physicscatalyst.com logo




Continuity and differentiability Class 12 Maths Important questions




Question 1
Determine the values of \(k\) for which the function \(f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 2k - x, & \text{if } x > 2 \end{cases}\) is continuous at \(x = 2\).

Answer

For the function \(f(x)\), we have:
1. \(f(2) = 4k\), which is defined.
2. \(\lim_{{x \to 2}} f(x)\) exists if and only if \(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x)\). Calculate the left and right limits:

\(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} kx^2 = 4k\)
\(\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (2k - x) = 2k - 2\)
For the limit to exist, we need \(4k = 2k - 2\).
Solving for \(k\): \(4k = 2k - 2\)
\(2k = -2\)
\(k = -1\)
3. Since \(k = -1\) makes the left and right limits equal, the limit \(\lim_{{x \to 2}} f(x)\) exists.
So, for \(k = -1\), the function \(f(x)\) is continuous at \(x = 2\).


Question 2
Is the function \(g(x) = \frac{1}{x^2 - 4}\) continuous at \(x = 2\)?

Answer

The function \(g(x) = \frac{1}{x^2 - 4}\) is continuous at \(x = 2\) if and only if \(g(2)\) is defined, which means the denominator \(x^2 - 4\) is not zero at \(x = 2\).
At \(x = 2\), we have:
\(x^2 - 4 = 2^2 - 4 = 0\)
Since the denominator is zero at \(x = 2\), the function is not defined at \(x = 2\), and therefore, it is not continuous at \(x = 2\).


Question 3
Determine the values of \(a\) and \(b\) that make the function \(h(x) = \begin{cases} -1, & \text{if } x \leq 0 \\ ax^2 + b, & \text{if } 0 < x < 1 \\ 3x - 2, & \text{if } x \geq 1 \end{cases}\) continuous at x=0 and x=1

Answer

for x=0
1. \(h(0) = -1 \), which is defined.
2. \(\lim_{{x \to 0}} h(x)\) exists if and only if \(\lim_{{x \to 0^-}} h(x) = \lim_{{x \to 0^+}} h(x)\). Calculate the left and right limits:
\(\lim_{{x \to 0^-}} h(x) = \lim_{{x \to 0^-}} -1 = -1\)
\(\lim_{{x \to 0^+}} h(x) = \lim_{{x \to 0^+}} ax^2 +b = b\)
For the limit to exist, we need \(b= -1\).

for x=1
1. \(h(1) = a(1)^2 + b = a + b\), which is defined.
2. \(\lim_{{x \to 1}} h(x)\) exists if and only if \(\lim_{{x \to 1^-}} h(x) = \lim_{{x \to 1^+}} h(x)\). Calculate the left and right limits:
\(\lim_{{x \to 1^-}} h(x) = \lim_{{x \to 1^-}} (ax^2 + b) = a + b\)
\(\lim_{{x \to 1^+}} h(x) = \lim_{{x \to 1^+}} (3x - 2) = 3 - 2 = 1\)
For the limit to exist, we need \(a + b = 1\).
Now we have b=-1, Therefore a=2


Question 4
Determine the interval on which the function \(f(x) = \frac{x}{x^2 + 1}\) is continuous.

Answer

The function \(f(x) = \frac{x}{x^2 + 1}\) is continuous on its entire domain. The domain of \(f(x)\) is all real numbers since the denominator \(x^2 + 1\) is always positive for real values of \(x\). Therefore, \(f(x)\) is continuous for all real numbers.


Question 5
Is the function \(g(x) = \sqrt{x - 1}\) continuous for all real numbers?

Answer

The function \(g(x) = \sqrt{x - 1}\) is continuous for all \(x\) values where the expression under the square root is non-negative. Therefore, it is continuous for \(x \geq 1\), but it is not defined for \(x < 1\). So, it is continuous for \(x \geq 1\) but not for all real numbers.


Question 6
Is the function \(h(x) = \frac{x^2 - 1}{x^2 + 1}\) continuous for all real numbers?

