A lamp of negligible height is placed on the ground '$l'_1$ metre away from a wall. A man ' $l_2$ ' metre tall is walking at a speed of $\frac{l_1}{10}$ m/sec. From the lamp to the nearest point on the wall. When he is mid-way between the lamp and the wall, the rate of change in the length of this shadow on the wall is
Answer & explanation
Correct answer: option 2
Clearly $\frac{h}{l_1}=\frac{l_2}{l_1-x}$
$\Rightarrow hl_1-hx=l_1 l_2$ (Since h is decreasing put a – ve sign)
$\Rightarrow l_1 \frac{d h}{d t}-h \frac{d x}{d t}-x \frac{d h}{d t}=0$
$\Rightarrow\left(l_1-x\right) \frac{d h}{d t}=\frac{h l_1}{10}$
at mid point $x_1=I_1 / 2, h=2 I_2$
$\Rightarrow \frac{d h}{d t}=\frac{2 l_2}{5}$
$f'(x)=2 e^x+a e^{-x}+2 a+1 \geq 0$ clear $a \geq 0$