Evaluate
Question: Evaluate $\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}$ Solution: To evaluate: $\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}$ Formula used: L'Hospital's rule Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where $\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$ then $\lim _{x ...
Read More →Mean and standard deviation of 100 items are 50 and 4,
Question: Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all items and the sum of the squares of the items. Solution: Given mean and standard deviation of 100 items are 50 and 4, respectively Now we have to find the sum of all items and the sum of the squares of the items As per given criteria, Number of items, $\mathrm{n}=100$ Mean of the given items, $\bar{x}=50$ But we know, $\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$ Substitut...
Read More →$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}$
[question] Question. $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{2}}$ We know that $\sin 2 x=2 \sin x \cos x$, using this formula we get Again by taking $2 \sin x$ common we get $\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^{3}}$ Now we have $\cos x=1-2 \sin ^{2}(x / 2)$ Using this identity above equation can be written as...
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Question: Evaluate $\lim _{h \rightarrow 0}\left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\right)$ Solution: To evaluate: $\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$ Formula used: L'Hospital's rule Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval $I$ except at a point a where $\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$ then $\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\m...
Read More →The mean life of a sample of 60 bulbs was 650
Question: The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation. Solution: Given the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours Now we have to find the overall standard deviation As per given criteria, in...
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Question: Evaluate $\lim _{x \rightarrow a}\left(\frac{\sqrt{x}-\sqrt{a}}{x-a}\right)$ Solution: To evaluate: $\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a}$ Formula used: We have, $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ As $\mathrm{x} \rightarrow \mathrm{a}$, we have $\lim _{x \rightarrow a} \frac{x^{\frac{1}{2}}-a^{\frac{1}{2}}}{x-a}=\frac{1}{2} a^{\frac{1}{2}-1}$ $\lim _{x \rightarrow a} \frac{x^{\frac{1}{2}}-a^{\frac{1}{2}}}{x-a}=\frac{1}{2 \sqrt{a}}$ Thus, the value of $\lim _{x \righta...
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Question: Evaluate $\lim _{x \rightarrow 1}\left(\frac{x^{n}-1}{x-1}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 1} \frac{x^{n}-1}{x-1}$ Formula used: We have, $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ As $\mathrm{x} \rightarrow \mathrm{a}$, we have $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ $\lim _{x \rightarrow a} \frac{x^{n}-1}{x-1}=n$ Thus, the value of $\lim _{x \rightarrow a} \frac{x^{n}-1}{x-1}$ is $n$....
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Question: Evaluate $\lim _{x \rightarrow a}\left\{\frac{(x+2)^{5 / 3}-(a+2)^{5 / 3}}{x-a}\right\}$ Solution: To evaluate: $\lim _{x \rightarrow a}\left\{\frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}\right\}$ Formula used: We have, $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ As $\mathrm{x} \rightarrow \mathrm{a}$, we have $\lim _{x \rightarrow a}\left\{\frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}\right\}=\lim _{x \rightarrow a}\left\{\frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{(x+2)-(a+2)...
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Question: Evaluate $\lim _{x \rightarrow a}\left(\frac{x^{5 / 2}-a^{5 / 2}}{x-a}\right)$ Solution: To evaluate: $\lim _{x \rightarrow a} \frac{x^{\frac{5}{2}}-a^{\frac{5}{2}}}{x-a}$ Formula used: We have, $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ As $x \rightarrow$ a, we have $\lim _{x \rightarrow a} \frac{x^{\frac{5}{2}}-a^{\frac{5}{2}}}{x-a}=\frac{5}{2} a^{\frac{5}{2}-1}$ $\lim _{x \rightarrow a} \frac{x^{\frac{5}{2}}-a^{\frac{5}{2}}}{x-a}=\frac{5}{2} a^{\frac{3}{2}}$ Thus, the value of $\lim _{x \r...
Read More →There are 60 students in a class.
