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

Regards,

Nalin Pithwa

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