\(\frac {1}{49}\)
\(\frac {3}{49}\)
\(\frac {43}{49}\)
\(\frac {91}{49}\)
\(ƒ(x) = 3x^2 + 1\)
\(ƒ^′(x)\) is bijective function
and \(ƒ(g(x)) = x⇒g(x\)) is inverse of \(ƒ(x)\)
\(g(ƒ(x)) = x\)
\(g^′(f(x)).f^′(x) = 1\)
\(g^′(f(x)) = \frac {1}{3x^2+1}\)
Put \(x = 4\) we get
\(g^′(63)=\frac {1}{49}\)
So, the correct option is (A): \(\frac {1}{49}\)
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.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
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.
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
