CONTINUITY
TYPE IV
Discuss the continuity of the function at a given point / on it’s domain / on the given interval.
1.f(x) = 3x – 4 for 0 x 2at pt
= 2x – 42 < x 5x = 2
2.f(x) = x2 – 2 for, 2 x 4at
= 2x + 64 < x 6x = 4
3.f(x) = 0 x 3at,
= 4x – 53 < x 6x = 3
4.f(x) = 0 x 4at
= 4 < x 6x = 4
5.f(x) = x2 + 4 0 x 2 on its domain
= 3x + 22 < x 4
= x + 14 < x 6
6.f(x) =0 < x 3on it’s domain
= 3x – 13 <x 5
= 5 < x 7
7.f(x) = 0 x 2
= 3x – 12 < x 4on it’s domain
= 4 < x 6
8.f(x) = sin x 0 < x on the interral
= tan x < x 3 (0, )
= cos x 3 < x
Type – V
Find the value k / a, b / , If the given function is continuous at a given point / on it’s domain / on its interral.
1. f(x) = 5x + 10x 1at, x = 1
= 3x + k1 < x 2
2. f(x) = 3x - 40 x 2at, x = 2
= 2x - k2 < x 4
3. f(x) = x > 0at, x = 0
x < 0
= x + 4 + bx = 0
= 1
4.f(x) = x2 + a x > 0at, x = 0
= x < 0
given f(x) = 2, find a, b.
5.f(x) = x > 0at, x = 0 find (a + b)
= x + 4 – bx < 0
6.f(x) = - 2 x 0
= 2x + 10 < x 1is continuous
= 2b 1 < x 2on (-2, 2)
7.f(x) = x2 + ax + b0 x 2is continuous
= 3x + 22 < x 4on (0, 6)
= 2ax + 5b4 < x 6
8.f(x) = 5b – 3a x -4 x - 2is continuous
= 4x – 1-2 < x 2on [-4, 4]
= ax2 + 17b2 < x 4
9.f(x) = -2 sin x - x - / 2is continuous
= sin x + - / 2 < x / 2on
= cos x /2 < x (-, )
Discuss the continuity of the function for all real ralues of x.
1.f(x) = on the interval (0, 6)
2.f(x) = on the interval (0, 6)
DERIVATIVE
Defn. –
If , = f(x) is a function of x then the derivative of with respect to ‘x’ denoted by OR f(x) is given by –
Question.
Q.Find the derivatives of the following from the 1st principle.
1)xn2) 2x2 + 33) 4x + 54) 5) 6) 7) 8) 9) 10) 11)
12)13) 14)sin x 15) cos x 16) tan x 17) cot x 18) sec x 19) cosec x 20) sin 2x 21) cos 3x 22) 23)
24) x sin x 25) x cos x 26) x2 sin x 27) x2 cos x28) ex29) e2x + 330) ax 31)72x+ 3 32) log x 33) log (4x + 5) 34) log ax
35) log10(4x + 5)
Standard Formulae :-
1.2.
3.4.
5.6.
7.8.
9.10.
11.12.
13.14.
15.16.
17.18.
19.20.
Derivative of composite function.
Thm. –
Statement :- If ‘’ is a differentiable function of ‘u’ & ‘u’ is a differentiable function. of ‘x’ then prove that,
Proof :-
Let x be the small increment in x, the corrousponding inctements in u are u respectively.
As x 0, , u 0
Since,
is a differentiable function of u u is a differentiable function x.
we have,
taking limit as x 0
as x 0 u 0
Here, the limit of R.H.S exists, hence, the limit of R.H.S. exists
Q.Find If,
1) = (4x2 + 3)7= 7(4x+2 + 3)6 (8x)
2) = (tan x + 3)4= 4 (tan x + 3)3(sec2x)
3) = (3x2 + 4)8= 8 (3x2+ 4)7(6x)
4) =(sin x + 4)8 =8 (sin x + 4)7(cos x)
5) = (log x + 3)7= 7(log x + 3)6
6) =
7) = (sin x)3= 3 (sin x)2(cos x)
8) = (cos x)3= 3 (cos x)2(-sin x)
9) = (tan x)3= 3 (tan x)2(sec2 x)
10) = sec3 x = 3 sec2 x. sec x tan x
11) = (4x2 + 3x + 7)8 = 8 (4x2 + 3x + 7)7(8x + 3)
12) = =
13) = =
14) = =
15) = =
1
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