\(f(x)=(2(1−\frac {x^{25}}{2})(2+x^{25}))^{\frac {1}{50}}\)
\(f(x)=(2(\frac {2−x^{25}}{{2}})(2+x^{25}))^{\frac {1}{50}}\)
\(=(4−x^{50})^{\frac {1}{50}}\)
\(f(f(x))=(4−((4−x^{50})^{\frac {1}{50}})^{50})^{\frac {1}{50}}=x\)
As \(f(f(x)) = x\) we have
\(g(x) = f(f(f(x))) + f(f(x)) = f(x) + x\)
\(g(x) = (4 – x^{50})^{\frac {1}{50}} + x\)
\(g(1) = 3^{\frac {1}{50}} + 1\)
\([g(1)] = 2\)
So, the answer is \(2\).
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
Let \(M = \{1, 2, 3, ....., 16\}\), if a relation R defined on set M such that R = \((x, y) : 4y = 5x – 3, x, y (\in) M\). How many elements should be added to R to make it symmetric.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
Which of the following best represents the temperature versus heat supplied graph for water, in the range of \(-20^\circ\text{C}\) to \(120^\circ\text{C}\)? 
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
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