5

Q1.

Solution

We have,                                        

Q2. Differentiate cosx.cos2x.cos3x w. r. to x.

Solution

Q3.

Solution

 

Q4.   Find the second derivative of y = e^2xsin3x

Solution

Q5.   Differentiate  with respect to 

Solution

      

Q6. If the function f(x) given by is continuous at x=1.Find the value of a and b.

Solution

   At                                                 We have                         This gives a=2,b=1

Q7.   Differentiate  with respect to x.

Solution

Put                                                           and y becomes                    

Q8.

Solution

 

Q9.

Solution

Q10.

Solution

     

Q11.

Solution

Q12. Verify the Rolle's theorem for the function f(x)=sin 2x-2 sin x in the interval [0,].

Solution

The sine function is continuous on its domain and is differentiable as well. We can see that the function has the same value on both the end points, that is 0 and . i.e. Now we have to find a point c such that  where f'(c)=0. f'(x)=2 cos 2x - 2cos x=2(cos 2x - cos x) now we must find a point c for which 2(cos 2x - cos x)=0 hence, cos 2c - cos c=0, on solving this trigonometric equation, we can see that Clearly, we can take the above two values on the given interval for the function. Of the two values, we cant consider c=0, as the value c must be in the range (0,) which is satisfied by the equation . Hence, Rolle's theorem is verified for this function

Q13. If x = cos t, y = cos 2t prove that (1 – x²)y₂ – xy₁ + 4y = 0.

Solution

Here                                               (1 – x²)y₁² = 4(1 – y²)           Differentiating again           (1 – x²) . 2y₁y₂ + (-2x)y₁² = -8y₁           (1 – x²)y₂ – xy₁ + 4y = 0

Q14. Discuss the continuity of the function  

Solution

  At                                                 So  is discontinuous at

Q15. Differentiate (log x)^{cot x} w. r. to x.

Solution

Q16.

Solution

Q17. If , verify the Rolle’s theorem on the interval [-1,1].

Solution

We know the exponential function is continuous over the set of all real numbers and polynomial function is continuous and differentiable over the set of all real numbers, R. Hence, by using the algebra of continuous functions, we can say that the given function is continuous over [-1,1] and differentiable for (-1,1).  is defined on the interval (-1,1).              Now, we have to find a point c such that and f'(c)=0  But , when c=0, which belongs to the interval (-1,1).

Q18.

Solution

Q19.

Solution

                                                       So,

Q20.

Solution

 

Q21.

Solution

Taking            

Q22.

Solution

Q23.   

Solution

Here                                                                                        

Q24. Discuss the applicability of Rolle’s theorem for the following function  on the indicated interval : f (x) = 3 + (x – 2)^2/3 on [1, 3]

Solution

We have, f (x) = 3 + (x – 2)^2/3, x Î [1, 3] Þ f ¢ (x) =   Clearly, So, f (x) is not differentiable at x = 2 Î (1, 3) Hence, Rolle’s theorem is not applicable to f (x) = 3 + (x – 2)^2/3 on the interval [1, 3]

Q25.

Solution

We have                          

Q26. Define   Discuss the continuity of the function at x=2.

Solution

At                          is discontinuous at

Q27. The polynomial function  is given. Verify the Rolle’s theorem for this function on the interval [-3,3].

Solution

The function being polynomial is continuous on [-3,3] and differentiable on (-3,3). We can see that the value of the function on the end points, that is –3 and 3 is 0, that is _. Now, we must find at least one point where  which is the statement of the theorem.  f'(c)=3c² - 9, whose value is zero at two points in the interval, . The following graph shows the geometrical inference of the Rolle’s theorem for the function f(x)=x³ - 9x  on the interval [-3,3]

Q28.

Solution

Differentiating                                    

Q29.

Solution

We have,                                         Differentiating again          

Q30. Discuss the continuity of the function at x=0

Solution

                                         is continuous at  

Q31.

Solution

                                                                       

Q32.  Show that the function  Is continuous but not differentiable at x=0.

Solution

Q33.

Solution

 We have             Differentiating with respect to            

Q34.

Solution

Q35.

Solution

             

Q36. Discuss the applicability of Rolle’s theorem on the function f (x) =

Solution

Since a polynomial function is everywhere continuous and differentiable. Therefore, f (x) is continuous and differentiable at all points except possibly at x = 1. Now, we consider the differentiability of f (x) at x = 1. We have, (LHD at x = 1) = [f (x) = x² + 1 for 0 ≤ x ≤ 1]   (RHD at x = 1) =   \ (LHD at x = 1) ¹ (RHD at x = 1) So, f (x) is not differentiable at x = 1. Thus, the condition of a differentiability at each point of the given interval is not satisfied. Hence, Rolle’s theorem is not applicable to the given function on the interval [0, 2].

Q37.      Find  if

Solution

Let           Then                                

Q38. Verify the mean value theorem for the function for the interval [2,5]

Solution

The function is continuous for the interval [2,5] as it is defined for all points where  and all points  and is differentiable on the interval (2,5), where , Now, b-a=3 now  which is equal to hence, we can see that  or   lies between the interval (2,5) and is approximately equal to 2.64. hence, mean value theorem is verified.

Q39.

Solution

Q40. If f(x) = prove that (x² + 2)y₂ + xy₁ = 2

Solution

Differentiating given above,                                              Differentiating again                     So, (x² + 2)y₂ + xy₁ = 2

Q41.

Solution

Q42. Show that the function f(x)=|1-x+|x|| where x is any real number is continuous function.

Solution

 Define Then hog(x)=h(g(x))=|1-x+|x||=f(x)             We know g(x),h(x) are continuous function f being a composite of two continuous function is continuous.            

Q43.

Solution