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Calculus - Calculus MCQ

36:  

The following function has a local manima at which value of x

local maxima in calculus

A.

equation

B.

under root 5

C.

under root 5 by 2

D.

minus root five by 2

 
 

Option: C

Explanation :

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37:  

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

A.

calculus mcq questions

B.

iota

C.

iota nad j

D.

5 iota and 8 j

 
 

Option: C

Explanation :

explanation for calculus questions

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38:  

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

A.

[0, 2]

B.

[-1, 1]

C.

[-1,0]

D.

[1, 2]

 
 

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(1)+ 5 = 4
       f(1) = 3(1)- 4(1)+ 5 = 4 
                              f(-1) = f(1)
Also            f ' (c) = 12c3 - 8c = 0 
⇒                     c = 0,
 
 c equals a 2 by 3
                          c ∈ [-1, 1]
Hence all the condition of Rolle's Theorem are satisfied in the interval [-1, 1] 

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39:  

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

A.

a right angled triangle with hypotenuse measuring 2r

B.

an equilateral triangle

C.

an isosceles triangle of height r

D.

doesnot exist

 
 

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.
Calculus questions
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 26 = 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.

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40:  

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

A.

0, 8

B.

0, -8 

C.

1,8

D.

1, -8

 
 

Option: D

Explanation :

f(x) = x4- 8x3 + 22x2 - 24x +1 , f(0) = 1
                                                   f(2) = 2- 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

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