88 views

I think all the options have at least one solution in $(0,1)$. First of all, $g'(x)=f(x)$. For option A, Let $F(x)=e^{-x}g(x)$. $\implies F(0)=F(1)=0$ $\implies \exists c \in (0,1)$ for which $F'(x)=0$...

61 views

Just to clarify, Fermat number is a number of the type $2^{2^n}+1$. About the part that no Fermat number ($F_n$) is a perfect square, I think the reasoning provided is correct. $F_n\equiv 7 \pmod {10}$...

56 views

Actually for (1), $n=1$ is a counterexample (trivial case, but counterexample nonetheless). $n^2 \equiv 0$ or $1 \pmod 3$. $\implies n^2+2\equiv 2$ or $0 \pmod 3$ and $n^2+2 \neq 0 \pmod 3$ (and here ...

94 views

I totally agree with the first answer. Here I'm just going to explain why it is corect. If we had $x_1 +x_2 +x_3+ \cdots +x_n =k$, then the number of integer solutions to this equation, where $x_i \... View answer 2 answers -2 votes 60 views Accepted answer 1 votes$\frac{dy}{dx}=\frac{x}{y}+\sqrt{\left(\frac{x}{y}\right)^2+1}$Let$y=tx$.$\implies \frac{dy}{dx}=t+x\frac{dt}{dx}\implies t+x\frac{dt}{dx}=\frac{1}{t} +\sqrt{\left(\frac{1}{t}\right)^2+1}\...

114 views

You need to substitute $\theta=\arcsin(\frac{x}{a})$, or as you said just find $\sin(4\theta)$ and $\sin (2\theta)$ using the double angle properties and put the values in the final answer.

56 views

You have almost solved it. You need to prove that $a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})\ge 0$. which is true due to the following : $a(\sqrt{a}-\sqrt{b})-b(\sqrt{a}-\sqrt{b})\ge 0$ $\implies (a-b)... View answer 1 answers 1 votes 101 views Accepted answer 1 votes Note: I'm going to introduce some integer variables like$k,n ,m $etc. They are only for the purpose of showing the remainder of a number when divided by a particular number. For example if I say$x=...

71 views

Q. Show that $16$ $|$ $19^{4n+1} +17^{3n+1} -4$ $\forall$ $n \in \mathbb N$. First of all, simplify the expression. Write $19=16+3$ and $17=16+1$, then use binomial theorem to expand it. That way ...

51 views

I don't think there is any value of $b$ for which it can be the vertex. Proof: $4=9-9b-5$ (Vertex lies on the parabola) $\implies b=0$ But the slope of tangent at the vertex should be zero, that is $\... View answer 5 answers 1 votes 72 views 1 votes Okay so you have a doubt in how$PM=y$and$PN=x$. When we write the equation$y^2=4ax$, the$PM$stands for axis of the parabola and$PN$stands for the tangent at the vertex. For the parabola$y^...

64 views

Assuming $(x,y,z,k) \in \mathbb Z$, $z^2 \equiv 0$ or $1\pmod 4$ and since $y^2 \equiv 0$ or $1 \pmod 4$ $kx^2 \not\equiv 2 $$\pmod 4 Now, as x^2 \equiv 0 or 1 \pmod 4, k \not\equiv ... View answer 3 answers 3 votes 80 views 0 votes If the given quadratic is ax^2+bx+c with \Delta=b^2-4ac, then ax^2+bx+c=\frac{1}{4a}\left( 2ax +b \right)^2 - \frac{\Delta}{4a}, which means the minimum or maximum value of the quadratic (- \... View answer 3 answers 2 votes 33 views 0 votes It's the same as proving that n^2-1 \equiv 0 \pmod 24. Since n \equiv 1 \pmod 2 \implies (n-1)(n+1) \equiv 0 \pmod 8 (as one of n-1, n+1 will be 2 \pmod 4 and the other will be 0 \pmod 4) ... View answer 1 answers 0 votes 107 views 0 votes This has to do with Gauss's Lemma for polynomials. It states that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers. So in the given ... View answer 5 answers 3 votes 142 views 0 votes I=\int \frac{dx}{(x^2+1)^2}=\frac{1}{2} \int \frac{1+x^2+1-x^2}{x^4+2x^2+1}.dx \implies I= \frac{1}{2} \left ( \int \frac{1+x^2}{x^4+2x^2+1}.dx +\int \frac{1-x^2}{x^4+2x^2+1}.dx\right). I_1=\int \... View answer 1 answers 0 votes 75 views 0 votes Partial progress : Here's the proof that n power numbers for any even n&gt;1 (i.e. squares, 4 th powers etc.) can't achieve all remainders modulo k for k \gt 2. Define sets A and B as ... View answer 4 answers 2 votes 340 views 0 votes Let f(x) be a function differentiable at x=0. Case 1: f is an even function.$$ f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}.$$Now find f'(-x).$$f'(-x)= ... View answer 2 answers 2 votes 81 views 0 votes If you have a polynomial$p(x)$of degree$n$, and you are given that$p(x)$has$n$roots, all of which are real then between any pair of consecutive roots (according to your graph, roots which are ... View answer 3 answers -1 votes 460 views 0 votes The method suggested in the first answer is easier than what I am going to use. We can find the coefficient of$x^{15}$in the expansion of$$(x^{-3} + x^{-2} + \cdots )(x^{0} + x^{1} + \cdots )(x^{-... View answer 2 answers 1 votes 69 views Accepted answer 0 votes You know that$x_n \lt x_{n+1}\implies x_n \lt \sqrt{3+2x_n}\implies 2x_n \lt 2 \sqrt{3+2x_n}\implies 3+ 2x_n \lt 3+ 2 \sqrt{3+2x_n}\implies \sqrt{3+ 2x_n} \lt \sqrt{3+ 2 \sqrt{3+2x_n}}...

48 views
$(1+x)^n = \binom{n}{0} + x \binom{n}{1}+ \cdots +x^n \binom{n}{n}$ ----(1) We need $S= 1\binom{n}{1} -2 \binom{n}{2} + \cdots + (-1)^{n-1} n \binom{n}{n}$. In equation (1), differentiate both ...
Let $f$ be a strictly increasing function. Consider two points $A$ and $B$ on the curve $y=f(x)$. Let $A \equiv (x_1 , f(x_1))$ and $B \equiv (x_1+k , f(x_1 +k))$ for some $k \gt 0$. Clearly, $x_1+... View answer 3 answers 1 votes 75 views 0 votes If you wish to find the number of ways to express a positive integer$n$as the sum of$m$positive integers ($m\le n$) and each of the positive integers must be less than or equal to a certain value$...