Engineering Maths - Calculus
- Option : B
- Explanation :
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,
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
- Option : B
- Explanation :
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.
If ∠ BCD = 0, where D is the mid.-point of BC, then ∠ BOD = 2ϑtherefore, 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 = 0Hence either cos 3ϑ = 0, or cos ϑ = 0, which is impossibleif 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
- Option : D
- Explanation : 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
