Question

(a) If x be real prove that the expression $\frac{x+2}{2x^2+3x+6}$ takes all values in the interval $[-{\frac{1}{13}}^{\prime }\frac{1}{3}]$.

(b) Show that the value of

$\frac{tanx}{tan3x}$ or $\frac{sinxcos3x}{cosxsin3x}$

Whenever defined never lies between $1/3$ and $3$.

(c) If x is real, the maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$ is

• A. $41$
  
• B. $1$
• C. $\frac{17}{7}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $41$

(a) If the given expression be y, then

$2x^{2}y+(3y-1)x+(6y-2)=0$

If $y\ne 0$ then $\Delta \ge 0$ for real x i.e. $B^2-4AC\ge 0$

or $-39y^2+10y+1\ge 0$

or $(13y+1)(3y-1)\le 0$

$\therefore$ $-\frac{1}{13}\le y\le \frac{1}{3}$

If $y=0$ then $x=-2$ which is real and this value of y is included in the above range.

(b) $y=\frac{tanx}{tan3x}=\frac{t(1-3t^2)}{3t-t^3}=\frac{1-3t^2}{3-t^3}$, as $t\ne 0$

$\because$ $t=0$ will make y indeterminate

$\therefore$ $y(3-t^2)=1-3t^2$

or $t^2=\frac{3y-1}{y-3}=+ive$

$=\frac{(3y-1)(y-3)}{{(y-3)}^2}=\frac{3(y-1/3)(y-3)}{{(y-3)}^2}$

Above will be $+ive$ if $y<1/3$ or $y>3$ as denominator is $+ive$

Hence we conclude that y cannot lie between $\frac{1}{3}$ and $3$.

(c) Ans (a).

Proceeding as in part (a), (b), $yϵ[1,41]$

$\therefore$ Max value of y is $41$.