ISI2015-DCG-7

Question

Let $x^2-2(4k-1)x+15k^2-2k-7>0$ for any real value of $x$. Then the integer value of $k$ is

• A. $2$
• B. $4$
• C. $3$
• D. $1$

Answer 1

$x^2 - 2(4k-1)x+15k^2 -2k-7 >0$

$(A)$

putting $k=2$

$x^2 - 6x+49>0$

Here we can't get a negative value since $x^2 +49$ will always be greater than $6x$.  so this is the correct option.

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$(B)$

putting $k=4$

$x^2 -30x +225 >0 \implies (x-15)^2 >0$

so on putting $x=15$ we get $0>0$  so this could not be the answer

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$(C)$

putting $k=3$

$x^2-22x-122>0$

On putting $x=1$ we are getting $1-22-122>0$ so this could not be the answer

$OR$

$x^{2}-22x-122=x^{2}-22x+121-121-122=(x-11)^{2}-243$ which is negative for any number $x\epsilon[-4,15]$

so this could not be the answer

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$(D)$

putting $k=1$

$x^2 - 6x+6>0$

On putting $x=2$ we are getting $4-12+6>0$ so this could not be the answer.

$OR$

$x^{2}-6x+6=x^{2}-6x+9-9+6=(x-3)^{2}-3$ which is negative for $x=4$. so this could not be the answer.

Comments

nice explanation @Satbir 

it would be better if equation in option c is written as$x^{2}-22x-122=x^{2}-22x+121-121-122=(x-11)^{2}-243$ which is negative for any number x$\epsilon$[-4,15]..Hence false

similarly, equation in option d can be written as $x^{2}-6x+6=x^{2}-6x+9-9+6=(x-3)^{2}-3$ which is negative for x=4.Hence false.

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check your solution for d, it is coming 4-3 = positive.

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thanks.. i have updated it now