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Posted by Komal Preet 5 years ago
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Shraddha ✨✰✰ 5 years ago
Posted by Shraddha ✨✰✰ 5 years ago
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Gaurav Seth 5 years ago
y = Sec ( Tan √x )
dy/dx = Sec ( Tan √x ) Tan ( Tan√x ) Sec² √x ( 1/ 2√x )
[Sec ( Tan √x ) Tan ( Tan√x ) Sec² √x ]/ 2√x
The same result implies for using logarithmic functions on both the sides ,
But it'll increase the payloads , hence , using Chain rule only
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Gaurav Seth 5 years ago
R = {(a, b) : 2 divides a – b}
where R is in the set Z of integers.
(i) a – a = 0 = 2 .0
∴ 2 divides a – a ⇒ (a, a) ∈ R ⇒ R is reflexive.
(ii) Let (a, a) ∈ R ∴ 2 divides a – b ⇒ a – b = 2 n for some n ∈ Z ⇒ b – a = 2 (–n)
⇒ 2 divides b – a ⇒ (b. a) ∈ R
(a, ft) G R ⇒ (b, a) ∈ R ∴ R is symmetric.
(iii) Let (a, b) and (b, c) ∈ R
2 divides a – b and b – c both ∴ a – b = 2 n1 and b – c = 2 n2 for some n1, n2 ∈ Z ∴ (a – b) + (b – c)= 2 n1 + 2 n2 ⇒ a – c = 2 (n1 + n2 )
⇒ 2 divides a – c
⇒ (a, c) ∈ R
∴ (a,b), (b,c) ∈ R ⇒ (a, c) ∈ R
∴ R is transitive
From (i), (ii), (iii) it follows that R is an equivalence relation.
Posted by Just Example 5 years ago
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Mohit Kashyap 5 years ago
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Gaurav Seth 5 years ago
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.
∴ By second derivative test, the volume of the cone is the maximum when
Posted by Aman Badhwar 5 years ago
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Gaurav Seth 5 years ago
x = a(cost + log tant/2)
differentiate with respect to t,
dx/dt = d{a(cost + logtan t/2)}/dt
= a[d(cost)/dt + d(logtan t/2)/dt]
= a[ -sint + 1/tan t/2 × sec²t/2× 1/2]
= a [ -sint + 1/{2sin t/2.cos t/2} ]
we know, 2sinA.cosA = sin2A , use it here
= a [ -sint + 1/sint]
= a { 1 - sin²t}/sint
= acos²t/sint
= acott. cost .......(1)
again, y = asint
dy/dt = acost .......(2)
now, dy/dx = {dy/dt}/{dx/dt}
= {acost}{acott.cost}
= 1/cott
hence, dy/dx = tan t
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