# Engineering mathematics set 1

By | November 21, 2018

# NTA NET SURE SUCCESS!!! Solve all MCQ and QUIZ on this site. You will get many common questions in Final exam. Do not forget to tell your friends

 Question 1 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
What's the general solution of $\frac{dy}{dx} = 3y^{2/3}$?
 A $y = (x+C)^3/27$. B $y = (3x+C)^3/27$. C $y = (2x+C)^3/27$. D $y = (x+C)^3/22$.
Question 1 Explanation:
On the left hand side we get, $3y^{\frac{1}{3}}$ and on the right hand side we get, $x+C$ $\Rightarrow 3y^{\frac{1}{3}}=x+C \Rightarrow y = (x+C)^3/27$.
 Question 2 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
Which of the following is true?
 A ${\sqrt 2}^{\sqrt 2}$ can be rational B There exists two irrational numbers $a, b$ such that $a^b$ is rational C Both A and B D None of the above
Question 2 Explanation:
Two irrational numbers $a, b$ such that $\\a^b\\$ is rational Let $\\a = \sqrt{2}^\sqrt{2}\\$ and $\\b = \sqrt{2}\\$, $\\a^b = \sqrt{2}^{\sqrt{2}^2} = \sqrt{2}^2 = 2\\$ which is clearly rational. If $\sqrt{2}^\sqrt{2}$ is irrational, then we have found our example. If $\sqrt{2}^\sqrt{2}$ is however rational, then of course $a=b=\sqrt{2}$ is an example of irrationals $a, b$ such that $a^b$ is rational. Henceforth, there exists two irrational numbers $\\a, b\\$ such that $\\a^b\\$ is rational.
 Question 3 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
If $\frac{1}{a} +\frac{1}{b}+ \frac{1}{c} =\frac{1}{a+b+c}$
 A $(\frac{1}{a} +\frac{1}{b}+ \frac{1}{c})^{2n+1}=\frac{1}{(a+b+c)^{2n+1}}$ B $\left(a^{2n+1}+b^{2n+1}\right)\left(a^{2n+1}+c^{2n+1}\right)\left(b^{2n+1}+c^{2n+1}\right)=0,$ C Both A and B D None of the above
Question 3 Explanation:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ it's just $(a+b)(a+c)(b+c)=0$ and we need to prove that $$\left(a^{2n+1}+b^{2n+1}\right)\left(a^{2n+1}+c^{2n+1}\right)\left(b^{2n+1}+c^{2n+1}\right)=0,$$ which is true because $a^{2n+1}+b^{2n+1}$ divided by $a+b$.
 Question 4 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
\begin{align} -20 &=-20\\ 16-36 &= 25-45\\ 4^2-4\times 9&=5^2-5\times 9\\ 4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\ 4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\ \end{align} Considering the formula $a^2+2ab+b^2=(a-b)^2$, one has \begin{align} (4-9/2)^2&=(5-9/2)^2\\ \sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\ 4-9/2&=5-9/2\\ 4&=5\\ 4-4&=5-4\\ 0&=1 \end{align}
 A $\sqrt {a^2}=|a|$ B cannot root the negative integer C Both A and B D None of the above
Question 4 Explanation:
\begin{align} (-2)^2 = 4 &\implies \sqrt{(-2)^2} = \sqrt{2^2} \\ &\implies-2 = 2 \\ &\implies-2 + 2 = 2 +2 \\ &\implies 0 = 4 \end{align} Can you see the mistake?
 Question 5 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
It took Marie 10 minutes to saw a board into 2 pieces. If she works just as fast, how long will it take for her to saw another board into 3 pieces?
 A 20 B 15 C 10 D All the above
Question 5 Explanation:
Cutting something into three pieces requires two cuts. hence 20 minutes

If you cut it differently into three pieces. You can cut it in 15 minutes.

One can cut in 10 minutes too. Saw must look like this


| |
| |
| |
| | <---cutting edges
| |
| |
+--+--+
| <---handle
|


 Question 6 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$1+2+3+\cdots =?$
 A $-\frac{1}{12}$ B $\infty$ C $n(n+1)/2$ D All the above
Question 6 Explanation:
Method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too. We will work with two sums, $$S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots$$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$\frac{1}{(1+x)^2} = 1-2x + 3x^2-4x^3 + \dots$$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$T = (1+2+3+\dots)-2\times (2+4+6+\dots) = (1+2+3+\dots) (1-4) =-3S$$ so $S=-T/3=-1/12$.

$1+2+3+4+\cdots=-1/12$, which Ramanujan sent to Hardy in 1913, admitting "If I tell you this you will at once point out to me the lunatic asylum as my goal."

 Question 7 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
What is the value of $0^0=$?
 A 1 B 0 C $\infty$ D Undefined. It is an Indeterminate form.
Question 7 Explanation:

