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New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Equation of tangent of P (2cosθ, sinθ) is
(cosθ)x + (2sinθ)y = 4
Solving equation of tangent with equation of tangents at major axis ends, i.e. x = -2 and x = 2
For point 'B' (at x=-2):
-2cosθ + 2sinθ y = 4 ⇒ y = (2+cosθ)/sinθ
B (-2, (2+cosθ)/sinθ)
For point 'C' (at x=2):
2cosθ + 2sinθ y = 4 ⇒ y = (2-cosθ)/sinθ
C (2, (2-cosθ)/sinθ)
Now BC is the diameter of circle
Equation of circle: (x+2) (x-2) + (y - (2+cosθ)/sinθ) (y - (2-cosθ)/sinθ) = 0
x²-4 + y² - (4/sinθ)y + (4-cos²θ)/sin²θ = 0
Check if (√3, 0) satisfies this:
(√3)²-4 + 0 - 0 + (4-cos²θ)/sin²θ = -1 + (3+sin²θ)/sin²θ = -1 + 3/sin²θ + 1 = 3/sin²

...more

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

ai+aj+ck,  i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab

New answer posted

a year ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

R? → R? -R? , R? → R? -R?
|sinx-cosx, cosx-sinx, 0; 0, sinx-cosx, cosx-sinx; cosx, sinx| = 0
(sinx-cosx)² |1, -1, 0; 0, 1, -1; cosx, sinx| = 0
(sinx-cosx)² (1 (sinx+cosx) + 1 (cosx) = 0
(sinx-cosx)² (sinx + 2cosx) = 0
sin x = cos x
tan x = 1 ⇒ x = π/4
or
sin x = -2cos x
tan x = -2
Not within given range.

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

∑? [ (-1)? n / 2]
= [8/2] + [-9/2] + [10/2] + [-11/2] + . + [-99/2] + [100/2]
= 4 - 5 + 5 - 6 + . + (-50) + 50
= 4

New question posted

a year ago

0 Follower 2 Views

New answer posted

a year ago

0 Follower 15 Views

R
Raj Pandey

Contributor-Level 9

C D = ( 1 0 + x 2 ) 2 − ( 1 0 − x 2 ) 2 = 2 1 0 | x |

Area 

= 1 2 * C D * A B = 1 2 * 2 1 0 | x | ( 2 0 − 2 x 2 )

⇒ 1 0 − x 2 = 2 x

3x2 = 10

 x = k

3k2 = 10

New answer posted

a year ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

f ( x ) = x 4 − 4 x + 1 = 0

f ' ( x ) = 4 x 3 − 4

= 4 ( x − 1 ) ( x 2 + 1 + x )

⇒ Two solution

New answer posted

a year ago

0 Follower 3 Views

J
Jaya Sharma

Contributor-Level 10

Some of the common mistakes that people usually make while using logarithmic differentiation have been mentioned below:

  • Not Multiplying by y: After logarithmic differentiation, it is mandatory to multiply by y to solve for dy/dx?
  • Incorrectly Applying the Chain Rule: Make sure that you have correctly used the chain rule whenever you are differentiating a logarithmic expression.
  • Using Wrong Logarithm: It is always advisable to use the natural logarithm (ln) instead of logarithms with other bases.
  • Ignoring Domain Restrictions: Natural logarithm is only defined for the positive real numbers; therefore, y>0 whenever you apply logarithmic diffe
...more

New answer posted

a year ago

0 Follower 3 Views

J
Jaya Sharma

Contributor-Level 10

Logarithmic differentiation is used in the following cases:

  • Logarithmic differentiation is used with functions that have a variable in both base and exponents. In such a case, standard differentiation rules do not apply directly to such functions. This differentiation converts exponentiation into multiplication.
  • Another area where logarithmic differentiation is used is with a function which is the product of a quotient of multiple terms. 
  • Whenever a function has a complex combination of multiplication, division and exponentiation, logarithmic differentiation is preferred. This differentiation eases the complexity by converting multip
...more

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