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Question

For any real x, the expression 2(kx)[x+x2+k2] cannot exceed

A
k2
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B
2k2
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C
3k2
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D
None of these
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Solution

The correct option is C 2k2
Let 2(kx)[x+x2+k2]=y
x2+k2=y2(kx)x
On squaring and rearranging, we get
4x2(yk2)4kx(y2k2)+y24k4=0
Since, x is real.
Therefore, D=b24ac=16[k2(y2+4k44yk2)(y34yk4y2k2+4k6)]0
y(2k2y)0
0y2k2
Ans: B

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