In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined. Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x, y)\to(0, 0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined. However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period. Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1, 2, \ldots, m\}$ to $\\{1, 2, \ldots, n\\}$" when interpreted appropriately, so it is again the same thing as in set theory). So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$). Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices. Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions. We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way. This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this: 1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work. 2) If $A, B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty). $0^{0}$ is just one instance of an empty product, which means it is the multiplicative identity 1. I'm surprised that no one has mentioned the IEEE standard for $0^0$. Many computer programs will give $0^0=1$ because of this. This isn't a mathematical answer per se, but it's worth pointing out because of the increasingly computational nature of modern mathematics, so that one doesn't run afoul of anything. The use of positive integer exponents appears in arithmetic as a shorthand notation for repeated multiplication. The notation is then extended in algebra to the case of zero exponent. The justification for such an extension is algebraic. Furthermore, in abstract algebra, if $G$ is a multiplicative monoid with identity $e$, and $x$ is an element of $G$, then $x^0$ is defined to be $e$. Now, the set of real numbers with multiplication is precisely such a monoid with $e=1$. Therefore, in the most abstract algebraic setting, $0^0=1$. Continuity of $x^y$ is irrelevant. While there are theorems that state that if $x_n \to x$ and $y_n \to y,$ then $(x_n + y_n) \to x+y$ and $(x_n)(y_n) \to xy$, there is no corresponding theorem that states that $(x_n)^{(y_n)} \to x^y$. I don't know why people keep beating this straw man to conclude that $0^0$ can't or shouldn't be defined. Maybe it's a good idea to put the problem into a broader perspective. The minimum we need to define a power is a multiplicative semigroup, which basically means we have a set with an associative operation which we write as multiplication. If $S$ is the semigroup, the power function is defined as $$S\times \mathbb Z^+ \to S, (x, n) \mapsto x^n = \begin{cases} x & n=1\\ x x^{n-1} & n>1 \end{cases}$$ This power function has the fundamental properties $$x^mx^n = x^{m+n}, \quad (x^m)^n = x^{mn}\tag{*}$$ Note that up to now, we have used absolutely nothing about the elements of $S$ except for the fact that we can multiply them. Now it is a natural question whether we can extend that definition from the positive integers to the non-negative integers, in other words, whether we can define $x^0$. Of course we would want to define it in a way that the power laws (*) still hold. This especially implies the following relations: $$(x^0)^2 = x^0 \tag{I}$$ (that is, $x^0$ is idempotent) and $$x^0 x = x x^0 = x.\tag{II}$$ Note that this does not imply that for any $x\ne y$, $x^0=y^0$. Now for a given $S$ and a given $x\in S$, there are three possible cases: There does not exist an element $z\in S$ that fulfils both (I) and (II). In that case, $x^0$ is undefined and undefinable without violating the power laws. For example, if you take $S$ as the set of positive even numbers, then $x^0$ is undefined for all $x\in S$ (because there's no positive even number that fulfils either (I) or (II) for any x) There exists exactly one element $z\in S$ that fulfils those conditions. In that case, the only reasonable definition is $x^0 = z$. For example, for non-zero real numbers, you get $x^0=1$ this way. There exist more than one element $z_i\in S$ fulfilling those equations. This is the case for $0^0$ because both $0$ and $1$ fulfil the equations. In that case, you have several choices: You can select one of the possible values and define $x^0$ as that value. Of course you'd not randomly choose any value, but choose the one which is most useful. Which may be different in different contexts. Note that each of the choices gives a different valid power function. You can leave $x^0$ undefined. In this case, $x^0$ is called unspecified because you could specify it (as in the previous bullet). In particular, the restriction of any of the power functions from the first bullet to $S\setminus\{x\}$ will be the power function from this bullet. Now one particularly interesting case is if you have a neutral element, that is, an element $e\in S$ so that $ex=xe=x$ for all $x\in S$. A semigroup with such an element is called a monoid. It is easy to check that in this case, $e$ fulfils both (I) and (II) for any element $x\in S$. Therefore you always get a valid power function by defining $x^0=e$. Indeed, by doing so, you get the general form $$x^n = \begin{cases}e & n=0\\ x x^{n-1} & n>0\end{cases}$$ which means that in a monoid, $x^0=e$ clearly is a distinguished, and therefore preferable choice of the power function, even for $x$ where other choices would be possible. Note that the real numbers form a monoid under multiplication, with $e=1$. Therefore this is a first hint that $0^0=1$ is a good definition. The definition $0^0=1$ turns out to be useful also in other areas, for example when considering the linear structure (that is, the distributive law with addition), as it makes sure that e.g. the binomial formula $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}$$ also holds if any of $a$, $b$ or $a-b$ is $0$. There's however one structure that doesn't favour $0^0=1$, and that is continuity: $x^y$ is discontinuous at $(0, 0)$, and every value can be obtained as limit of a suitably chosen sequence of arguments approaching the origin. Therefore from the point of continuity, $0^0$ can be considered unspecified. However, since $x^y$ is discontinuous at the origin whether or not one defines it there, and independent of the value one chooses, that is not really an argument against choosing $0^0=1$, but more an argument against using that definition blindly. I'm not aware of a context where the definition $0^0=0$ would be more useful, but I wouldn't want to exclude that it exists. Any other value of $0^0$ would violate condition (I) above, and therefore likely not be useful at all. So in summary, the definition $0^0=1$ is the most useful in most situations, and not harmful in situations where one would otherwise leave $0^0$ unspecified, and if any context exists where another definition would be more useful, it's arather unusual context. Therefore the definition $0^0=1$ is the most reasonable one.

 Question 8 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
If you choose an answer to this question at random, what is the probability that you will be correct?
 A $25\%$ B $50\%$ C $0\%$ D $60\%$
Question 8 Explanation:
This puzzle is roaming on internet in many variations. One explanation is: If all of the answers are wrong, the possibility of getting a right one by choosing randomly is always 0%.
 Question 9 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
Let $\, A, B, C\in M_{n}(\mathbb C)\,$ be Hermitian and positive-definite matrices such that $A+B+C=I_{n},$ where $I_{n}$ is the identity matrix. Then which of the following is true?
 A $\det\left(4(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$ B $\det\left(5(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$ C $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$ D None of the above
There are 9 questions to complete.