Let for a ne {a₁} ne 0, fleft( x right) = a{x^2} + bx + c,gleft( x right) = {a₁}{x^2} + {b₁}x

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1.Relation and Function

hard

Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) - g\left( x \right),$ If $p\left( x \right) = 0$ only for $ x=-1 $ and $p\left( { - 2} \right) = 2$ then value of $p\left( 2 \right)$ is

$9$

$6$

$3$

$18$

(AIEEE-2011)

Solution

Given that $P(x)=F(x)-g(x)$ has only one root $-1$

$\Rightarrow p(x)=\left(a-a_{1}\right) x^{2}+\left(b-b_{1}\right) x+\left(c-c_{1}\right)$ has

one root, $-1$ only

$\Rightarrow p^{\prime}(x)$ will also has root as $-1$ $\Rightarrow p^{\prime}(x)=0$ at $x=-1$

$\Rightarrow 2\left(a-a_{1}\right) x+\left(b-b_{1}\right)=0$ at $x=-1$

$\Rightarrow-2\left(a-a_{1}\right)+\left(b-b_{1}\right)=0$

$\Rightarrow \quad \frac{-\left(b-b_{1}\right)}{\left(a-a_{1}\right)}=-2$ ……$(i)$

Now, $p(x)=\left(a-a_{1}\right) x^{2}+\left(b-b_{1}\right) x+\left(c-c_{1}\right)$

$\Rightarrow \quad \frac{p(x)}{a-a_{1}}=x^{2}+\frac{b-b_{1}}{a-a_{1}} x+\frac{\left(c-c_{1}\right)}{a-a_{1}}$

$\because \quad p(-1)=0$

$\therefore \quad 0=(-1)^{2}-\frac{\left(b-b_{1}\right)}{a-a_{1}}+\frac{\left(c-c_{1}\right)}{a-a_{1}}$

$\Rightarrow \quad 0=1-2+\frac{\left(c-c_{1}\right)}{a-a_{1}} $ $\quad [{\rm{ using Eq}}{\rm{.}}\left( i \right)]$

$\Rightarrow \frac{\mathrm{c}-\mathrm{c}_{1}}{\mathrm{a}-\mathrm{a}_{1}}=1$ ……$(ii)$

Also, given that $p(-2)=2$

$\Rightarrow 4\left(a-a_{1}\right)-2\left(b-b_{1}\right)+\left(c-c_{1}\right)=2 \quad \ldots$ $(iii)$

From Eqs. $(i), (ii)$ and $(iii),$ we have

${4\left(a-a_{1}\right)-4\left(a-a_{1}\right)+\left(a-a_{1}\right)=2} $

$\Rightarrow$ ${a-a_{1}=2}$

On substituting $a-a_{1}=2$ in Eq. $(ii),$ we get

$c-c_{1}=2$

On substituting $a-a_{1}=2$ in Eq. $(i),$ we get

$b-b_{1}=4$

$ \text { Now, } p(2) =4\left(a-a_{1}\right)+2\left(b-b_{1}\right)+\left(c-c_{1}\right) $

$=4 \times 2+2 \times 4+2 $

$=8+8+2=18 $

Standard 12

Mathematics

A real valued function $f(x)$ satisfies the function equation $f(x – y) = f(x)f(y) – f(a – x)f(a + y)$ where a is a given constant and $f(0) = 1$, $f(2a – x)$ is equal to

hard

(AIEEE-2005)

Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.

medium

The range of $f(x) = [\cos x + \sin x]$ is (Where $[.]$ is $G.I.F.$)

normal

If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt 3}{4}$ , then $g(x)$ equals :-

normal

Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f : A \rightarrow A$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to $……………$.