Mathematics - Functions Question with Solution | TestHub

Given that: $kf\left(k\right)+2=0$ for $k=2,3,4,5$ which means $(k-2),(k-3),(k-4),(k-5)$ are the factors of this expression.

Let $kf\left(k\right)+2=a(k-2)(k-3)(k-4)(k-5)$ $....\left(\mathrm{i}\right)$

$Put k=0$

$2=a(-2)(-3)(-4)(-5)$

$a=\frac{1}{60}$

Put $a=\frac{1}{60}$ in $\left(\mathrm{i}\right)$, we get

$kf\left(k\right)+2=\frac{1}{60}(k-2)(k-3)(k-4)(k-5)$

Now, put $k=10$

$10f\left(10\right)+2=\frac{1}{60}\times 8\times 7\times 6\times 5$

$10 f\left(10\right)=26$

So, $52-10 f\left(10\right)=26$