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Evaluate
Since at numerator and denominator becomes zero.
∴ it is of form.
∴ By factor theorem, is a factor of both numerator and denominator.
is also a factor of both numerator and denominator because irrational roots occur in pairs.
So, x2 -2 is a factor of both numerator and denominator.
By long division method, dividing numerator and denominator by x2 -2.
x9 - 3x8 + x6 - 9x4 - 4x2 - 16x + 84 = (x2 -2) (x7 - 3x6 + 2x5 -5x4 + 4x3 -19x2 +8x - 42) and x5-3x4 -4x + 12 = (x2 -2) (x3 - 3x2 + 2x -6)
∴
Evaluate
L.H.L = [∵ here x < 0]
R.H.L. [∵ here x > 0]
[ ∵ for x >0,
]
Thus, L.H.L R.H.L.
∴ does not exist.
Suppose that
and if , what are possible values of a and b?
L.H.L. = [∵ here x <1]
Put x = 1-4, h>0 and h0
∴ L.H.L =
R.H.L. = [∵ here x > 1]
Put x = 1-h, h>0 and h0
∴ R.H.L =
Also, f(1) = 4
Now,
a + b = 4 adn b - a = 4
Adding these equations, we get 2b = 8 b = 4
∴ a = 4 - b = 4 - 4 = 0
Hence, a = 0, b = 4
Let f be a function defined by
Evaluate what can you say about
?
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Let f(x) = x2-2
Now, f (10) denote the derivative of x2-2 at x = 10.
∴ f' (10)
Hence, f(10)=20
Let f(x) = (x-1)(x-2) = x2 -3x + 2.
∴
= 0 + 2x-3 = 2x -3
Let f(x) find Lf'(0) and Rf'(0)
Here, f(0) = 1-02 = 1 (Note that here h < 0)
(Note that here h > 0)
Clearly, Lf'(0) = 0 = Rf'(0) f is derivable at x =0
∴ derivative of function of f(x) = 99x at x = 100 is 99.
Let
find a so that f may be derivable at x =1.
Here, f(1) = 1+a
(here h < 0
1+h < 1)
(here h >0
1 + h > 1)
Since f is derivable at x =1
∴ Lf'(1) = Rf'(1) 1 = 2a
If f is derivable at x=a, evaluate .
Since f is derivable at x =a
∴
Now,
[By putting x =a +h as xa, h
0]
For the function : Prove that f(1) = 100 f'(0)
Here,
L.H.S. = f(1) = (1)99 + (1)98 + ....... +(1)1 + 1 = 1+1+ ....+1+1 = 100
R.H.S. = [100f'(0) = 100(0)99 +(0)98 + ..... + (0)1 +1] = 100
∴ L.H.S =R.H.S.
Hence, f'(1) = 100 f'(0)
Let y = x2 - (a+b)x + ab
∴
= 2x-(a+b).1=2x-(a+b)
y = (ax2 +b)2
Let y = (5x3 +3x -1) (x-1)
differentiating both sides w.r.t. x, we get
=(5x3 +3x -1) (1-0)+(x-1)(5x2 +3 -0)
= 5x3 +3x -1+15x3 +3x -3
= 20x3 -15x2 + 6x -4
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Let y = x-4 (3 - 4x-5) y = 3 x-4 -4x-9
Differentiating both sides w.r.t. x, we get
Let y = (ax+b)n (cx+ d)n
differentiating both sides w.r.t. x, we get
Derivative of (ax + b)n, n being a rational number.
Let f(x) = (ax+b)n
Hence,
Corollary : Put a = 1m, b = 0
Find the derivative of (2x+3)(5x2-7x+1)
Let f(x) = (2x+3)(5x2-7x+1)
(Byn Product rule)
= (2x+3) (10x-7)+(5x2 -7x+1)(2)
= 20x2 -14x + 30x -21 + 10x2 -14x + 2
= 30x2 + 2x - 19
Find the derivatives of the following functions :
5 sin x - 6 cos x +7
Let y = 5 sin x - 6 cos x +7
Differntiating both sides w.r.t. x, we get.
= 5cos x - 6(-sin x)+0 - 5 cos x + 6 sin x
Find the derivatives of the following functions :
5 sec x + 4 cos x
Let y = 5 sec x + 4 cos x
Diffentiating both sides w.r.t. x, we get
= 5 (sec x tan x) + (-sin x) = 5 sec x tan x -4 sin x
Find the derivatives of the following functions :
sin x cos x
Let y sin x cos x
Differentiating both sides w.r.t. x, we get
Find the derivatives of the following functions :
2 tan x - 7 sec x
Let y = 2tan x-7 sec x
Differentiating both sides w.r.t. x, we get
Let
Differentiating both sides w.r.t. s, we get
Let y
Differentiating both sides w.r.t. x, we get
= cos a sec2x +0 = cos a sec2x
Let y = (ax2+sin x) (p + q cos x)
Differentiating both sides w.r.t. x, we get
Let
Differentiating both sides w.r.t. x, we get
Let y = (x + sec x)(x - tan x)
Differentiating bioth sides w.r.t. x, we get
Let
Differentiating both sides w.r.t. x, we get
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Find the derivatives of followings functions from first principle.
(i) sin 2x (ii) cot (3x+1)
(i) 2 cos 2x (ii) -3 cosec2 (3x + 1)
Find the left and right derivatives of
Find the left and that f is not differentiate at x = 1.
Lf'(1)=3, Rf'(1)=-1
Differentiate following functions w.r.t. x
(i) x cosec x (ii) (1-2 tan x) (5 +4 sin x) (iii) x4 tan x (iv) x sin x + cos x
(I) (1 - x cot x) cosec x (ii) 4 (1 - 2 tan x) cos x -2 (5 +4 sin x) sec2x (iii) x3 (x xec2x + 4 tan x)
The equation of the tangent to the curve y = x +4/x2, that is parallel to the x-axis, is
y= 0
y= 1
y= 2
y= 3
D.
y= 3
We have,
On differentiating w.r.t x, we get
since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x3 = 8
⇒ x = 2 abd y =3
Let cos (α + β) = 4/5 and let sin (α - β) = 5/13, where 0 ≤α,β ≤ π/4. Then tan 2α is equal to
25/16
56/33
19/12
20/7
B.
56/33
Cos (α + β) = 4/5
⇒ (α + β) ∈ 1st quadrant
and sin (α - β) = 5/13
⇒ (α - β) ∈ 1st quadrant
⇒ 2α =(α + β) + (α - β)
∴
If 5(tan2x – cos2x) = 2cos 2x + 9, then the value of cos4x is
-7/9
-3/5
1/3
2/9
A.
-7/9
⇒5(1 – t – t2) = t(4t + 7)
⇒ 9t2 + 12t – 5 = 0
⇒ 9t2 + 15t – 3t – 5 = 0
⇒ (3t – 1) (3t + 5) = 0
⇒ t = t/3 as t≠-5/3.
cos2x = 2(1/3)-1 = -1/3
The differential equation which represents the family of curves y=c1ec2xe, where c1 and c2 are arbitrary constants, is
y' =y2
y″ = y′ y
yy″ = y′
yy″ = (y′)
D.
yy″ = (y′)
y c1ec2x = …..(i)
y' = c1c2ec2x
y' = c2y.....(from (i) ....(ii)
y" = c2y' ....... (iii)
from (ii) & (iii)
The solution of the differential equation satisfying the condition y (1) = 1 is
y = ln x + x
y = x ln x + x2
y = xe(x−1)
y = x ln x + x
D.
y = x ln x + x
y = vx
Since, y (1) = 1, we have y = x log x + x
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