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QUESTION: 1

As xx varies from −1 to +3, which one of the following describes the behaviour of the function f(x) = x^{3}–3x^{2 }+ 1?

Solution:

Given x various from –1 to +3

f(x) = x^{3} – 3x^{2} + 1

f′(x) = 3x^{2} – 6x

⇒ f′(−1) = 3(−1) 2 – 6 (−1) = 3 + 6 = 9 > 0,f′(1) = 3 – 6 < 0

f′(0) = 0,f′(2) = 3(4) – 6(2) = 0,

f′(3) = 3x^{2} – 6x = 3(a) – 6(3) = 0

∴ f(x) increases, then decreases and increases again

QUESTION: 2

Given the following statements about a function f : R→R, select the right option:

P: If f(x) is continuous at x = x0, then it is also differentiable at x = x_{0}

Q: If f(x) is continuous at x = x_{0}, then it may not be differentiable at x = x_{0}

R: If f(x) is differentiable at x = x_{0}, then it is also continuous at x = x_{0}

Solution:

If the function is differentiable then always it is continuous but vice – versa is not true. If the function is continuous then it may (or) may not differentiable.

*Answer can only contain numeric values

QUESTION: 3

Given f(x) = x^{3} + ax^{2} + bx + c and y-intercept for the function is 1. Also f has local extrema at x = -4 and x = 2. Then a + b + c = ____

Solution:

(i) As y-intercept is 1, c = 1

(ii) Also at x = −4,f′(x) = 0

3(−4)^{2}+2a(−4)+b+1=0

48 – 8a + b + 1 = 0

49 – 8a + b = 0

(iii) Similarly 3(2)^{2} + 2a(2) + b + 1 = 0

12 + 4a + b + 1 = 0

13 + 4a + b = 0

∴ a = 3, b = -24

QUESTION: 4

Find C of Rolle’s theorem for f(x) = e^{x}(sin x - cos x) in

Solution:

Given function f(x) = e^{x}(sin x - cos x)

f(x) is continuous in 3π/4

f(x) is differentiable in

f(a) = f(b)

According to Rolle’s theorem,

there exists at least one value C ∈ (a,b)

such that f’(c) = 0

f’(c) = 0

f’(x) = e^{x} (sin x – cos x) + e^{x} (cos x + sin x)

f’(c) = e^{c} (sin c – cos c) + e^{c} (cos c + sin c) = 0

⇒ 2 e^{c} sin c = 0

QUESTION: 5

If f(x) is differentiable and g’(x) ≠ 0 such that f(1) = 4, f(2) = 16, f’(x)= 8g’(x) and g(2) = 4 then what is the value of g(1) ?

Solution:

By using Cauchy’s Mean value theorem

QUESTION: 6

An open box is to be made out of square cardboard of 18 cm by cutting offs equal squares from the corners and turning up the sides. What is the maximum volume of the box in cm^{3}?

Solution:

Let x be the corner side which is cut off and v be volume.

v = l × b × h

v = (18 − 2x) × (18 − 2x) × x

v = 4x^{3} − 72x^{2} + 324x

v′ = 12x^{2 }− 144x + 324v′

v′′ = 24x − 144

v′ = 12x^{2} − 144x + 324 = 0

x2 − 12x + 27=0

(x−3) (x−9) = 0

(x=3) or (x = 9)

atx = 3

v′= 24x −144 = 24(3) −144 = −72<0

at x = 3 we have volume maximum

v = 4(3)^{3} − 72(2)^{2} + 324(2)

∴ v = 432

QUESTION: 7

What are the minimum and maximum value of the below-given function respectively?

f(x) = 3x^{3} − 9x^{2} − 27x + 30

Solution:

f(x) = 3x^{3} − 9x^{2} − 27x + 30

f′(x) = 9x^{2} − 18x − 27

f′′(x) = 18x − 18

Points at which maximum or minimum exists

f′(x) = 0

9x^{2} − 18x − 27 = 0

x2 − 2x − 3 = 0

(x − 3) (x + 1 ) = 0

(x = 3) or (x = −1)

Substitute the value in (x = 3) or (x = −1)

f′′(−1) = 18(−1) − 18 = − 36 < 0

∴ at x = −1,maximum value exists

f(−1) = 3(−1)^{3 }−9(−1)^{2 }−27(−1) + 30

f (−1) = 45

f′′ (3) = 18(3) − 18 = 36 > 0

∴ at x = 3, mimimum value exists

f(3) = 3(3)^{3 }− 9(3)^{2 }− 27(3) + 30

f (3) = −51

QUESTION: 8

Find the point at which maximum value of f(x) = 2x3−3x^{2} occurs in interval [-2, 2]?

Solution:

f(x) = 2x^{3} − 3x^{2}

f′(x) = 6x^{2} − 6x

f′′(x) = 12x − 6

f′(x) = 6x^{2} − 6x = 0

6x (x−1) = 0

x = 0 or x = 1

Now check f(x) at x = 0, x = 1 and the end points of given interval since at end point tangent cannot be drawn so, above concept cannot be used to find maximum or minimum value for end points.

QUESTION: 9

If f(x) = x^{3} − 3x−1 is continuous in the closed interval and f’(x) exists in the open interval then find the value of c such that it lies in

Solution:

Since given function is continuous and differentiable then by Lagrange’s Mean-Value Theorem.

3c^{2} − 3 = 0

c^{2} = 1

∴ c = ±1

QUESTION: 10

Find the maximum and minimum values of f(x) = sin x + cos 2x where 0≤ x ≤2π

Solution:

f(x) = sin x + cos 2x in [0, 2π]

f’(x) = cos x – 2 sin 2x = cos x [1 – 4 sinx] = 0

⇒ cos x = 0; 1 – 4sinx = 0

⇒ x = π/2, 3π/2 in [0, 2π]

Sin x = 1/4, Also f’(x) exists for all x in [0, 2π]

Now f(0) = 1; f(2π) = 1; f(π/2) = 0; f(3π/2) = -2

At sinx = 1/4 , f(x) = 1/2 + (1 – 2(1/4)^{2}) = 9/8

Therefore the maximum and minimum values of f(x) are **9/8 and -2 o**ccurred at

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