Domain and range of a function in Hindi. Example  part- 1
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Domain and range of a function in Hindi. Example part- 1


Hello friends, today we will find the domain of these four functions to find the domain, we need to remember two things first, the denominator term will never be zero second, the term inside square root will always be positive so keeping these two points in mind, let’s find the domain of these four functions. the first function over here is f of x equals one by x minus 2 now we know that the denominator term will never be zero i.e x minus 2 is not equal to zero i.e x minus 2 is not equal to zero lets visualize this on number line. so here we have zero, one and two. But 2 will not be included here because on substituting two in the given function, the denominator will become zero which is undefined so, domain of this function includes all the numbers from minus infinity to plus infinity excluding 2 hence, domain is from minus infinity to 2, where 2 will not be included and hence we put open bracket on two for infinity we always put open bracket which is also known as small bracket union all the numbers from two to plus infinity within open brackets so, this is the domain for this function now for the second question we have 1 by x square minus x minus 6 again we will see that denominator should not be zero. i.e x square minus x minus 6 should not be equals to 0 hence solving for x square minus x minus 6 not equals to 0 which gives that x should not be equal to three and x should also not be equal to minus two we can see on number line -2 and 3 are the points on number line since after solving this equation we have two roots: -2 and 3 so all the numbers except these two points included in domain from -∞ to -2 and from -2 to 3 and from 3 to ∞ hence domain will be (-∞,-2) U (-2,3) U(3,∞) ∞ always comes with small bracket and here -2 & 3 are excluded from the domain so we use small bracket In third question we have to check for both the conditions because we have square root as well. In denominator, we have x square plus four which means that whatever number we substitute for x,
it will lead denominator to positive value only hence no possibility for zero in denominator. no further check required we need to check the condition in numerator if we look carefully at numerator, the term inside square root is x minus 1, so we need to ensure that this term is always greater than or equal to zero. assume same root in denominator, In that case we cant put equal to sign, it will create divide by zero problem So x is greater than or equal to 1 lets visualize this on number line. points are zero and one we can clearly see that the required domain becomes any number greater than or equal to one. so domain=[1,∞) since one is included in the domain so use [ bracket In next problem, we need to ensure that denominator is not zero and term inside square root is positive hence we calculate x square minus four is greater than zero. we can not write equal to zero since it will create divide by zero problem So factoring this term as x minus two and
x plus two greater than zero lets visualize this on number line. -2 and 2 are the two points -2 and 2 will not come under domain since expression is strictly greater than zero Now will check the sign of each interval take any value between -2 and 2 e.g x=1 by putting x=1, equation becomes negative. so sign between -2 and 2 becomes negative since this is a quadratic equation other two interval’s sign become positive we need only positive value since equation is strictly positive so domain becomes (-∞,-2)U(2,∞)

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