Limits and Derivatives - Class 11 Mathematics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Limits and Derivatives | NCERT | Mathematics | Class 11
Let $[k]$ denotes the greatest integer less than or equal to $k$. Then the number of positive integral solutions of the equation $\left[\frac{x}{\left[\pi^{2}\right]}\right]=\left[\frac{x}{\left[11 \frac{1}{2}\right]}\right]$
A) 29
B) 24
C) 21
D) 34
The correct option is B) 24
To solve the equation
$$\left[\frac{x}{\left[\pi^{2}\right]}\right]=\left[\frac{x}{\left[11 \frac{1}{2}\right]}\right]$$
it is necessary to determine the floor values of $\pi^2$ and $11 \frac{1}{2}$.
The calculations show that:
$\left[\pi^{2}\right] = \left[9.8696\right] = 9$
$\left[11 \frac{1}{2}\right] = [11.5] = 11$
The equation hence simplifies to: $$ \left[\frac{x}{9}\right] = \left[\frac{x}{11}\right] $$
Case I: $0 \leq \frac{x}{9} < 1$ and $0 \leq \frac{x}{11} < 1$
$\Rightarrow 0 \leq x < 9 \text{ and } 0 \leq x < 11$
Common values of $x$ are ${1, 2, 3, \ldots, 8}$ (8 values)
Case II: $1 \leq \frac{x}{9} < 2$ and $1 \leq \frac{x}{11} < 2$
$\Rightarrow 9 \leq x < 18 \text{ and } 11 \leq x < 22$
$\Rightarrow x \in {11, 12, \ldots, 17}$ (7 values)
Case III: $2 \leq \frac{x}{9} < 3$ and $2 \leq \frac{x}{11} < 3$
$\Rightarrow 18 \leq x < 27 \text{ and } 22 \leq x < 33$
$\Rightarrow x \in {22, 23, \ldots, 26}$ (5 values)
Case IV: $3 \leq \frac{x}{9} < 4$ and $3 \leq \frac{x}{11} < 4$
$\Rightarrow 27 \leq x < 36 \text{ and } 33 \leq x < 44$
$\Rightarrow x \in {33, 34, 35}$ (3 values)
Case V: $4 \leq \frac{x}{9} < 5$ and $4 \leq \frac{x}{11} < 5$
$\Rightarrow 36 \leq x < 45 \text{ and } 44 \leq x < 55$
$\Rightarrow x \in {44}$ (1 value)
The total number of positive integers satisfying the conditions across the cases is: $$ 8 + 7 + 5 + 3 + 1 = 24 $$
Therefore, the number of positive integral solutions to the given equation is 24.
The value of $\lim_{x\rightarrow\infty}\left(\frac{x^{2}-1}{x^{2}+1}\right)^{x^{2}}$ is
A) 1
B) $\mathrm{e}^{-1}$
C) $\mathrm{e}^{-2}$
D) $\mathrm{e}^{-3}$
Solution
Correct Answer: C) $e^{-2}$
Here's a step-by-step approach to solve the limit: $$ \lim _{x \rightarrow \infty}\left(\frac{x^2-1}{x^2+1}\right)^{x^2} $$
First, simplify the inner fraction: $$ \frac{x^2-1}{x^2+1} = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1} $$
Next, substitute the simplified fraction back into the original limit, adjusting the exponent accordingly: $$ \lim_{x \rightarrow \infty} \left(1 - \frac{2}{x^2 + 1}\right)^{x^2} $$
To handle the power term as $x \rightarrow \infty$, use the expansion associated with exponentials: $\lim_{x \rightarrow \infty} \left(1 + \frac{a}{x}\right)^x = e^a$. Here, $a = -\frac{2}{x^2+1} \approx -\frac{2}{x^2}$, where the approximation holds since $x^2 + 1 \approx x^2$ as $x$ becomes large.
Hence the limit turns into: $$ \lim_{x \rightarrow \infty} \exp\left(-\frac{2}{x^2+1} \cdot x^2\right) = \exp\left(-2\right) = e^{-2} $$
Thus, the value of the given limit is $e^{-2}$.
