Let function f(x) = {x^2} + x + sin x - cos x + log (1 + |x|) be defined over interval [0, 1] . odd

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1.Relation and Function

hard

Let function $f(x) = {x^2} + x + \sin x - \cos x + \log (1 + |x|)$ be defined over the interval $[0, 1]$. The odd extensions of $f(x)$ to interval $[-1, 1]$ is

${x^2} + x + \sin x + \cos x - \log (1 + |x|)$

$ - {x^2} + x + \sin x + \cos x - \log (1 + |x|)$

$ - {x^2} + x + \sin x - \cos x + \log (1 + |x|)$

None of these

Solution

(b) Odd extension from $[0, 1]$ to $[-1, 1]$ means we have to choose the function out of $(a), (b), (c), (d)$ which satisfies the condition $f( – x) = – f(x).$
Now $| – x|\,\, = \,\,|x|$
$f( – x) = {x^2} – x – \sin x – \cos x + \log \,(1 + |x|)$
$ = – \,[{\rm{function given in (b)]}}$
$\therefore \,\,\,({\rm{b}})$ is correct. No other given function satisfies this criteria of $f( – x) = – f(x).$

Standard 12

Mathematics

Which of the following function is surjective but not injective

normal

Let $2{\sin ^2}x + 3\sin x – 2 > 0$ and ${x^2} – x – 2 < 0$ ($x$ is measured in radians). Then $x$ lies in the interval

hard

(IIT-1994)

Let $f\left( n \right) = \left[ {\frac{1}{3} + \frac{{3n}}{{100}}} \right]n$ , where $[n]$ denotes the greatest integer less than or equal to $n$. Then $\sum\limits_{n = 1}^{56} {f\left( n \right)} $ is equal to

hard

(JEE MAIN-2014)

Let $R _{1}$ and $R _{2}$ be two relations defined as follows :

$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$

where $Q$ is the set of all rational numbers. Then

hard

(JEE MAIN-2020)

If $f(x) = \cos (\log x)$, then $f({x^2})f({y^2}) – \frac{1}{2}\left[ {f\,\left( {\frac{{{x^2}}}{2}} \right) + f\left( {\frac{{{x^2}}}{{{y^2}}}} \right)} \right]$ has the value

medium

Let function $f(x) = {x^2} + x + \sin x - \cos x + \log (1 + |x|)$ be defined over the interval $[0, 1]$. The odd extensions of $f(x)$ to interval $[-1, 1]$ is

Solution

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Let $R _{1}$ and $R _{2}$ be two relations defined as follows :

$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$

where $Q$ is the set of all rational numbers. Then