Let OA = x m, OB = y m
Then x² + y² = 25

Differentiating, \(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dx } =-\frac { x }{ y } \frac { dx }{ dt } \)
When \(\frac { dx }{ dt } =\frac { 1 }{ 2 } ,\frac { dy }{ dt } =\frac { -x }{ 2y } \)
When y = 4, x² = 25-y²
⇒ x =\(\sqrt { 25-16 } \) = 3
Thus \(\frac { dy }{ dt } =-\frac { 3 }{ 2\times 4 } =\frac { -3 }{ 8 } \).

y = sin²x
\(\frac { dy }{ dx } \) = 2 sin x cos x = sin 2x
∴ m = \(\left( \frac { dy }{ dx } \right) _{ \left( \frac { \pi }{ 3 } ,\frac { 3 }{ 4 } \right) }=sin\frac { 2\pi }{ 3 } =\frac { \sqrt { 3 } }{ 2 } \)
∴ Slope of the normal = \(-\frac { 1 }{ m } =-\frac { 2 }{ \sqrt { 3 } } \)
∴ Equation of normal is y-y₁ = \(-\frac { 1 }{ m } \)(x-x₁)
⇒ \(y-\frac { 3 }{ 4 } =-\frac { 2 }{ \sqrt { 3 } } \left( x-\frac { \pi }{ 3 } \right) \)
⇒ 12\(\sqrt { 3 } \)y-9\(\sqrt { 3 } \) = -24x + 8π [ multiply 12\(\sqrt { 3 } \)]
∴ 24x + 12\(\sqrt { 3 } \)y = 8π + 9\(\sqrt { 3 } \).

Let f(x) = log(1+x)-\(\frac { x }{ 1+x } \)
f'(x) =\(\frac { 1 }{ 1+x } -\frac { (1+x)-x }{ (1+x)^{ 2 } } \)
=\(\frac { 1 }{ 1+x } -\frac { 1 }{ (1+x)^{ 2 } } \)
=\(\frac { x }{ (1+x)^{ 2 } } \) > 0 for x>0
∴ f(x) is strictly increasing in (0, ∞)
∴ x>0 ⇒ f(x) > f(0)
⇒ log (1+x) - \(\frac { x }{ 1+x } \) >0
⇒ log(1+x) > \(\frac { x }{ 1+x } \)
⇒ \(\frac { x }{ 1+x } \) < log (1+x).

We have, f (x) is continuous in \([\frac{1}{2},2 ]\) and differentiable in \((\frac{1}{2},2 )\) with  \(f(\frac{1}{2})=\frac{5}{2}=f(2) \). 
By  the Rolle’s theorem there must exist a \(c \in (\frac{1}{2},2 )\) such that, \(f'(c)=1-\frac{1}{c^{2}}=0 \Rightarrow c^{2}=1 \) gives \(\Rightarrow c=\pm1, \) As \(1\in(\frac{1}{2},2)\) we choose c = 1.

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec  
Instantaneous Velocity at t = 6
= V(6) = 27m/sec

Let r be the radius of the base, h be the height of the cone, I be the slant height and θ be the semi vertical angle.
In ΔOAB l² = h²+r²
⇒ r² =l²-h²
⇒ V =\(\frac { 1 }{ 3 } { \pi }r^{ 2 }h=\frac { \pi }{ 3 } \)(l²-h²)h
=\(\frac { \pi }{ 3 } \) (l²h - h³)
\(\frac { dv }{ dh } =\frac { \pi }{ 3 } \)(l²-3h²)

\(\frac { dv }{ dh } \)=0
⇒ \(\frac { \pi }{ 3 } \)(l² - 3h²) = 0
⇒ l² - 3h² = 0 ⇒ \(\frac { { l }^{ 2 } }{ { h }^{ 2 } } \) =3
⇒ \(\frac { l }{ h } \) = \(\sqrt { 3 } \)  ⇒ h=\(\frac { 1 }{ \sqrt { 3 } } \)
Now, \(\frac { d^{ 2 }V }{ dh^{ 2 } } =\frac { \pi }{ 3 } \)(-6h) = -2πh
at h =\(\frac { l }{ \sqrt { 3 } } ,\frac { d^{ 2 }V }{ dh^{ 2 } } =-2\pi \left( \frac { l }{ \sqrt { 3 } } \right) \) < 0
V is max at h =\(\frac { l }{ \sqrt { 3 } } \)
∴ r² =l²-\(\frac { { l }^{ 2 } }{ 3 } =\frac { 2{ l }^{ 2 } }{ 3 } \) =2h²
\(\\ \left[ \because h=\frac { l }{ \sqrt { 3 } } \Rightarrow h^{ 2 }=\frac { { l }^{ 2 } }{ 3 } \right] \)
∴ \(\frac { { r }^{ 2 } }{ { h }^{ 2 } } \) =2
⇒ \(\frac { r }{ h } =\sqrt { 2 } \)
⇒ tanθ =\(\sqrt { 2 } \)
⇒ θ = tan^-1(\(\sqrt { 2 } \))

f(x) = a log x + bx² + x
f'(x) = \(\frac { a }{ x } \) + 2bx +1
Since f (x) has extreme values at x = -1 and x = 2,
f'(-1) = 0 and f'(2) = 0
f'(-1) = 0
⇒ -a - 2b + 1 = 0.....(1)
f'(2) = 0
⇒ \(\frac { a }{ 2 } \) + 4b + 1 = 0
⇒ a + 8b + 2 = 0....(2)
(1) + (2) ⟶
6b + 3 = 0
⇒ b = \(\frac { -1 }{ 2 } \)
Substituting b = \(\frac { -1 }{ 2 } \) in (1) ⇒ -a - 2\(\left( -\frac { 1 }{ 2 } \right) \)+1 = 0
⇒ -a + 1 + 1 = 0
⇒ a = 2
∴ a = 2, b = \(\frac { -1 }{ 2 } \)

Given s (t) = 2t³ − 9t² + 12t ≥ 0

f(x) = x³+ 2x²-1
f'(x) = 3x² + 4x = 0
⇒ x (3x + 4) = 0
⇒ x = 0 or \(\frac { 4 }{ 3 } \)
The possible intervals are \(\left( -\infty ,-\frac { 4 }{ 3 } \right) \left( -\frac { 4 }{ 3 } ,0 \right) \) and (0, ∞).

x =\(\sqrt { t } \)
V = \(\frac { dx }{ dt } =\frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } }\) ..(1)
Acceleration = \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } =\frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } t^{ -\frac { 3 }{ 2 } } \right) =\frac { -t^{ -\frac { 3 }{ 2 } } }{ 4 } \)
∴ Acceleration is negative
Acceleration = \(-\frac { 1 }{ 4 } \left( { t }^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\)
= \(-2\left( \frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\) = 2V³    [using (1)]
Hence, acceleration is negative proportional to the cube of the velocity.

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.