Question: There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test: Where x is a positive integer. Determine the mean and standard deviation of the marks. Solution: Given there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given. Now we have to find the mean and standard deviation of the marks. It is given there are 60 students in the class, so fi=60 ⇒(x 2) + x + ...
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Question: Evaluate $\lim _{x \rightarrow 2}\left(\frac{x^{5}-32}{x^{3}-8}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}$ Formula used: We have, $\frac{x^{m}-y^{m}}{x-y}=m y^{m-1}$ As $x \rightarrow 4$, we have $\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}=\lim _{x \rightarrow 2} \frac{x^{5}-2^{5}}{x^{3}-2^{3}}$ $\lim _{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}=\lim _{x \rightarrow 2} \frac{\frac{x^{5}-2^{5}}{x-2}}{\frac{x^{3}-2^{3}}{x-2}}$ $\lim _{x \righta...
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Question: Evaluate $\lim _{x \rightarrow 4}\left(\frac{x^{3}-64}{x^{2}-16}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ and $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ As $x \rightarrow 4$, we have $\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=\lim _{x \rightarrow 4} \frac{(x-4)\left(x^{2}+4 x+16\right)}{(x+4)(x-4)}$ $\lim _{x \rightarrow 4} \frac{x^{3}-64}{x^{2}-16}=\lim _{x \rightarrow 4} ...
Read More →For the frequency distribution:
Question: For the frequency distribution: Find the standard distribution. Solution: Given frequency distribution table Now we have to find the standard deviation Let us make a table of the given data and append other columns after calculations And we know standard deviation is $\sigma=\sqrt{\frac{\sum f_{i} x_{i}^{2}}{n}-\left(\frac{\sum f_{i} x_{i}}{n}\right)^{2}}$ Substituting values from above table, we get $\sigma=\sqrt{\frac{1393}{60}-\left(\frac{277}{60}\right)^{2}}$ $\sigma=\sqrt{23.23-(4...
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Question: Evaluate $\lim _{x \rightarrow \frac{1}{2}}\left(\frac{4 x^{2}-1}{2 x-1}\right)$ Solution: To evaluate: $\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ As $x \rightarrow \frac{1}{2}$, we have $\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x+1)(2 x-1)}{2 x-1}$ $\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}}(2 x+1)$ $\l...
Read More →The frequency distribution:
Question: The frequency distribution: Where A is a positive integer, has a variance of 160. Determine the value of A. Solution: Given frequency distribution table, where variance =160 Now we have to find the value of A, where A is a positive number Now we have to construct a table of the given data And we know variance is $\sigma^{2}=\frac{\sum f_{i} x_{i}^{2}}{n}-\left(\frac{\sum f_{i} x_{i}}{n}\right)^{2}$ Substituting values from above table and also given variance $=160$, we get $160=\frac{9...
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Question: Evaluate $\lim _{x \rightarrow 3}\left(\frac{x^{2}-4 x+3}{x^{2}-2 x-3}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 3} \frac{x^{2}-4 x+3}{x^{2}-2 x-3}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ and As $\mathrm{X} \rightarrow 3$, we have $\lim _{x \rightarrow 3} \frac{x^{2}-4 x+3}{x^{2}-2 x-3}=\lim _{x \rightarrow 3} \frac{(x-3)(x-1)}{(x-3)(x+2)}$ $\lim _{x \rightarrow 3} \frac{x^{2}-4 x+3}{x^{2}-2 x-3}=\lim _{x \rightarrow 3} \frac{(x-1)}{(x+2)}$ $\lim _{x \rig...
Read More →Two sets each of 20 observations have the same
Question: Two sets each of 20 observations have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets. Solution: Given two sets each of 20 observations, have the same standard derivation 5 . The first set has a mean 17 and the second a mean 22 . Now we have to show that the standard deviation of the set obtained by combining the given two sets As per given criteria, for first set ...
