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 \(x = 2\).
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 \(x = 2\).
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
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.
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.
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.
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 \(x = 0\).
$y =x^{\frac {1}{x}}$
$log y = \frac {1}{x} log x$
$x log y = log x$
$\frac {x}{y} \frac {dy}{dx} + log y = \frac {1}{x}$
$\frac {dy}{dx}=\frac {y(1-x log y)}{x^2}$
At x=1, y=1
Therefore
$\frac {dy}{dx}=1$
$\frac {dx}{dt}=2 cos 2t$
$\frac {dy}{dt}=-2 sin 2t + \frac {1}{tan t} sec^2 t$
$=-2 sin 2t + \frac {2}{sin 2t}$
$= \frac {2 cos^2 2t}{sin 2t}$
Now
$\frac {dy}{dx} = \frac {dy}{dt} \div \frac {dx}{dt} = cot 2t$
Answer is (d)
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
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.
This is continous and differentiable at x=0
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
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, \( f'(x) = 3x^2 \cos(x^3) \).
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, \( f'(1) = 5 \).
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, \( g'(4) = \frac{1}{4} \).
Since \( f(x) \) is a polynomial, it's continuous at all points, including \( x = 0 \).
\( f'(x) = \frac{1}{x} \). So, \( f'(e) = \frac{1}{e} \).
Using the chain rule, \( f'(x) = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4} \).
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).
Using the chain rule, \( f'(x) = 5(3x^2 - 2)^4 \cdot 6x = 30x(3x^2 - 2)^4 \).
First derivative \( f'(x) = \frac{-2x}{(1 + x^2)^2} \) and second derivative \( f''(x) = \frac{2(3x^2 - 1)}{(1 + x^2)^3} \).
Using the product rule, \( f'(x) = 2x \ln(x) + x \). Therefore, \( f'(1) = 1 \).