$\lim_{x \to a} \frac{a \sin x - x \sin a}{a x^{2} - x a^{2}}$
To address the given limit: $$ \lim_{x \to a} \frac{a \sin x - x \sin a}{a x^{2} - x a^{2}} $$
First, observe and simplify the denominator: $$ a x^2 - x a^2 = ax(x - a) $$
Substituting back in: $$ \lim_{x \to a} \frac{a \sin x - x \sin a}{ax(x - a)} $$
Introduce the substitution, with $t = x - a$. Hence, as $x \to a$, then $t \to 0$, and thus $x = t + a$. We rewrite the limit using this substitution: $$ \lim_{t \to 0} \frac{a \sin (t + a) - (t + a) \sin a}{a(t + a)t} $$
Apply the angle sum identity on $\sin (t+a)$: $$ \sin(t + a) = \sin t \cos a + \cos t \sin a $$
Now replacing $\sin(t + a)$: $$ \lim_{t \to 0} \frac{a(\sin t \cos a + \cos t \sin a) - (t+a)\sin a}{a(t+a)t} $$
Distribute and simplify the numerator: $$ = \lim_{t \to 0} \frac{a\sin t \cos a + a \cos t \sin a - t \sin a - a \sin a}{at(t+a)} $$
We can cancel and further simplify: $$ = \lim_{t \to 0} \frac{a\sin t \cos a + (a\cos t - a - t) \sin a}{at(t+a)} $$
Let $t \approx 0$ for approximation simplification on $\cos t \approx 1$ and $\sin t \approx t$: $$ = \lim_{t \to 0} \frac{at \cos a + (a - a - t) \sin a}{a^2 t} $$
The numerator simplifies to: $$ = \lim_{t \to 0} \frac{at \cos a - t \sin a}{a^2 t} $$ $$ = \lim_{t \to 0} \frac{t(a \cos a - \sin a)}{a^2 t} $$
Canceling $t$: $$ = \frac{a \cos a - \sin a}{a^2} $$
Conclusively, the simplified limit is: $$ \frac{\cos a - \frac{\sin a}{a}}{a} $$
This final expression represents the limit as $x \to a$ for the given function. The key derivations were simplifying the original trigonometric and polynomial expressions, applying angle identities, and strategic cancellation. The final limit, $\frac{\cos a - \frac{\sin a}{a}}{a}$ clearly encompasses both trigonometric relations and the effect of approaching $a$.
The two ends of a class are called:
A) range
B) observation
C) class width
D) class limits
The correct answer is D) class limits.
Example: In the class interval $10-15$, the numbers $10$ and $15$ are referred to as the class limits.
$\lim_{x \to a} f(x)$ exists if and only if
- $\lim_{x \to a} + f(x)$ and $\lim_{x \to a} - f(x)$ exist finitely.
- $\lim_{x \to a} + f(x) = \lim_{x \to a} - f(x) = f(a)$.
A True
B False
Solution
The correct answer is B False.
The existence of the limit of a function $f(x)$ as $x$ approaches $a$ (denoted as $\lim_{x \to a} f(x)$) requires two conditions:
- Both the right-hand limit ( $\lim_{x \to a^+} f(x)$) and the left-hand limit ($\lim_{x \to a^-} f(x)$) must exist and be finite.
- These limits must be equal to each other.
The presence of $\lim_{x \to a} f(x)$ does not depend on whether these limits are equal to the function's value at that point, $f(a)$. Thus, stating $\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = f(a)$ is not generally correct since the limit $\lim_{x \to a} f(x)$ can exist even if it does not equal $f(a)$. This diverges from the conditions provided in the question, which erroneously includes equivalence to $f(a)$.
Therefore, the statement giving criteria for the existence of $\lim_{x \to a} f(x)$ is not accurately reflected in the given choices, making option B (False) the right answer.
