I like commenter @YuriyGalanter's idea of choosing items randomly until all are taken and only then repeating, so here's an implementation:

function randomNoRepeats(array) {
  var copy = array.slice(0);
  return function() {
    if (copy.length < 1) { copy = array.slice(0); }
    var index = Math.floor(Math.random() * copy.length);
    var item = copy[index];
    copy.splice(index, 1);
    return item;
  };
}

var chooser = randomNoRepeats(['Foo', 'Bar', 'Gah']);
chooser(); // => "Bar"
chooser(); // => "Foo"
chooser(); // => "Gah"
chooser(); // => "Foo" -- only repeats once all items are exhausted.

var a = ["Mango","Orange","Banana","Apple","Grapes","Berry","Peach"];

function searchRandom(count, arr){
  let answer = [], counter = 0;
 
  while(counter < count){
    let rand = arr[Math.floor(Math.random() * arr.length)];
    if(!answer.some(an => an === rand)){
      answer.push(rand);
      counter++;
    }
  }
  
  return answer;
}

console.log(searchRandom(3,a))

Making it flexible to support any count you want and ensured uniqueness

1. Your check method is inefficient for large datasets. I'd recommend using a hash instead. This will eliminate the need of iterating over the array every time you want to check if a number is there.
2. It is possible for your code to enter an infinite loop if count is greater than max - min. You need to test for this.
3. The code can be simplified a bit as well.
4. The recursion slows down the generation dramatically.

I've take then liberty of performing some optimizations: http://jsfiddle.net/FMSpG/2/

function Randoms(c, l, h) {
    var count = c,
        min = l,
        max = h,
        nums = {},
        out = [],
        r, len = 0;

    if (max - min < count) count = max - min;

    var getRand = function() {
        return Math.floor(Math.random() * (max - min) + min);
    },  check = function(a) {
        return nums[a];
    },  add = function(a) {
        nums[a] = 1;
        out.push(a);
        len++;
    };

    (function init() {
        while (len < count){
            if (len == 0) r = getRand();
            else while (check(r = getRand()));
            add(r);
        }
    })();

    this.get = function() {
        return out;
    };
}

And here's a jsPerf showing the huge difference in performance for large datasets:

http://jsperf.com/unique-randomization

Dyalog APL, 1 byte

?

Just a builtin. Try it here.

If I understand right then you're just looking for a permutation (i.e. the numbers randomised with no repeats) of the numbers 1-10? Maybe try generating a randomised list of those numbers, once, at the start, and then just working your way through those?

This will calculate a random permutation of the numbers in nums:

var nums = [1,2,3,4,5,6,7,8,9,10],
    ranNums = [],
    i = nums.length,
    j = 0;

while (i--) {
    j = Math.floor(Math.random() * (i+1));
    ranNums.push(nums[j]);
    nums.splice(j,1);
}

So, for example, if you were looking for random numbers between 1 - 20 that were also even, then you could use:

nums = [2,4,6,8,10,12,14,16,18,20];

Then just read through ranNums in order to recall the random numbers.

This runs no risk of it taking increasingly longer to find unused numbers, as you were finding in your approach.

EDIT: After reading this and running a test on jsperf, it seems like a much better way of doing this is a Fisher–Yates Shuffle:

function shuffle(array) {
    var i = array.length,
        j = 0,
        temp;

    while (i--) {

        j = Math.floor(Math.random() * (i+1));

        // swap randomly chosen element with current element
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;

    }

    return array;
}

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

Basically, it's more efficient by avoiding the use of 'expensive' array operations.

BONUS EDIT: Another possibility is using generators (assuming you have support):

function* shuffle(array) {

    var i = array.length;

    while (i--) {
        yield array.splice(Math.floor(Math.random() * (i+1)), 1)[0];
    }

}

Then to use:

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

ranNums.next().value;    // first random number from array
ranNums.next().value;    // second random number from array
ranNums.next().value;    // etc.

where ranNums.next().value will eventually evaluate to undefined once you've run through all the elements in the shuffled array.

Overall this won't be as efficient as the Fisher–Yates Shuffle because you're still splice-ing an array. But the difference is that you're now doing that work only when you need it rather than doing it all upfront, so depending upon your use case, this might be better.

Try this (splice removes the selected element from the source array) :

var r = [];
for (var i = 0; i < 5; i++) {
    r.push(itemp.splice(
        Math.floor(Math.random() * itemp.length), 1
    )[0]);
}
alert(r);

Generate a range of numbers:

var numbers = [1, 2, 3, 4];

And then shuffle it:

function shuffle(o) {
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

var random = shuffle(numbers);