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# 6. p, q, r are given real numbers. Consider the polynomials (q-r)(x-q)(x-r) + (r-p)(x-r)(x-p) + +(p-q)(x-p)(x-q) + k for all real k. Then [ ]A) there exists a k such that the degree of the polynomial is 2B) there exists a k such that the degree of the polynomial is 1C) there exists no k such that the degree of the polynomial is 0D) there exists no k such that the degree of the polynomial is undefinedE) none of thesesir plzzzzzzzz      solve    .... i will surely approve..... 10 years ago

(q-r)(x-q)(x-r) + (r-p)(x-r)(x-p) + (p-q)(x-p)(x-q) + k

-->first let us solve

 (q-r)(x-q)(x-r)

we will get,

qx^2-q^2x-rx^2+rqx-rqx+rq^2+r^2x-r^2q

x^2(q-r)+(r^2-q^2)x+rq(q-r)

Now solve

 (r-p)(x-r)(x-p)

Here we can see the similar pattern i.e.Herein place of "q" it is "r" and in place of "r" it is "p".So,we can write

x^2(r-p)+(p^2-r^2)x+rp(r-p)

Similarly,3rd erm also folows similar pattern So

 x^2(p-q)+(q^2-p^2)x+qp(p-q) When we will add all the 3 terms it will sum out to 0and only 'k' is left As 'k' is a constant so the possible degree of polynomial is  0 It is satisfying only fifth option so answer is none of these