$\lim_{x \to 0} \frac{(a + h)^2 \sin(a + h) - a^2 \sin a}{h}$
Solution Reframed
To solve the limit $$ \lim_{x \to 0} \frac{(a + h)^2 \sin(a + h) - a^2 \sin a}{h}, $$ we first expand the squared term and distribute:
$$ (a+h)^2 \sin(a+h) = (a^2 + 2ah + h^2)\sin(a+h). $$
Substitute this into the original limit expression:
$$ \lim_{x \to 0} \frac{a^2 \sin(a+h) + 2ah \sin(a+h) + h^2 \sin(a+h) - a^2 \sin a}{h}. $$
Separate this into three terms using properties of limits and simplify:
$$ \lim_{x \to 0} \left[ \frac{a^2(\sin(a+h) - \sin a)}{h} + \frac{2ah \sin(a+h)}{h} + \frac{h^2 \sin(a+h)}{h} \right]. $$
The first term evaluates using the definition of the derivative of $\sin a$. The second term simplifies directly as the $\sin(a + h)$ part converges to $\sin a$, and the $h$ in the denominator and numerator cancel out. The last term goes to 0 as $h \to 0$. So, the expression becomes:
$$ a^2 \lim_{x \to 0} \frac{\sin(a+h) - \sin a}{h} + 2a \sin a + \lim_{x \to 0} h \sin(a+h). $$
Apply the small angle approximation to $\sin(\frac{h}{2}) \approx \frac{h}{2}$ in the derivative approximation:
$$ \sin(a+h) - \sin a \approx \cos\left(a + \frac{h}{2}\right) 2\sin\left(\frac{h}{2}\right), $$ and hence, $$ \lim_{x \to 0} \frac{\sin(a+h) - \sin a}{h} = \cos a \cdot \lim_{x \to 0} \frac{2\sin\left(\frac{h}{2}\right)}{h} = \cos a. $$
Substituting back, the simplified limit expression is:
$$ a^2 \cos a + 2a \sin a. $$
This yields the final simplified limit as:
$$ \boxed{a^2 \cos a + 2a \sin a}. $$
$$\lim _{x \to \infty}\left(\frac{x^{2} + 5x + 3}{x^{2} + x + 2}\right)^{x} =$$
A) 1
B) $\mathrm{e}^{2}$
C) $\mathrm{e}$
D) $\mathrm{e}^{4}$
The correct answer is D) $e^4$
This solution involves calculating the limit: $$\lim _{x \to \infty}\left(\frac{x^{2} + 5x + 3}{x^{2} + x + 2}\right)^{x}$$
To solve this limit, start by simplifying the inner fraction as $x$ approaches infinity. Notice that the leading terms in the numerator and denominator are $x^2$, so the higher-order terms will dominate the behavior of the function. By factoring out $x^2$ from the numerator and denominator, the fraction simplifies to: $$\frac{x^2(1 + 5/x + 3/x^2)}{x^2(1+1/x + 2/x^2)} = \frac{1 + 5/x + 3/x^2}{1 + 1/x + 2/x^2}$$
As $x \to \infty$, the terms $5/x$, $3/x^2$, $1/x$, and $2/x^2$ all approach zero. So, the fraction simplifies to: $$\frac{1 + 0 + 0}{1 + 0 + 0} = 1$$
The limit then becomes: $$\lim _{x \to \infty} 1^x = 1$$
However, the original expression requires a further detailed look. Consider expanding the numerator and denominator for a close $x$. Suppose: $$y = \left(\frac{x^2 + 5x + 3}{x^2 + x + 2}\right) = 1 + \frac{4x + 1}{x^2 + x + 2}$$
Approximate for a very large $x$: $$y \approx 1 + \frac{4}{x}$$
Now, evaluate $\lim_{x \to \infty} y^x$: $$\lim _{x \to \infty}\left(1 + \frac{4}{x} \right)^{x}$$
Using the limit definition of the exponential function: $$\lim _{x \to \infty}\left(1 + \frac{4}{x} \right)^{x} = e^{4}$$
Thus, the correct answer to the limit is $e^4$, making the option D) $e^4$ correct.
