2621) i am getting two values for a2+b2 as 317 and 325
2) getting A as 2/3
3) 96
1)If the range of the function f(x) = x2+ax+bx2+2x+3 is [-5,4],a,b are natural nos,then find the value of a2 + b2 ?
2)Given,x,y belong to reals and x2 +y2 > 0.If the max and min value of the expression x2 +y2x2+xy+4y2 are M and m and A denotes the average value of M and m,compute (2007)A ?
3)Two roots of a biquadratic equation x4 - 18x3 + kx2 + 200x - 1984=0 have their product equal to (-32).Find the value of k?
4)Let P(x)=x2+bx+c,where b and c are integers.If P(x) is a factor of both x4 + 6x2 +25 and 3x4 + 4x2 + 28x + 5,find the value of P(1) ?
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9 Answers
262
11) 377 i.e. (a,b)=(14,9)
3) 65847
4) p(1) = 4 i.e. P(x) = x2-2x+5 (this ones' the easiest)
1708\hspace{-16}$Here $\mathbf{x,y\in\mathbb{R}}$ and $\mathbf{x^2+y^2>0}$\\\\ Let $\mathbf{\frac{y}{x}=t,x\neq 0}$. Then $\mathbf{x^2+y^2>0\Leftrightarrow t^2+1>0\forall x\in\mathbb{R}}$\\\\ Now Let $\mathbf{\mathbb{K}=\frac{x^2+y^2}{x^2+xy+4y^2}=\frac{1+t^2}{1+t+4t^2}}$\\\\ $\mathbf{4\mathbb{K}t^2+\mathbb{K}t+\mathbb{K}=1+t^2}$\\\\ $\mathbf{(4\mathbb{K}-1).t^2+\mathbb{K}.t+(\mathbb{K}-1)=0}$\\\\ Now $\mathbf{(4\mathbb{K}-1)>0}$ for real and equal roots.\\\\ So its Discriminant must be greater then or equal to $\mathbf{\mathbb{Z}}$ero\\\\ $\mathbf{\mathbb{K}^2-4.(4\mathbb{K}-1).(\mathbb{K}-1)}\geq 0$\\\\ After rearranging, We Get\\\\ $\mathbf{15.\mathbb{K}^2-20.\mathbb{K}+4\leq 0}$\\\\ After Solving We Get\\\\ $\mathbf{\frac{10-2\sqrt{10}}{15}\leq \mathbb{K} \leq \frac{10+2\sqrt{10}}{15}}$\\\\ Now Average of Max. and Min. value of $\mathbf{\frac{10-2\sqrt{10}}{15}}$ and $\mathbf{\frac{10+2\sqrt{10}}{15}}$ is\\\\
\hspace{-16}\mathbf{\mathbb{A}=\frac{\frac{10-2\sqrt{10}}{15}+\frac{10+2\sqrt{10}}{15}}{2}=\frac{10}{15}=\frac{2}{3}}$\\\\ So $\mathbf{2007.\mathbb{A}=2007.\frac{2}{3}=1338}$
1057@koni... in q1.if (a,b) =(14,9) then a2 + b2 is 277 and not 377... :P nd plz expain how u got the answer...
@aditya....question 3 answer is 86 and not 96 and also explain.....
1708\hspace{-16}$Let $\mathbf{\alpha,\beta,\gamma,\delta}$ be the roots of the equation. and Let \\\\ $\mathbf{p=-(\alpha+\beta)\;\;\;,\alpha.\beta=-32}$(bcz product of two roots)\\\\ $\mathbf{q=-(\gamma+\delta)\;\;\;,\gamma.\delta =62 }$(bcz $\mathbf{\alpha.\beta.\gamma.\delta = -1984}$)\\\\ So $\mathbf{x^4-18x^3+kx^2+200x-1984=\left\{x^2-(\alpha+\beta).x+\alpha.\beta\right\}.\left\{x^2-(\gamma+\delta).x+\gamma.\delta\right\}}$\\\\ $\mathbf{x^4-18x^3+kx^2+200x-1984=(x^2+px-32).(x^2+qx+62)}$\\\\ After Camparing Coefficients, We Get\\\\ $\begin{Bmatrix} \bold{p+q=-18 } \\\\ \bold{pq+30=k } \\\\ \bold{62p-32q=200 } \end{Bmatrix}$\\\\\\ after Solving For $\mathbf{p\;,q}\;,$ We Get \\\\ $\mathbf{p=-4\;,q=-14}$\\\\ So $\mathbf{k=pq+30=56+30=86}$\\\\ So $\mathbf{\boxed{\boxed{\bold{K=86}}}}$
1@ketan,
1) Since we want -5,4 to be extremum values we need the equations f(x)=-5 and f(x)=4 to have double roots i.e. their discriminants must be zero each. using this we get (a,b) ≡ (-10,-15) and (14,9). As it is mentioned that a,b are naturals .:. (a,b) = (14,9)
and yeah it's 277 :P
14) If p(x) = x2 + bx + c | x4 + 6x2 + 25,
and p(x) = x2 + bx + c | 3x4 + 4x2 + 28x + 5,
then also p(x) = x2 + bx + c | 3(x4 + 6x2 + 25) - (3x4 + 4x2 + 28x + 5) = 14(x2 - 2x + 5).
Therefore p(x) = x2 - 2x + 5, so (b,c) = (-2,5)