JEE Main 2023 (Online) 29th January Evening ShiftQuestion 1
Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f''(x)=g''(x)+6x$
$f'(1)=4g'(1)-3=9$
$f(2)=3g(2)=12$
Then which of the following is NOT true ?
Differentiation and its Application
Solution
Concept. Let $h=f-g$. Then $h''=6x$; integrate twice using the given values. Formulas used. (1) $h''=6x$ $\Rightarrow$ $h'=3x^2+c_1$, $h=x^3+c_1x+c_2$Step 1 - Data. $f'(1)=9$ and $4g'(1)-3=9$ give $g'(1)=3$, so $h'(1)=6$. $f(2)=12$ and $3g(2)=12$ give $g(2)=4$, so $h(2)=8$. Step 2 - Constants. By (1): $3+c_1=6$, so $c_1=3$; $8+6+c_2=8$, so $c_2=-6$. So $$h(x)=x^3+3x-6,\qquad h'(x)=3x^2+3.$$ Step 3 - Check each. (A) $g(-2)-f(-2)=-h(-2)=-(-8-6-6)=20$. True. (B) $h$ is increasing and $h(-1)=-10$, so $|h|$ is close to $10$ just to the right of $-1$. Not true. (C) $|3x^2+3|<6\iff x^2<1$. True. (D) $h(1)=-2<0$ and $h\left(\frac32\right)=\frac{15}{8}>0$, so $h$ has a zero in $\left(1,\frac32\right)$. True. Answer: (B)