The value of $\lim _{x \rightarrow 0} \frac{(1+x)^{1/x}-e}{x}$ is
A $-\frac{e}{2}$
B $\frac{e}{2}$
C $-\frac{2}{e}$
D $\frac{2}{e}$
The correct answer is A $-\frac{e}{2}$. The problem involves evaluating the limit: $$ \lim _{x \rightarrow 0} \frac{(1+x)^{1/x}-e}{x} $$
By transforming the expression using Taylor expansion for logarithm, we begin from: $$ \lim _{x \rightarrow 0} \frac{e^{1/x \log(1+x)}-e}{x} $$ Using the expansion $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots$, we reframe the exponential function: $$ \lim _{x \rightarrow 0} \frac{e^{1 - \frac{x}{2} + \frac{x^2}{3} - \ldots}-e}{x} $$
Then, rewriting using the approximation and standard limit properties, we have: $$ e \lim _{x \rightarrow 0} \frac{e^{-\frac{x}{2} + \frac{x^2}{3} - \ldots} - 1}{x} = e \lim _{x \rightarrow 0} \left(-\frac{1}{2}\right) = -\frac{e}{2} $$
For a clear step-by-step re-calculation using L'Hopital's Rule, here's the alternative approach: $$ \begin{aligned} \lim _{x \rightarrow 0} \frac{(1+x)^{1/x} - e}{x} &= \lim _{x \rightarrow 0} \frac{e^{\frac{1}{x} \ln(1+x)} - e}{x} \ &= \lim _{x \rightarrow 0} \frac{e^{\frac{1}{x} \ln(1+x)} \left(\frac{1}{x} \cdot \frac{1}{1+x} - \frac{\ln(1+x)}{x^2}\right)}{1} \ &= e \times \lim _{x \rightarrow 0} \frac{-\frac{1}{2(1+x)^2}}{1} \ &= -\frac{e}{2} \end{aligned} $$
Hence, the solution remains the same using both the Taylor series expansion and L'Hopital's Rule, and the answer is A $-\frac{e}{2}$.
$\lim_{x \rightarrow 0} \frac{\log(a+x) - \log(a)}{x}$
We begin with the given limit expression:
$$ \lim_{x \rightarrow 0} \frac{\log(a + x) - \log(a)}{x} $$
First, utilize the properties of logarithms to simplify the numerator:
$$ \log(a + x) - \log(a) = \log\left( \frac{a + x}{a} \right) $$
Then, the limit expression becomes:
$$ \lim_{x \rightarrow 0} \frac{\log\left( \frac{a + x}{a} \right)}{x} $$
Simplify the fraction inside the logarithm:
$$ \frac{a + x}{a} = 1 + \frac{x}{a} $$
So, the limit now is:
$$ \lim_{x \rightarrow 0} \frac{\log\left( 1 + \frac{x}{a} \right)}{x} $$
To progress, observe that for very small ( x ):
$$ \log\left(1 + y\right) \approx y \quad \text{when} \quad y \rightarrow 0 $$
Here ( y = \frac{x}{a} ), thus:
$$ \log\left(1 + \frac{x}{a}\right) \approx \frac{x}{a} $$
Using this approximation:
$$ \lim_{x \rightarrow 0} \frac{\frac{x}{a}}{x} = \frac{1}{a} $$
Therefore, the limit is:
$$ \boxed{\frac{1}{a}} $$
The expression $2x^{2}+4x+7$ has minimum value $m$ at $x=\alpha$. The ordered pair $(\alpha, m)$ is:
A) $(1, -5)$
B) $(-1, -5)$
C) $(-1, 5)$
D) $(1, 5)$
To determine the ordered pair $(\alpha, m)$ where the expression $2x^2 + 4x + 7$ has its minimum value, we need to rewrite the expression in a more convenient form.
Given: $$2x^2 + 4x + 7$$
First, let's complete the square for the quadratic expression. Start by factoring out the common factor, which is 2:
$$2(x^2 + 2x) + 7$$
Next, we complete the square inside the parentheses. We add and subtract 1 inside the parentheses:
$$2(x^2 + 2x + 1 - 1) + 7$$
This simplifies to:
$$2((x + 1)^2 - 1) + 7$$
Distribute the 2:
$$2(x + 1)^2 - 2 + 7$$
Combine the constants:
$$2(x + 1)^2 + 5$$
In this form, it is clear that the expression $2(x + 1)^2 + 5$ reaches its minimum when $(x + 1)^2$ is 0 because the square of any real number is non-negative. Thus, set (x + 1 = 0):
$$x = -1$$
When $x = -1$, the expression simplifies to:
$$2(0) + 5 = 5$$
Thus, the minimum value ( m ) is 5, and it occurs at $\alpha = -1$. Therefore, the ordered pair is:
Answer: $(\alpha, m) = \mathbf{(-1, 5)}$
So, the correct choice is:
C) $(-1, 5)$
The maximum value of $8x - x^{2}$ is _____
To find the maximum value of the given expression $8x - x^{2}$, we can approach it using the method of completing the square.
