(i) 1
(ii) 2
(iii) 1

\(\frac { 1 }{ \sqrt { 2 } } \left[ 1-\frac { \left( x-\frac { \pi }{ 4 } \right) }{ 1! } -\frac { \left( x-\frac { \pi }{ 4 } \right) ^{ 2 } }{ 2! } \right] \)

1−

\(2\pm \frac { 2 }{ \sqrt { 3 } } \)

2.89

\(C=\frac { 3\pi }{ 4 } \)

(π,-2)

\(\theta =\frac { \pi }{ 2 } \) 

4x + 3y = 7a

\(\frac { 8 }{ 3 } { cm }^{ 2 }/sec\)

2:π

s = 13.8 t - 4.9 t²
v = \(\frac { ds }{ dt } \) =13.8 - 4.9 (2t)
=13.8-9.8t
When t = 1, v = 13.8 - 9.8(1)
4 m/sec.
Acceleration = \(\frac { d^{ 2 }x }{ { dt }^{ 2 } } \) = -9.8 m/sec²
At maximum height, v = 0
∴ 13.8 - 9.8 t = 0
⇒ 13.8 = 9.8 t
⇒ t = \(\frac { 13.8 }{ 9.8 } \) = 1.40 sec
At t = 1.4 sec,
distance (s) = 13.8 (1.40) - 4.9 (1.40)²
= 19.32 - 9.604 = 9.716 m

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 \(\frac { dy }{ dt } =-\frac { 3 }{ 2\times 4 } =\frac { -3 }{ 8 } \).

f(x) = x³-2x²-x+3
f'(x) = 3x² - 4x - 1
f'(c) = 3c² - 4c- 1
f(a) = f(0) = 3
f(b) = f(1) = 1³-2-1+3 = 1
Then, if atleast one C \(\in \) (0, 1) such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ 3c²-4c-1 = \(\frac { 1-3 }{ 1-0 } \)
⇒ 3c²-4c+1 = 0
⇒ c = \(\frac { 4\pm \sqrt { 16-4(3) } }{ 2(3) } \)
⇒ c = \(\frac { 4\pm 2 }{ 6 } =\frac { 6 }{ 6 } \) or \(\frac { 2 }{ 6 } \)
⇒ 1 or \(\frac { 1 }{ 3 } \).

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\(\sqrt { 3 } \).