Test: Engineering Mathematics- 5 - Computer Science Engineering (CSE) MCQ

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

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.

(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

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

By using Cauchy’s Mean value theorem

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′ = 12x² − 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

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)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

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.

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

3c² − 3 = 0

c² = 1

∴ c = ±1

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