Read More →The mean and standard deviation of a set of n1 observations are
Question: The mean and standard deviation of a set of n1observations areand s1, respectively while the mean and standard deviation of another set of n2observations areand s2, respectively. Show that the standard deviation of the combined set of (n1+ n2) observations is given by S.D. $=\sqrt{\frac{\mathrm{n}_{1}\left(\mathrm{~s}_{1}\right)^{2}+\mathrm{n}_{2}\left(\mathrm{~s}_{2}\right)^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}\left(\overline{\mathrm{x}}_{1}-\overline...
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Question: Evaluate $\lim _{x \rightarrow 3}\left(\frac{x^{4}-81}{x-3}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 3} \frac{x^{4}-81}{x-3}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ and As $\mathrm{x} \rightarrow 3$, we have $\lim _{x \rightarrow 3} \frac{x^{4}-81}{x-3}=\lim _{x \rightarrow 3} \frac{\left(x^{2}+9\right)\left(x^{2}-9\right)}{x-3}$ $\lim _{x \rightarrow 3} \frac{x^{4}-81}{x-3}=\lim _{x \rightarrow 3} \frac{\left(x^{2}+9\right)(x+3)(x-3)}{x-3}$ $\lim _{x \r...
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Question: Evaluate $\lim _{x \rightarrow-2}\left(\frac{x^{3}+8}{x+2}\right)$ Solution: To evaluate: $\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ and $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$ As $x \rightarrow-2$, we have $\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^{2}-2 x+4\right)}{x+2}$ $\lim _{x \rightarrow-2} \frac{x^{3}+8}{x+2}=\lim _{x \rightarrow-2}\left(x^{2}-2 x+4\right)$ $\lim _...
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Question: Evaluate $\lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ and $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ As $\mathrm{X} \rightarrow 1$, we have $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^{2}+x+1\right)}{x-1}$ $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}=\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)$ $\...
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Question: Evaluate $\lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{x-5}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ As $x \rightarrow 5$, we have $\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=\lim _{x \rightarrow 5} \frac{(x+5)(x-5)}{x-5}$ $\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=x+5$ $\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=5+5$ $\lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5}=10$ Thus, the value o...
Read More →The mean and standard deviation of some
Question: The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: Number of observations $=25$, mean $=18.2$ seconds, standard deviation $=3.25$ seconds. Further, another set of 15 observations $x_{1}, x_{2, \ldots,} x_{15}$, also in seconds, is now available and we have Calculate the standard derivation based on all 40 observations. Solution: Given: Number of observations $=25$, mean $=18.2$ seconds, standard deviation = $3.2...
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Question: Evaluate $\lim _{x \rightarrow 3}\left(\frac{x^{2}-4 x}{x-2}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 3} \frac{x^{2}-4 x}{x-2}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ As $\mathrm{x} \rightarrow 3$, we have $\lim _{x \rightarrow 3} \frac{x^{2}-4 x}{x-2}=\frac{3^{2}-4(3)}{3-2}$ $\lim _{x \rightarrow 3} \frac{x^{2}-4 x}{x-2}=\frac{3^{2}-4(3)}{3-2}$ $\lim _{x \rightarrow 3} \frac{x^{2}-4 x}{x-2}=-3$ Thus, the value of $\lim _{x \rightarrow 3} \frac{x^{2}-4 x...
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Question: Evaluate $\lim _{x \rightarrow 3}\left(\frac{x^{2}+9}{x+3}\right)$ Solution: To evaluate: $\lim _{x \rightarrow 3} \frac{x^{2}+9}{x+3}$ Formula used: We have, $\lim _{x \rightarrow a} f(x)=f(a)$ As $x \rightarrow 3$, we have $\lim _{x \rightarrow 3} \frac{x^{2}+9}{x+3}=\frac{3^{2}+9}{3+3}$ $\lim _{x \rightarrow 3} \frac{x^{2}+9}{x+3}=\frac{18}{6}$ $\lim _{x \rightarrow 3} \frac{x^{2}+9}{x+3}=3$ Thus, the value of $\lim _{x \rightarrow 3} \frac{x^{2}+9}{x+3}$ is $3 .$...
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