Test: Engineering Mathematics- 5


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10 Questions MCQ Test RRB JE for Computer Science Engineering | Test: Engineering Mathematics- 5

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Test: Engineering Mathematics- 5 - Question 1

As xx varies from −1 to +3, which one of the following describes the behaviour of the function f(x) = x3–3x+ 1?

Detailed Solution for Test: Engineering Mathematics- 5 - Question 1

 

Given x various from –1 to +3
f(x) = x3 – 3x2 + 1

f′(x) = 3x2 – 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) = 3x2 – 6x = 3(a) – 6(3) = 0

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

Test: Engineering Mathematics- 5 - 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 = x0

Q: If f(x) is continuous at x = x0, then it may not be differentiable at x = x0

R: If f(x) is differentiable at x = x0, then it is also continuous at x = x0

Detailed Solution for Test: Engineering Mathematics- 5 - Question 2

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
Test: Engineering Mathematics- 5 - Question 3

Given f(x) = x3 + ax2 + 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 = ____


Detailed Solution for Test: Engineering Mathematics- 5 - Question 3

(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

Test: Engineering Mathematics- 5 - Question 4

 

Find C of Rolle’s theorem for f(x) = ex(sin x - cos x) in

Detailed Solution for Test: Engineering Mathematics- 5 - Question 4

Given function f(x) = ex(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) = ex (sin x – cos x) + ex (cos x + sin x)

f’(c) = ec (sin c – cos c) + ec (cos c + sin c) = 0

⇒ 2 ec sin c = 0


Test: Engineering Mathematics- 5 - 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) ?

Detailed Solution for Test: Engineering Mathematics- 5 - Question 5

By using Cauchy’s Mean value theorem

Test: Engineering Mathematics- 5 - 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 cm3?

Detailed Solution for Test: Engineering Mathematics- 5 - Question 6

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 = 4x3 − 72x2 + 324x

v′ = 12x− 144x + 324v′

v′′ = 24x − 144

v′ = 12x2 − 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

Test: Engineering Mathematics- 5 - Question 7

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

f(x) = 3x3 − 9x2 − 27x + 30

Detailed Solution for Test: Engineering Mathematics- 5 - Question 7

f(x) = 3x3 − 9x2 − 27x + 30

f′(x) = 9x2 − 18x − 27

f′′(x) = 18x − 18
Points at which maximum or minimum exists
f′(x) = 0

9x2 − 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)−9(−1)−27(−1) + 30

f (−1) = 45

f′′ (3) = 18(3) − 18 = 36 > 0

∴ at x = 3, mimimum value exists

f(3) = 3(3)− 9(3)− 27(3) + 30

f (3) = −51

Test: Engineering Mathematics- 5 - Question 8

Find the point at which maximum value of f(x) = 2x3−3x2 occurs in interval [-2, 2]?

Detailed Solution for Test: Engineering Mathematics- 5 - Question 8

f(x) = 2x3 − 3x2

f′(x) = 6x2 − 6x

f′′(x) = 12x − 6

f′(x) = 6x2 − 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.

Test: Engineering Mathematics- 5 - Question 9

If f(x) = x3 − 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

Detailed Solution for Test: Engineering Mathematics- 5 - Question 9

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

3c2 − 3 = 0

c2 = 1

∴ c = ±1

Test: Engineering Mathematics- 5 - Question 10

 

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

Detailed Solution for Test: Engineering Mathematics- 5 - Question 10

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 occurred at

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