Solve for x & y$\sqrt x + y = 7$$x + \sqrt y = 11 Putting y = {(11 - x)^2} obtained from the 2nd equation into the 1st equation, We have, \sqrt x + {(11 - x)^2} = 7 \Rightarrow x = {[7 - {(11 - x)^2}]^2} = 49 + {(11 - x)^4} - 14{(11 - x)^2} Let, 11 - x = t So, 11 - t = 49 + {t^4} - 14{t^2} \Rightarrow {t^4} - 14{t^2} + t + 38 = 0 t=2 satisfies the above equation. So, using factor theorem {t^3}(t - 2) + 2{t^2}(t - 2) - 10t(t - 2) - 19(t - 2) = 0 \Rightarrow (t - 2)({t^3} + 2{t^2} - 10t - 19) = 0 ........... (A) Let, f(t) = {t^3} + 2{t^2} - 10t - 19 t = 11 - x = \sqrt y \ge 0 y = 7 - \sqrt x , \Rightarrow y \le 7 So, t = \sqrt y \le \sqrt 7 Thus, 0 \le t \le \sqrt 7 Or, 0 \le t < 3 Let us investigate the behaviour of f in the interval [0, 3]. f(0) = -19 f(1) = -26 f(2) = -23 f(3) = -4 The function decreases initially, then increases but is unable to cut the t-axis in the interval [0, 3). t = 2 is the only solution as obtained earlier as per (A). Now, 11 - x = 2 or x = 9 y = {t^2} = 4 Popular posts from this blog f(x)=x^6+2x^4+x^3+2x+3$$\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, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$