Start with the given expression: $$ 8x - x^2 $$
Rewrite the expression in a more familiar quadratic form: $$ -(x^2 - 8x) $$
Complete the square inside the parentheses: $$ -(x^2 - 8x) = -\left(x^2 - 8x + 16 - 16\right) $$
Notice that $8x$ has been split into $2 \cdot 4 \cdot x$ and $16$ is added and subtracted.
Group the perfect square trinomial and the constant separately: $$ -(x - 4)^2 + 16 $$
Now we have the expression in the form of $a - b(x - h)^2$, which makes it easy to identify the maximum value.
Identify the nature of the quadratic function: $$ -(x - 4)^2 + 16 $$
Here, $-(x - 4)^2$ represents a parabola opening downwards. The maximum value of $-(x - 4)^2$ occurs when $(x - 4)^2 = 0$.Solve for when $(x - 4)^2 = 0$: $$ (x - 4)^2 = 0 \quad \Rightarrow \quad x = 4 $$
Substitute $x = 4$ back into the expression to find the maximum value: $$ -(4 - 4)^2 + 16 = 0 + 16 = 16 $$
Thus, the maximum value of the given expression $8x - x^2$ is $ \boxed{16} $.
The maximum value of the expression $\frac{x^{2}+x+1}{2 x^{2}-x+1}$, for $x \in \mathbb{R}$, is:
A $\frac{7+2 \sqrt{7}}{7}$
B $\frac{7-2 \sqrt{7}}{7}$
C $\frac{7}{3}$
D $\frac{14+2 \sqrt{7}}{7}$
To find the maximum value of the expression
$$ \frac{x^2 + x + 1}{2x^2 - x + 1} \text{ for } x \in \mathbb{R}, $$
we start by setting
$$ y = \frac{x^2 + x + 1}{2x^2 - x + 1}. $$
Upon cross-multiplying, we get:
$$ y(2x^2 - x + 1) = x^2 + x + 1. $$
Rearranging terms, the equation becomes:
$$ 2yx^2 - xy + y = x^2 + x + 1. $$
Further simplifying, we get:
$$ 2yx^2 - x^2 - xy - x + y - 1 = 0. $$
Group similar terms:
$$ (2y - 1)x^2 - (y + 1)x + (y - 1) = 0. $$
We can compare this to the standard quadratic form $ax^2 + bx + c = 0$, where we identify the coefficients as:
$a = 2y - 1$
$b = -(y + 1)$
$c = y - 1$
For the roots of this quadratic equation to be real, the discriminant must be non-negative:
$$ b^2 - 4ac \geq 0. $$
Substitute the values of $a$, $b$, and $c$:
$$ [-(y+1)]^2 - 4(2y-1)(y-1) \geq 0. $$
This simplifies to:
$$ (y + 1)^2 - 4(2y - 1)(y - 1) \geq 0. $$
Expanding and simplifying:
$$ y^2 + 2y + 1 - 4(2y^2 - 3y + 1) \geq 0, $$ $$ y^2 + 2y + 1 - 8y^2 + 12y - 4 \geq 0, $$ $$ -7y^2 + 14y - 3 \geq 0. $$
Rewriting, we get:
$$ 7y^2 - 14y + 3 \leq 0. $$
Next, using the quadratic formula to solve for $y$, we have:
$$ y = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 7 \cdot 3}}{2 \cdot 7}, $$
which simplifies to:
$$ y = \frac{14 \pm \sqrt{196 - 84}}{14}, $$ $$ y = \frac{14 \pm \sqrt{112}}{14}, $$ $$ y = \frac{14 \pm 4\sqrt{7}}{14}, $$ $$ y = 1 \pm \frac{2\sqrt{7}}{7}. $$
Thus, the possible values of $y$ are:
$$ y = 1 + \frac{2\sqrt{7}}{7} \quad \text{and} \quad y = 1 - \frac{2\sqrt{7}}{7}. $$
Out of these, the maximum value is:
$$ \boxed{\frac{7 + 2\sqrt{7}}{7}}. $$
Hence, the correct answer is Option A: $\frac{7 + 2\sqrt{7}}{7}$.
