Practicing Success
Match List - I with List - II
Choose the correct answer from the options given below : |
(A)-(IV), (B)-(II), (C)-(I), (D)-(III) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (A)-(I), (B)-(IV), (C)-(III), (D)-(II) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) |
(A)-(IV), (B)-(II), (C)-(I), (D)-(III) |
$|x-1|= \begin{cases}x-1 & x \geq 1 \\ -x+1 & x<0\end{cases}$ $|x-3|=\left\{\begin{array}{l}x-3, x \geq 3 \\ -x+3, x<3\end{array}\right.$ 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) |