=ax³+b²

Then dy/dx = 3ax² + 2b

At x=2, \(\frac{dy}{dx}\) = 3(a)(2)² + 2(2)

12a + 4

Ans: (a) 2 option is correct

Explanation: A function is said to be continuous and differentiable if it satisfies the below conditions:

if \(lim_{x→b}f(x)=f(b)\).

if \(lim_{x→b}f′(b)=\frac{(f(b+h)−f(b))}{h}\)

So, now solve it for x = 2

\(lim_{x→b}f(x)\)=f(b) = \(lim_{x→2-}f(x)\) = \(lim_{x→2-}\) - (x-2) = 0

\(lim_{x→b}f′(b)=\frac{(f(b+h)−f(b))}{h} \\= lim_{x→0}f′(0)=\frac{(f(0+0)−f(0))}{0} = 0\)

Ans: (c) t

Explanation: x = 2t³ ⇒\(\frac{dx}{dt}\) = 3=6t²

y=3t⁴⇒\(\frac{dy}{dt}\) = 12t³

⇒\(\frac{dy}{dx}\) = \(\frac{12t^3}{6t^2}\) = t

Hence, \(\frac{d^2y}{dx^2}\) = t

Answer: (d) 81a + 30

Explanation: Given that, y=3ax³+5b²

Then \(\frac{dy}{dx}\) = 9ax² + 10b

At x=3, \(\frac{dy}{dx}\) = 9(a)(3)² + 10(3)

81a + 30

Ans: (a) Rocket

Explanation: Rocket is the application of continuity and differentiability as it requires a large amount of force to push the rocket higher in the sky. The force applied is known as thrust.

Ans: (a) 13

Explanation: The given function is \(f(x) =\frac{(3x^2 - 2x) }{ (x)}\), and its value at the point x = 5 is f(5) = 13.

Let us find the limit of the function at the point x = 5.

\(lim_{x→5}f(x)\)

\(lim_{x→5}\)\(\frac{(3x^2 - 2x) }{ (x)}\)

\(=\frac{3(5)^2- 2(5)}{5}\\=f(5)\)

Ans: (c) 2 cot² x cos x

Explanation: x = 2sin³x

⇒\(\frac{dx}{dt}\) = 6sin²x

y=3cos⁴

x⇒\(\frac{dy}{dt}\) = 12cos³ x

⇒\(\frac{dy}{dx}\) = 12cos³ x/6sin²x

= 2 cot² x cos x

Hence, \(\frac{d^2y}{dx^2}\) =2 cot² x cos x

Ans: (a) f’(c) = \(\frac{[f(b) - f(a)]}{ (b - a)}\)

Explanation: The mean value theorem states that if a function is defined at closed interval [a, b] and it is both continuous and differentiable at c. So the value of c at closed interval [a,b] is given as: f’(c) = \(\frac{[f(b) - f(a)]}{ (b - a)}\)