Answer

The function \(h(x) = \frac{x^2 - 1}{x^2 + 1}\) is continuous for all real numbers. The denominator \(x^2 + 1\) is always positive for real values of \(x\), so there are no values of \(x\) for which the function is discontinuous.


Question 7
Investigate the continuity of the function \(f(x) = \frac{|x|}{x}\) at \(x = 0\).

Answer

For the function \(f(x)\), we have:
1. \(f(0) = \frac{|0|}{0} = \frac{0}{0}\), which is an indeterminate form. Therefore, \(f(0)\) is not defined.
2. To investigate the limit, we can consider the left and right limits separately:
\(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} \frac{|x|}{x} = \lim_{{x \to 0^-}} \frac{-x}{x} = \lim_{{x \to 0^-}} (-1) = -1\)
\(\lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} \frac{|x|}{x} = \lim_{{x \to 0^+}} \frac{x}{x} = \lim_{{x \to 0^+}} (1) = 1\)
Since the left and right limits are different, the limit \(\lim_{{x \to 0}} f(x)\) does not exist.
3. Since \(f(0)\) is not defined, the function is not continuous at \(x = 0\).


Question 8
if $y =x^{\frac {1}{x}}$ then find $\frac {dy}{dx}$ at x=1

Answer

$y =x^{\frac {1}{x}}$
$log y = \frac {1}{x} log x$
$x log y = log x$
$\frac {x}{y} \frac {dy}{dx} + log y = \frac {1}{x}$
$\frac {dy}{dx}=\frac {y(1-x log y)}{x^2}$
At x=1, y=1
Therefore
$\frac {dy}{dx}=1$


Question 9
if x= sin 2t
y=a(cos 2t + log tan t)
then find $\frac {dy}{dx}$

Answer

$\frac {dx}{dt}=2 cos 2t$
$\frac {dy}{dt}=-2 sin 2t + \frac {1}{tan t} sec^2 t$
$=-2 sin 2t + \frac {2}{sin 2t}$ $= \frac {2 cos^2 2t}{sin 2t}$ Now $\frac {dy}{dx} = \frac {dy}{dt} \div \frac {dx}{dt} = cot 2t$


Question 10
The value of k for which function
\(f(x) = \begin{cases} x^2, & \text{if } x \geq 0 \\ kx, & \text{if } x < 0 \end{cases}\) is differentiable at x = 0 is

Question 11
The function f(x) = [x] where [x] denotes the greatest integer function is continous at
(a) 0
(b) 1
(c) -1
(d) 1.3

Answer

Answer is (d)


Question 12
Determine the values of \(k\) for which the function \(f(x) = \begin{cases} \frac {1-cosx}{2x^2}, & \text{if } x \ne 0 \\ k, & \text{if } x =0 \end{cases}\) is continuous at \(x = 0\).

Answer

For the function \(f(x)\), we have:
1. \(f(0) = k\), which is defined.
2. \(\lim_{{x \to 0}} f(x)\) exists if and only if \(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x)\). Calculate the left and right limits:
\(\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} \frac {1-cosx}{2x^2} = \lim_{{x \to 0^-}} \frac {2sin^2 (x/2)}{2x^2} = 1/4 \)
\(\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} \frac {2sin^2 (x/2)}{2x^2} = 1/4 \)
Since the function is continous
$f(0) =\lim_{{x \to 0}} f(x) $
or k=1/4


Question 13
Check if the function \( h(x) = |x - 2| \) is differentiable at \( x = 2 \).

Answer

The function \( h(x) \) is not differentiable at \( x = 2 \) as the left-hand derivative and right-hand derivative at \( x = 2 \) are not equal.


Question 14
The function f(x) =x|x|. Check the continuity and differentialbility x=0

Answer

This is continous and differentiable at x=0


Question 15
Determine the values of a and b for which the function \(f(x) = \begin{cases} ax +b, & \text{if } 0< x \leq 1 \\ 2x^2 - x, & \text{if } 1 < x < 2 \end{cases}\) is a differentiable in (0,2)

Answer

Since the function differentiable in (0,2), it must be differetiable at x=1
Consider left hand limit
\(\lim_{{h \to 0^-}} \frac {f(1+h) - f(1)}{h} = \lim_{{h \to 0^-}} \frac {a(1+h) + b - (a+b)}{h} =a \)
Right Hand limit
\(\lim_{{h \to 0^+}} \frac {f(1+h) - f(1)}{h} = \lim_{{h \to 0^-}} \frac {2(1+h)^2 -(1+h) - (2-1)}{h} =\lim_{{h \to 0^-}} (2h+3) =3 \)
Since both the limit should be equal a=3
Since it is differetiable at x=1,it must be continous also
\(\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x)\)
a+ b =1
Now a=3
Therefore b=-2


Question 16
Show that the function \( f(x) = x^2 - 3x + 2 \) is continuous at \( x = 1 \).

Answer

To prove continuity at \( x = 1 \), we need to show that \( \lim_{x \to 1} f(x) = f(1) \). Calculating the limit and the function value at \( x = 1 \) gives the same result, 0. Hence, \( f(x) \) is continuous at \( x = 1 \).
Differentiate \( f(x) = \sin(x^3) \) with respect to \( x \).
Using the chain rule, \( f'(x) = 3x^2 \cos(x^3) \).


Question 17
Find the derivative of \( g(x) = e^{2x} \ln(x) \).

Answer

Applying the product rule, \( g'(x) = e^{2x} \ln(x) \cdot 2 + \frac{e^{2x}}{x} \).
If \( f(x) = x^3 + 2x + 1 \), find \( f'(1) \).
\( f'(x) = 3x^2 + 2 \). So, \( f'(1) = 5 \).


Question 18
Find the second derivative of \( f(x) = \cos(x) \).

Answer

First derivative \( f'(x) = -\sin(x) \), and second derivative \( f''(x) = -\cos(x) \).
If \( g(x) = \sqrt{x} \), find the derivative at \( x = 4 \).
\( g'(x) = \frac{1}{2\sqrt{x}} \). Therefore, \( g'(4) = \frac{1}{4} \).


Question 19
Prove that the function \( f(x) = x^3 - 3x + 2 \) is continuous at \( x = 0 \).

Answer

Since \( f(x) \) is a polynomial, it's continuous at all points, including \( x = 0 \).


Question 20
For the function \( f(x) = \ln(x) \), find the derivative at \( x = e \).

Answer

\( f'(x) = \frac{1}{x} \). So, \( f'(e) = \frac{1}{e} \).


Question 21
Find the derivative of the function \( f(x) = \tan^{-1}(x^2) \).

Answer

Using the chain rule, \( f'(x) = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4} \).


Question 22
Determine the differentiability of \( f(x) = \sqrt{|x|} \) at \( x = 0 \).

Answer

The function \( f(x) \) is not differentiable at \( x = 0 \) as the derivative is not defined (the slope is infinite or undefined at this point).


Question 23
Differentiate \( f(x) = (3x^2 - 2)^5 \).

Answer

Using the chain rule, \( f'(x) = 5(3x^2 - 2)^4 \cdot 6x = 30x(3x^2 - 2)^4 \).


Question 24
Find the second derivative of \( f(x) = \frac{1}{1 + x^2} \).

Answer

First derivative \( f'(x) = \frac{-2x}{(1 + x^2)^2} \) and second derivative \( f''(x) = \frac{2(3x^2 - 1)}{(1 + x^2)^3} \).


Question 25
If \( f(x) = x^2 \ln(x) \), find the derivative at \( x = 1 \).

Answer

Using the product rule, \( f'(x) = 2x \ln(x) + x \). Therefore, \( f'(1) = 1 \).



Also Read




Go back to Class 12 Main Page using below links
Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology


Latest Updates
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7