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Find dydx of each of the functions expressed in parametric form in Exercises from 44 to 48.

x=t+1t,   y=t−1t

0 6 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Short Answer Type Questions as classified in NCERT Exemplar

    Sol:

    G i v e n     t h a t                     x = t + 1 t ,       y = t − 1 t D i f f e r e n t i a t i n g     b o t h     t h e     g i v e n     p a r a m e t r i c     f u n c t i o n s     w . r . t .     t                                   d x d t = 1 − 1 t 2 ,         d y d t = 1 + 1 t 2 ∴                               d y d x = d y d t d x d t = 1 + 1 t 2 1 − 1 t 2 = t 2 + 1 t 2 − 1 H e n c e ,     d y d x = t 2 + 1 t 2 − 1

Similar Questions for you

Let f and g be twice differentiable even functions on (-2, 2) such that f ( 1 4 ) = 0 ,     f ( 1 2 ) = 0 ,     f ( 1 ) = 1 and f ( 3 4 ) = 0 ,     g ( 1 ) = 2  

Then, the minimum number of solutions of f ( x ) g " ( x ) + f ' ( x ) g ' ( x ) = 0     i n     ( − 2 , 2 )  is equal to……….
V
Vishal Baghel

f (x) is an even function

f ( − 1 4 ) = f ( − 1 2 ) = f ( 1 2 ) = f ( 1 4 ) = 0  

So, f (x) has at least four roots in (-2, 2)

g ( − 3 4 ) = g ( 3 4 ) = 0         

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) ⋅ g " ( x ) = f ' ( x ) ⋅ g ' ( x ) = 0  

It is same as number of roots of d d x ( f ( x ) ⋅ g ' ( x ) ) = 0 will have atleast 4 roots in (-2, 2)

Let f be a real valued continuous function on [0, 1] and f(x) = x + ∫ 0 1 ( x − t ) f ( t ) d t .  Then, which of the following points (x, y) lies on the curve y = f(x)?
V
Vishal Baghel

f ( x ) = x + x ∫ 0 1 f ( t ) d t − ∫ 0 1 t 0 f ( t ) d t

Let 1 + ∫ 0 1 f ( t ) d t = α

∫ 0 1 t     f ( t ) d t = β

So, f(x) = x

Now, α = ∫ 0 1 f ( t ) d t + 1

α = ∫ 0 1 ( a t − β ) d t + 1

β = ∫ 0 1 t ⋅ f ( t ) d t

β = 4 1 3 , α = 1 8 1 3

f(x) = αx – b

= 1 8 x − 4 1 3

option (D) satisfies

Let f(x) be non-constant thrice differentiable function defined on ( − ∞ ,     ∞ )  such that f(x) = f(6 – x) and f’(0) = 0 = f’(2) = f’(5). If ‘n’ is the minimum number of roots of (f”(x))2 + f(x)f”(x) = 0 in the interval x ∈  [0, 6] then sum of digits of n equals
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Vishal Baghel

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

? f ' ( 3 ) = 0

So f' (x) = 0 has minimum 7 roots in x ? [ 0 , 6 ] ? f ' ' ( x )  has min 6 roots in   x ? [ 0 , 6 ]

h (x) = f' (x) . f' (x)

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Vishal Baghel

1 + x? - x? = a? (1+x)? + a? (1+x) + a? (1+x)² . + a? (1+x)?
Differentiate
4x³ - 5x? = a? + 2a? (1+x) + 3a? (1+x)².
12x² - 20x³ = 2a? + 6a? (1+x).
Put x = -1
12 + 20 = 2a? ⇒ a? = 16

Let f be a real valued function, defined on R - {-1, 1} and given by f(x) = 3logₑ|(x-1)/(x+1)| - 2/(x-1). Then in which of the following intervals, function f(x) is increasing?
A
alok kumar singh

Kindly go through the solution 

 

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