# Relations and Functions - Class 12 - Mathematics

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## Extra Questions - Relations and Functions | NCERT | Mathematics | Class 12

If $f(x)$ is a polynomial function satisfying $f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x})$ and $f(1)=2$, then the number of such functions possible is/are

Option A) 1

Option B) 2

Option C) more than 2 but finite

Option D) infinitely many

To solve the question, we need to analyze the functional equation:

$$ f(x)f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) $$

One approach to simplify this equation is to see if there's a way to factorize or rearrange it to find a generalized form for $f(x)$. Let's first rearrange it slightly,

$$ f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 $$

This can be rewritten as:

$$ (f(x) - 1)(f\left(\frac{1}{x}\right) - 1) = 1 $$

From this rearrangement, we can see that if we pick any function $f(x)$ such that $f(x)-1$ and $f(\frac{1}{x})-1$ multiply to 1, then it satisfies the equation. Some observations can give us more insights:

1. **Choosing Simple Polynomials**: Consider the simplest non-constant polynomial, say $f(x) = x + 1$, we see immediately that this satisfies the condition $(x+1)(1/x+1)=1+1/x+x+1=x+1+1/x+1$.

2. **Testing Other Polynomials**: Let’s test $f(x) = x^2 + 1$, leading to $(x^2+1)(1/x^2+1) = 1 + x^2 + 1/x^2 + 1$ which doesn't satisfy the condition. Thus, not all polynomials of a simple form work.

3. **Compounded Functional Forms**: Since $(f(x)-1)(f(\frac{1}{x})-1)=1$, functions like $f(x)=kx+1$ where $k$ is any constant will satisfy our original equation because plugging any linear form $kx + 1$ into the equation $(kx+1)(\frac{k}{x}+1) = 1 + k(x + \frac{1}{x}) + k^2$ which indeed neatly resolves. However, $k = -1$ is a particularly interesting choice, as this gives $f(x) = -x + 1$. Check yields $(−x+1)(−\frac{1}{x}+1)=1−x−\frac{1}{x}+1 = 2- x−\frac{1}{x}$ which correctly simplifies.

4. **Differentiating with $f(1)=2$**: $f(1)=2$, so both choices $kx+1$ where $k=1$ (leading to $f(x)=x+1$) or $k=-1$ (leading to $f(x)=-x+1$) satisfy $f(1)=2$.

Overall, these examples show that multiple polynomials can satisfy the condition, and since both $x+1$ and $-x+1$ are solutions, and potentially other linear transformations or possibly more complex functions could also work. Hence, without additional restrictions, there are infinitely many potential solutions, matching with Option D) infinitely many.

Verify Rolle's theorem for the function $f(x) = x^{2} + 2x - 8$, $x \in [-4,2]$.

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Find the intervals in which the function $f$ given by $f(x) = 2x^{3} - 3x^{2} - 36x + 7$ is a) strictly increasing b) strictly decreasing

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The terms right to the symbol: ":" have the same relationship as the two terms to the left symbol "::" out of the four terms. Figure one is missing, which is shown in bold. Four alternatives are given for each question. Find out the correct alternative and write its number against the corresponding question on your answer sheet -

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**Solve the inequality:**

$\sqrt{3x-8} < -2$

A. $\phi$

B. $[1,2]$

C. $[12, \infty)$

D. $(1,2]$