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### $x^2+4y^2+16z^2=48$$xy+4yz+2zx=24$$\{x,y,z\}\equiv$

(A) $\{ \mp 4, \pm 2, \pm 1\}$
(B) $\{ \pm 4, \mp 2, \pm 1\}$
(C) $\{ \pm 4, \pm 2, \mp 1\}$
(D) $\{ \pm 4, \pm 2, \pm 1\}$

Solution

$(x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8zx+16yz$

$=x^2+4y^2+16z^2+4(xy+2zx+4yz)$

$=48+4\times 24 =144$

$\Rightarrow x+2y+4z=\pm 12$ ......(*)

Now, $x^2+4y^2+16z^2=48=2\times 24 =2(xy+4yz+2zx)$

$\Rightarrow x^2+(2y)^2+(4z)^2-x(2y)-(2y)(4z)-(4z)x=0$

$\Rightarrow \frac{1}{2}\left[ {{{(x - 2y)}^2} + {{(2y - 4z)}^2} + {{(4z - x)}^2}} \right] = 0$

$\Rightarrow x=2y=4z$

From (*), $x+x+x=\pm 12$

$\Rightarrow x=\pm 4$, $y=\pm 2$, $z=\pm 1$

Answer: Option (D)

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$