Application of Integrals

More Topic from Mathematics

Question 1

integral fraction numerator dx over denominator cos space straight x space plus space square root of 3 space sin space straight x end fraction equals
  • 1 half space log space tan space open parentheses straight x over 2 plus straight pi over 12 close parentheses space plus straight C
  • 1 half space log space tan space open parentheses straight x over 2 minus straight pi over 12 close parentheses plus straight c
  • log space tan space open parentheses straight x over 2 minus straight pi over 12 close parentheses plus straight c
  • log space tan space open parentheses straight x over 2 plus straight pi over 12 close parentheses plus straight c

Solution

A.

1 half space log space tan space open parentheses straight x over 2 plus straight pi over 12 close parentheses space plus straight C integral fraction numerator dx over denominator cos space straight x space space plus space square root of 3 space sin space straight x end fraction
space equals space 1 half space integral sec space open parentheses straight x minus straight pi over 3 close parentheses dx
equals space 1 half space log space tan space open parentheses straight x over 2 minus straight pi over 6 plus straight pi over 4 close parentheses space plus straight C
space equals space 1 half space log space tan space open parentheses straight x over 2 plus straight pi over 12 close parentheses space plus straight C
Question 2

limit as straight n rightwards arrow infinity of space sum from straight r equals 1 to straight n of space 1 over straight n straight e to the power of straight r over straight n end exponent space is space
  • e

  • e+1

  • e-1
  • 1-e

Solution

C.

e-1
limit as straight x rightwards arrow infinity of space sum from straight r space equals 1 to straight n of space 1 over straight n straight e to the power of straight r divided by straight n end exponent
space equals space integral subscript 0 superscript 1 space straight e to the power of straight x space dx
space equals space open square brackets straight e to the power of straight x close square brackets subscript 0 superscript 1
straight e minus 1
Question 3

For x ∈, (0, 5π/2) define f(x). Then f (x) = integral subscript 0 superscript straight x space square root of straight t space sin space straight t space dt has

  • local maximum at π and 2π.

  • local minimum at π and 2π

  • local minimum at π and the local maximum at 2π.

  • local maximum at π and local minimum at 2π.

Solution

D.

local maximum at π and local minimum at 2π.

straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x square root of straight t space sin space straight t space dt
straight f apostrophe left parenthesis straight x right parenthesis space equals space square root of straight x space sin space straight x

local maximum at π
and local minimum at 2π

Question 4

If =-1 and x =2 are extreme points of f(x) =α log|x| + βx2 +x, then

  • α = -6, β = 1/2

  • α = -6, β = -1/2

  • α = 2, β = -1/2

  • α = 2, β = 1/2

Solution

C.

α = 2, β = -1/2

Here, x =-1 and x = 2 are extreme points of f(x) = α log|x| +βx2 +x then,
f'(x) = α/x +2βx + 1
f'(-1) = -α -2β +1 = 0  .... (i)
[At extreme point f'(x) = 0]
f'(2) = α/x +4βx + 1 = 0 .. (ii)
On solving Eqs (i) and (ii), we get
α = 2 and β = -1/2

Question 5

If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:

  • 5 square root of 2
  • 5 square root of 3
  • 5

  • 10

Solution

B.

5 square root of 3

Given equation of a circle is x2 + y2 -4x +6y -12 = 0, whose centre is (2,-3) and radius
space equals space square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 12 end root
equals square root of 4 plus 9 plus 12 space equals space 5 end root
Now, according to given information, we have the following figure.

x2+y2-4x +6y-12 =0
Clearly, AO perpendicular to BC, as O is mid-point of the chord.
Now in ΔAOB,. we have
OA space equals space square root of left parenthesis negative 3 minus 2 right parenthesis squared plus left parenthesis 2 plus 3 right parenthesis squared end root
equals space square root of 25 plus 25 end root space equals space square root of 50 space end root space equals space 5 square root of 2
and space OB space equals space 5
therefore comma
OB space equals space 5
AB space equals space square root of OA squared plus OB squared end root space equals space square root of 50 plus 25 end root space equals space square root of 75 space equals space 5 square root of 3