Match List - I with List - II List – I List - II (A) Derivative of |x…

$|x-1|= \begin{cases}x-1 & x \geq 1 \\ -x+1 & x<0\end{cases}$

$|x-3|=\left\{\begin{array}{l}x-3, x \geq 3 \\ -x+3, x<3\end{array}\right.$

A.  |x - 1| + |x - 3|

at x = 2

so at x = 2

equation become

x - 1 + (-x) + 3

= 2

so derivative  $\frac{d}{d x}(2)=0$          (IV)

B.  $y=\log \sqrt{\tan x} \Rightarrow y =\frac{1}{2} \log \tan x$

so  $\frac{d y}{d x} =\frac{1}{2} \frac{\sec ^2 x}{\tan x}$

at $\frac{\pi}{4}$

$\tan x=1$

$\sec ^2 x=2$

so  $\frac{d y}{d x}=\frac{2}{2}=1$          (II)

C.  $\sin (x+y)=\log (x+y)$

differentiating both sides wrt x

so  $\cos (x+y)\left[1+\frac{d y}{d x}\right]=\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]$

so  $\left(+\frac{d y}{d x}\right)\left(\cos (x+y)-\frac{1}{(x+y)}\right)=0$

$\Rightarrow \frac{d y}{d x}=-1$           (I)

D. In langrage's mean value theorem 

for x ∈ [0, 4]

$\Rightarrow f(x)=x^2+x+1$

$f'(x)=2 x+1$

$\Rightarrow \frac{f(4)-f(0)}{4-0}=f'(c)$  for atleast some C

$\frac{4^2+4+1-0^2-0-1}{4-0}=2 c+1$

$\Rightarrow \frac{16+4}{4}=2 c+1$

$\Rightarrow 5=2 c+1 \Rightarrow 2 c=4 \Rightarrow c=2$         (III)