5

Q1. To verify Rolle's Theorem for f(x) defined in [a, b], we need to show

• f(a) = 0
• f(b) = 0
• f(a) = f(b)
• f(a) + f(b) = 0

Solution

f(a) = f(b)

Q2. The greatest integer function is:

• continuous everywhere
• discontinuous everywhere
• continuous except at the integral values of x
• discontinuous except at end points.

Solution

The graph of  greatest integer function [x] breaks at integral values of x

Q3. If f(x) = e^x, then the value of f'(-3) is

• e³
• e^-3
• log (3)
• log (-3)

Solution

Q4. Geometrically Rolle's theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is

• Parallel to the x axis
• Parallel to the y axis
• Parallel to the  line y = x
• Parallel to the  line joining the end points of the curve

Solution

Q5. If the function f (x) = x² - 8x + 12 satisfies the condition of Rolle's Theorem on (2, 6), find the value of c such that f '(c) = 0

• 4
• 8
• 2
• 6

Solution

Q6. To verify Rolle's Theorem which one is essential?

• Continous in closed interval and differentiable in open interval.
• Continous in open interval and differentiable in closed interval.
• Continous in closed interval and differentiable in closed interval.
• Continous in open interval and differentiable in open interval.

Solution

Continous in closed interval and differentiable in open interval.

Q7. Find the second derivative of e^xcosx

• -e^xsinx 
• -2e^xcosx
• -2e^xsinx
• e^x(sinx + cosx)

Solution

        Let y = e^x cosx                                       = -2e^x sinx

Q8. Rolle's theorem is verified for the function y = x² + 2,in the interval[a,b] and a = -2. Find the smallest integral value of b

• 1
• 2
• 3
• 4

Solution

Q9. For the curve corresponding to f(x) = x² - 8x + 12, find the point at which the tangent is parallel to the x axis

• (-1,1)
• (-4,4)
• (4,-4)
• (1,1)

Solution

Q10. Differentiate sin x³ with respect to x³.

• - cos x³
• cos x³
• sin x³
• - sin x³

Solution

Q11.  Discuss applicability of Rolle's Theorem for x + |x| in interval [-1, 1]

• Continous and differentiable
• Not continous and differentiable
• Continous and non-differentiable
• Not continous and not differentiable

Solution

Let f(x) = x + |x|Then, f'(x) = 0 at x < 0 And f'(x) = 2 at x > 0 So, it is not differentiable. But as x and |x| are continuous functions so their sum is also continuous.

Q12. Geometrically the Mean Value theorem ensures that there is at least one point on the curve f(x) , whose abscissa lies in (a, b) at which the tangent is

• Parallel to the x axis
• Parallel to the y axis
• Parallel to the  line y = x
• Parallel to the  line joining the end points of the curve

Solution

Q13. Find the derivative of cot² x³

• 2 cot x³
• -2 cot x³ cosec² x³
• -6x² cosec² x³. cot x³
• -6x² cosec² x3.

Solution

Q14. If f(x) = e^2x-5, then f'(x) is

• 2 e^{2x - 5}
• 5 e^{2x - 5}
• log (2x - 5)
• 2 log (2x - 5)

Solution

Q15. The differential coefficient of the function f(x) = a^{sin x}, where a is positive constant is:

• cos x. log sin x
• a^{sin x}. cos x. log sin x
• a^{sin x} log sin x
• a^{sin x}. cos x. log a

Solution

Let y = a^sinx, then log y = sin x log a. Now, differentiating w. r. to x, we get

Q16. If f(x) = |x - 2|, then at x = 2, f'(x) is

• Continous but not differentiable
• Differentiable but not continous
• Continous and differentiable both
• Neither continous nor differentiable

Solution

f'(x) = -1 at x < 2f'(x) = 1 at x > 2Therefore, not differentiable But is continous as it is composition of two functions i.e., polynomial and modulus

Q17. The derivative of 2^x tan x is

• 2^x [sec² x + tan x]
• 2^x tan x [sec x + log 2]
• 2^x log 2 [ sec² x + tan x]
• 2^x [sec² x + log 2 tan x]

Solution

Q18. To verify Lagrange's Mean value Theorem which one is essential.

• Continous in closed interval and differentiable in open interval
• Continous in open interval and differentiable in closed interval
• Continous in closed interval and differentiable in closed interval
• Continous in open interval and differentiable in open interval

Solution

Continous in closed interval and differentiable in open interval.

Q19. If f (x) = (x - 4) (x - 6) (x - 8) satisfies mean value theorem in [4, 10] , find the value/s of c

• 4, 8
• 8, 2
• 4
• Mean value theorem does not hold good for this function

Solution

Since the  function f(x) satisfies, Mean Value Theorem Therefore, there exists c (4, 10) such that,     f'(c)    f '(c) = 3c² - 36c + 104   Substituting these values in (1), we have              

Q20. Which of the following function is not differentiable at x = 0?

• x²
• |x|
• cos x
• ax + b

Take Another Test