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 \(x = 2\).

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 \(x = 2\).

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

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.

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.

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.

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 \(x = 0\).

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

$\frac {dy}{dx}=1$

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
$\frac {dy}{dx} = \frac {dy}{dt} \div \frac {dx}{dt} = cot 2t$

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

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 is (d)

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

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.

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

This is continous and differentiable at x=0

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

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, \( f'(x) = 3x^2 \cos(x^3) \).

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, \( f'(1) = 5 \).

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, \( g'(4) = \frac{1}{4} \).

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 \( x = 0 \).

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

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

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

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

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).

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

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

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

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

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

**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

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