Out of the two roots of $x^{2}-2 \lambda x+\lambda^{2}-1=0$, one is greater than 4 and the other root is less than 4, then the limits of $\lambda$ are:
$ 3 < \lambda < 5$
$\lambda<3$
$ \lambda < 5$
$ 5 < \lambda < 6$
To find the limits of $\lambda$ given that one root of the equation $x^2 - 2 \lambda x + \lambda^2 - 1 = 0$ is greater than 4 and the other is less than 4, follow these steps:
Given the quadratic equation: $$ x^2 - 2\lambda x + \lambda^2 - 1 = 0 $$
Given conditions:
One root is greater than 4
One root is less than 4
To find the roots, use the property of quadratic equations. Suppose the roots are $r_1$ and $r_2$. Since $r_1 + r_2 = 2\lambda$ and $r_1 \cdot r_2 = \lambda^2 - 1$, and knowing that one root is greater than 4 and the other is less than 4, let ( r_1 > 4 ) and ( r_2 < 4 ).
Applying the condition: Substitute $x = 4$ in the quadratic equation to find the point where the function changes direction: $$ 4^2 - 2\lambda(4) + \lambda^2 - 1 < 0 $$
Simplify the inequality: $$ 16 - 8\lambda + \lambda^2 - 1 < 0 \implies \lambda^2 - 8\lambda + 15 < 0 $$
Factorize the quadratic expression: $$ \lambda^2 - 8\lambda + 15 = (\lambda - 5)(\lambda - 3) < 0 $$
Determine the intervals: For the product $(\lambda - 5)(\lambda - 3) < 0$ to be true, $\lambda$ must lie between 3 and 5: $$ 3 < \lambda < 5 $$
Conclusion: Given that $\lambda$ must be between 3 and 5 for one root to be greater than 4 and the other less than 4, $\boxed{3 < \lambda < 5}$ is the required range.
Hence, the limits of $\lambda$ are: $$ \boxed{3 < \lambda < 5} $$
Which of the following is the correct expression for lower class limit? (m = class midpoint, u = upper class limit, I = lower class limit)
A) I = m - 2u
B) I = m - u
C) I = 2m + u
D) I = 2m - u
The correct option is D: $I = 2m - u $.
The class midpoint is the middle value of each data class. To calculate the class midpoint, you take the average of the upper and lower class limits:
$$ \text{Class Midpoint}(m) = \frac{\text{Upper class limit}(u) + \text{Lower class limit}(I)}{2} $$
Simplifying this, we get:
$$ \begin{aligned} m & = \frac{u + I}{2} \ \Rightarrow 2m & = u + I \ \Rightarrow I & = 2m - u \end{aligned} $$
Thus, the expression for the lower class limit $I $ in terms of the class midpoint $ m$ and the upper class limit $ u $ is $ \mathbf{I = 2m - u} $.
Let $$ f(x) = \begin{array}{ll} x^{3} - x^{2} + 10x - 7, & x \leq 1 \\ -2x + \log_{2}(k^{2} - 4), & x > 1 \end{array} $$
The set of values of $k$ for which $f(x)$ has the greatest value at $x = 1) is:
A. $[-6, 6]$
B. $[-6, -2] \cup [2, 6]$
C. $[-6, -2) \cup (2, 6]$
D. $(2, 6)$
The given function is defined as:
$$ f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & \text{for } x \leq 1 \ -2x + \log_2(k^2 - 4), & \text{for } x > 1 \end{cases} $$
To determine the values of $k$ for which $f(x)$ has the greatest value at $x = 1$, we need to consider the right-hand limit of $f$ at $x = 1$.
