Let x and y be two 2- digit numbers such that y is obtained by reversing digits of x . Suppose they

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4-2.Quadratic Equations and Inequations

normal

Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

$88$

$112$

$144$

$154$

(KVPY-2014)

Solution

(d)

We have,

$x$ and $y$ be two-digit numbers.

Let $x=10 a+b$, where $b$ is units place and $a$ is ten's place.

$\because y=10 b +a$

$x^2-y^2 =m^2$

$\because (10 a+b)^2-(10 b+a)^2 =m^2$

$\Rightarrow(10 a+b+10 b+a)$

$\Rightarrow \quad \begin{aligned}(10 a+b-10 b-a) &=m^2 \\ \Rightarrow \quad 11(a+b) \cdot 9(a-b) &=m^2 \end{aligned}$

$\begin{aligned} \Rightarrow & & 11(a+b) \cdot 9(a-b) &=m^2 \\ \Rightarrow & & 99\left(a^2-b^2\right) &=m^2 \end{aligned}$

$\Rightarrow \quad 3^2 \times 11\left(a^2-b^2\right)=m^2$

Now, $3^2 \times 11\left(a^2-b^2\right)$ is a perfect square.

$\because a^2-b^2=11 \Rightarrow(a+b)(a-b)=11 \times 1$

$a+b=11$ and $a-b=11$

On solving, we get $a=6, b=5$

$\because$ Number $x=60+5=65$

$y=56$

and $m^2=(65)^2-(56)^2=(65+56)(65-56)$

$\Rightarrow m^2=121 \times 9 \Rightarrow m=33$

$\because \quad x+y+m=65+56+33=154$

Standard 11

Mathematics

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normal

(KVPY-2016)

lf $2 + 3i$ is one of the roots of the equation $2x^3 -9x^2 + kx- 13 = 0,$ $k \in R,$ then the real root of this equation

hard

(JEE MAIN-2015)

Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$, with $\alpha>\beta$. For all positive integers $n$, define

$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$

$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$

Then which of the following options is/are correct?

$(1)$ $a_1+a_2+a_3+\ldots . .+a_n=a_{n+2}-1$ for all $n \geq 1$

$(2)$ $\sum_{n=1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$

$(3)$ $\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\frac{8}{89}$

$(4)$ $b=\alpha^n+\beta^n$ for all $n>1$

normal

(IIT-2019)

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medium

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The number of real roots of the equation, $\mathrm{e}^{4 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{\mathrm{x}}+1=0$ is

hard

(JEE MAIN-2020)

Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

Solution

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