Engineering Maths MCQ - Calculus

36 If temperature field in a body varies according to the equation T(x, y)= x² + 4xy, then direction of fastest variation in temperature at the point (1, 0) is given by

37 If f(x) = 3x⁴ -4x² +5, then the interval for which f(x) satisfied all the condition of Rolle's Theorem is

Given function is, f(x) = 3x⁴ - 4x² + 5
(i) f (x) is continuous in every real interval
(ii) f ' (x) exist in any real interval
(iii) f(-1) = 3(-1)⁴- 4(1)²+ 5 = 4
f(1) = 3(1)⁴- 4(1)²+ 5 = 4
f(-1) = f(1)
Also f ' (c) = 12c³ - 8c = 0
⇒ c = 0,

Calculus

c ∈ [-1, 1]

Hence all the condition of Rolle's Theorem are satisfied in the interval [-1, 1]

38 The triangle of maximum area inscribed in a circle of radius r is

Let ABC be a triangle inscribed in the circle with centre 0 and radius r. If area of this triangle is maximum, then vertex C should be at a maximum distance from the base AB i.e. , CD must be perpendicular to AB. Hence ABC is an isosceles triangle. Calculus If ∠ BCD = 0, where D is the mid.-point of BC, then ∠ BOD = 2ϑ

therefore, AB= 2BD = 2r sin 2ϑ CD = CO + OD = r + r cos 2ϑ If S be the area of the triangle ABC, then S = 1/2 AB x CD = 1/2 x 2r sin 2 ϑ(r + r cos 2ϑ)

ds /dϑ = r² [sin 2ϑ (-2 sin 2ϑ) + (1 + cos 2ϑ) (2 cos 2ϑ)] = 2r² [cos ²2ϑ - sin² 2ϑ + cos 2ϑ] = 2r² (cos 4ϑ + cos 2ϑ) For maximum and minimum, ds / dϑ = 0 cos 4ϑ + cos 2ϑ = 0 2 cos 3ϑ cos ϑ = 0

Hence either cos 3ϑ = 0, or cos ϑ = 0, which is impossible

if cos ϑ = 0, then ϑ = Π / 2 if cos 3ϑ = 0, then 3 ϑ = Π / 2 ϑ = Π / 6 ( a²s / dϑ ²) _{6 =}Π/6 is negative therefore S is maximum for ϑ = 1 / 6 * Π ∠ ACB = 2ϑ = 2( Π / 6 ) = Π / 3 = ∠ ABC = ∠ BAC Hence ABC is an equilateral triangle.

39 The greatest and least value of f(x) = x⁴- 8x³ + 22x² - 24x +1 in [0, 2] are

f(x) = x⁴- 8x³ + 22x² - 24x +1 , f(0) = 1
f(2) = 2⁴- 8.2³ + 22.2² - 24.2 + 1 = -7
Now f ' (x) = 4x³-24x² + 44x - 24
For maximum and minimum, f ' (x) = 0
⇒ 4x³-24x² + 44x - 24 = 0
⇒ 4(x - 1)(x - 2)(x - 3) = 0
therefore x = 1, 2, 3
Since x = 3 doesnot lie in [0, 2]
therefore, consider only x = 1 and x = 2
We have f(1) = 1⁴ - 8.1³ +22.1² - 24.1 + 1 = -8
Greatest of f(x) = largest of {1, -7, -8} = 1
Least of f(x) = smallest of {1, -7, -8} = -8

40 The value of ξ in the mean value theorem of f(b) - f(a) = (b - a)f' (ξ) for f(x) = Ax² + bx + C in (a, b) is