Steps:
Evaluate $f(1)$: $$ f(1) = 1^3 - 1^2 + 10 \cdot 1 - 7 = 1 - 1 + 10 - 7 = 3 $$
Consider the right-hand limit: $$ \lim_{h \rightarrow 0^+} f(1 + h) = \lim_{h \rightarrow 0^+} (-2(1 + h) + \log_2(k^2 - 4)) $$ This simplifies to: $$ -2 - 2h + \log_2(k^2 - 4) \quad \text{as} \quad h \rightarrow 0 $$ So, $$ \lim_{h \rightarrow 0^+} f(1 + h) = -2 + \log_2(k^2 - 4) $$
Set up the inequality for the maxima: $$ -2 + \log_2(k^2 - 4) \leq 3 $$
Solve the inequality: $$ \log_2(k^2 - 4) \leq 5 $$ Converting to exponential form: $$ k^2 - 4 \leq 2^5 $$ This simplifies to: $$ k^2 - 4 \leq 32 $$ $$ k^2 \leq 36 $$ Additionally, since the argument inside the logarithm must be positive: $$ k^2 - 4 > 0 $$ Which means: $$ k^2 > 4 $$
Combine the constraints to find the solution set for $k$: $$ 4 < k^2 \leq 36 $$ This leads to: $$ k \in (-6, -2) \cup (2, 6) $$
Therefore, the set of values for $k$ for which $f(x)$ has the greatest value at $x = 1$ is given by option C:
$$ \boxed{[-6, -2) \cup (2, 6]} $$
Let $f(x) = \frac{1}{3}x \sin x - (1-\cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is
(A) 4
(B) 3
(C) 2
(D) 1
The correct option is C: 2
Given the function $f(x) = \frac{1}{3}x \sin x - (1 - \cos x)$, we need to find the smallest positive integer $k$ such that
$$\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$$
First, consider the limit we are working with:
$$\lim_{x \rightarrow 0} \frac{f(x)}{x^k} = \lim_{x \rightarrow 0} \frac{\frac{x \sin x}{3} - 1 + \cos x}{x^k}$$
This initially results in a $\frac{0}{0}$ indeterminate form. To resolve this, we apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator until the limit can be evaluated. Differentiating the numerator and the denominator:
First Derivative: $$ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}\left(\frac{x \sin x}{3} - 1 + \cos x\right)}{\frac{d}{dx}(x^k)} = \lim_{x \rightarrow 0} \frac{\frac{\sin x}{3} + \frac{x \cos x}{3} - \sin x}{k x^{k-1}}$$
This results again in a $\frac{0}{0}$ form.
Second Derivative: $$ \lim_{x \rightarrow 0} \frac{\frac{\cos x}{3} + \frac{\cos x}{3} - \frac{x \sin x}{3} - \cos x}{k(k-1) x^{k-2}}$$
Now, evaluate the numerator at $x = 0$:
$\cos 0 = 1$
$\sin 0 = 0$
Therefore, the numerator becomes: $$ \frac{1}{3} + \frac{1}{3} - \frac{0}{3} - 1 = \frac{1}{3} + \frac{1}{3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3} $$
This limit will be non-zero if and only if the power of $x$ in the denominator matches the leading power of $x$ in the numerator. It's clear that for the limit to not be zero or undefined, $k$ must be 2.
Hence, the smallest positive integer $k$ that satisfies the condition is 2.
The value of
$$ \lim _{n \rightarrow \infty} \frac{n!}{(n+1)!-n!} $$
is:
Let's evaluate the given limit step-by-step:
$$ \lim _{n \rightarrow \infty} \frac{n!}{(n+1)! - n!} $$
First, recall that $ (n+1)! $ can be expanded as $ (n+1) \cdot n! $. Substituting this into the expression, we get:
$$ \lim _{n \rightarrow \infty} \frac{n!}{(n+1) \cdot n! - n!} $$
Next, factor out $ n! $ from the denominator:
$$ \lim _{n \rightarrow \infty} \frac{n!}{n![(n+1) - 1]} $$
Simplify the denominator inside the brackets:
$$ \lim _{n \rightarrow \infty} \frac{n!}{n! \cdot n} $$
Since $ n! $ in the numerator and denominator cancel each other out, the expression simplifies to:
$$ \lim _{n \rightarrow \infty} \frac{1}{n} $$
As $ n $ approaches infinity, $ \frac{1}{n} $ approaches 0. Therefore, the value of the limit is:
$$ \boxed{0} $$
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