Limits and Continuity: part 11: IITJEE Maths tutorial problems for practice

Problem 1:

A function f(x) is defined as follows:

f(x) = \frac{(e^{2x}-1)(1-\cos{x})}{\tan^{2}{(x)}\log{(1+2x)}} when x \neq 0

f(0) = \log{a} is continuous at x=0.

The value of a should be

(i) \frac{e}{2} (b) \frac{1}{2e} (c) 2 (d) none

Problem 2:

If f(x) = \frac{e^{(2x)} + e^{(-2x)} -2}{1-\cos{(4x)}} when x \neq 0 is continuous at x=0, then what is the value of f(0)

Problem 3:

Given f(x) = x + a, when -1  \leq x \leq 0

and f(x) = x + b when 0 < x \leq 1

and f(x) = c -x when 1 < x \leq 2

if f is continuous at x=0 and x=1 and f(2)=1, then the value of 3a+b-2c=

(i) 0 (ii) 1 (iii) 2 (iv) 3

Problem 4:

If the function f(x) is continuous on its domain where

f(x) = x^{2} + ax + b for 0 \leq x < 2

f(x)=4x-1 for 2 \leq x < 4

f(x)=ax^{2+17b} for 4 \leq x \leq 6

then the quadratic equation whose roots are 2a and 2b is:

(i) x^{2}+2x-8 (b) x^{2}-2x-8=0 (c) x^{2}+2x+8 (d) x^{2}-2x+8=0

Problem 5:

The value of c for which the function

f(x) = \frac{\sin{(x)} + \sin{((a+1)x)}}{x} when x<0

f(x) = c when x=0

f(x) = \frac{(x+bx^{2})^{\frac{1}{2}}-x^{\frac{1}{2}}}{bx^{\frac{3}{2}}}

is continuous at x=0 is

(i) 1/2 (ii) -1/2 (iii) 2 (iv) -2

Problem 6:

If f(x) = \frac{\sin{x\pi}}{x-1}+a when x<1

f(x) = 2x, when x=1

f(x)= \frac{1+\cos{x\pi}}{\pi (1-x)^{2}} + b when x>1

is continuous at x=1, then a and b have the values:

(i) 3\pi, 3\frac{\pi}{2} (ii) 3\pi, \frac{\pi}{2} (iii) \pi, \frac{\pi}{2} (iv) \pi, 3\frac{\pi}{2}

Problem 7:

If f(x) = \frac{(\sin{x} - \cos{x})^{2}}{\sqrt{2}-\sin{x}-\cos{x}}, when x \neq \frac{\pi}{4} is continuous at x=\frac{\pi}{4} then f(\frac{\pi}{4})=

(a) 1/2 (b) -1/2 (c) 2 (d) none of these

Problem 8:

If f(x)= \frac{x+1}{x+2} and g(x)=\frac{1}{x}, then \lim_{x \rightarrow 2} (g+f)(x)=

(i) 4/3 (b) 5/3 (c) 2 (d) 7/3

Problem 9:

Evaluate the following: \lim_{x \rightarrow 4} \frac{(x^{2}-x-12)^{18}}{(x^{3}-8x^{2}+16x)^{9}}


Nalin Pithwa

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: