# 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 $x = 2$.

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 $x = 2$?

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

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.

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?

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?

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 $x = 0$.

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

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

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

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 $x = 0$.

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 $x = 2$.

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

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)

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 $x = 1$.

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 $g(x) = e^{2x} \ln(x)$.

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 $f(x) = \cos(x)$.

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 $x = 0$.

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 $x = e$.

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

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

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 $x = 0$.

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 $f(x) = (3x^2 - 2)^5$.

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 $f(x) = \frac{1}{1 + x^2}$.

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 $x = 1$.

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

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