1. f(x) = x³  6x²+11x6 in the interval [1,3] 2. f(x) = x²4x+8 in the interval [1,3] 
As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3] Also, f(1) = f(3) = 0 Now we find f’(x) = 0 3x²  12x +11 = 0 We get, x = 2+ and 2  Hence both of them lie in (1,3). Hence the theorem holds good for the given function in interval [1,3]
(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3] Also, f(1) = 1 4 +8 = 5 and f(3) = 9 – 12 + 8 = 5 Hence f(1) = f(3) Now the first derivative of the function, f’(x) = 0 2x – 4 = 0 , gives X = 2 We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0. This means that the Rolle’s theorem holds good for the given function and given interval. 
As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4). f(x) = (x1)(x2)(x3) f(x) = x6x²+11x6 now at x = 0, we get f(0) = 6 and at x = 4, we get. f(4) = 6 diff. the function w.r.t.x , we get f’(x) = 3x²6x+11 suppose x = c, we get f’(c) = 3c²6c+11 by Lagrange’s mean value theorem, f’(c) = = = = 3 now we get, 3c²6c+11 = 3 3c²6c+8 = 0 On solving the quadratic equation, we get C = 2 Here we see that the value of c lies between 0 and 4 Therefore the given function is verified. 
We are given, f(x) = x⁴ and g(x) = x Derivative of these fucntions , f’(x) = 4x³ and g’(x) = 2x put these values in Cauchy’s formula, we get 2c² = c² = c = now put the values of a = 1 and b = 2 ,we get c = = = (approx) Hence the Cauchy’s theorem is verified. 
Let f(x) = sin x Then, =
By using Taylor’s theorem
+ ……. (1) Here f(x) = sin x and a = π/2
f’(x) = cosx , f’’(x) =  sin x , f’’’(x) =  cos x and so on.
Putting x = π/2 , we get f(x) = sin x = = 1 f’(x) = cos x = = 0 f’’(x) = sin x = = 1 f’’’(x) = cos x = = 0 from equation (1) put a = and substitute these values, we get + ……. = ……………………….. 
Let
Put these values in Maclaurin’s series we get 
Here we notice that it is an indeterminate form of . So that , we can apply L’Hospital rule 
Let … By L – Hospital rule … … … (1) … But From equation (1) 
We can see that this is an indeterminate form of type 0/0. Apply L’Hospital’s rule, we get But this is again an indeterminate form, so that we will again apply L’Hospital’s rule We get = 
Apply L’Hospital rule as we can see that this is the form of =
Note In some cases like above example, we can not apply L’Hospital’s rule. 
Let … Taking log on both sides, … By L – Hospital rule, i.e. 
Let Then first derivative So that Now So that So that Therefore f(x) has a maximum at . 
. 
Here we notice that f:x→cos x is a decreasing function on [a , b], Therefore by the definition of the definite integrals Then Now, Here Thus 


= = = ) ....................... = = 
Here limits are given a = 1 , b = 4, We know that, area under the curve, A = = = = = (16 – 7/4) = 57/4 So that the area of the region is 57/4 unit square. 
Here, F = kx So that, 1200 = 2k K = 600 N/cm In that case, F = 600x We know that, W = W = , which gives W = 3600 N.cm 
We get the following figure by using the equations of three straight lines y = 4 – x, y = 3x and 3y = x
Area of shaded region 
and g(x) = 6 – x 
We get the following figure by using these two equations
To find the intersection points of two functions f(x) and g(x) f(x) = g(x) On factorizing, we get x = 6, 2 Now Then, area under the curve A =
Therefore the area under the curve is 64/3 square unit. 
The graph of the function f(x) = 1/x will look like The volume of the solid of revolution generated by revolving R(violet region) about the yaxis over the interval [1 , 3] Then the volume of the solid will be 