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Prabjeet Singh 7 years, 5 months ago
{tex}\text {Given:}{/tex} {tex}p(x)=kx^3+3x^2-3 \text { and } q(x) = 2x^3-5x+k{/tex}
{tex}\text {Now, when } p(x) \text { and } q(x) \text { are divided by } g(x), \text { leaves same remainder in both cases.}{/tex}
{tex}\text {And, } g(x) = x-4{/tex}
{tex}\text {Putting } x= 4, \text { in } p(x), \text { we get remainder for } p(x), {/tex}
{tex}p(4) = (k)(4)^3+3(4)2-3{/tex}
{tex}\Rightarrow p(4) = 64k + 48-3{/tex}
{tex}\Rightarrow p(4) = 64k + 45{/tex}
{tex}\text {Putting } x= 4, \text { in } q(x), \text { we get remainder for } q(x), {/tex}
{tex}q(4)=2(4)^3-5(4)+k{/tex}
{tex}\Rightarrow q(4) = 2(64) - 20 + k{/tex}
{tex}\Rightarrow q(4) = 128 -20 + k{/tex}
{tex}\Rightarrow q(4) = 108 + k{/tex}
{tex}\text {Since, remainders are equal, so }p(4) = q(4),{/tex}
{tex}\therefore 64k+45=108+k{/tex}
{tex}\Rightarrow 64k-k=108-45{/tex}
{tex}\Rightarrow 63k = 63{/tex}
{tex}\Rightarrow k= \cfrac {63}{63}{/tex}
{tex}\Rightarrow k = 1{/tex}
{tex}\text {Thus, the required value of } k \text { is 1.}{/tex}
Posted by Prateek Patil 7 years, 5 months ago
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