Download App

5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

Gujarat Boardગુજરાતી માધ્યમધોરણ 12 સાયન્સગણિત5. continuity and differentiation1 Mark
MCQ
${d \over {dx}}[|x - 1| + |x - 5|]$ એ $x = 3$ આગળ મેળવો.
  • A
    $-2$
  • ✓
    $0$
  • C
    $2$
  • D
    $4$
✓

Answer

Correct option: B.
$0$
b
(b) $f(x) = |x - 1| + |x - 5|$

$f(x) = \left\{ \begin{array}{l} - (x - 1) - (x - 5),\,\,\,x < 1\\\,\,\,\,(x - 1) - (x - 5),\,\,\,1 < x < 5\\\,\,\,\,\,\,\,x - 1 + x - 5,\,\,\,\,x > 5\end{array} \right.$

$f(x) = \left\{ {\begin{array}{*{20}{c}}{6 - 2x,}&{x < 1}\\{4\,\,\,\,\,\,,}&{1 < x < 5}\\{2x - 6,}&{x > 5}\end{array}} \right.$

$\because$ $x = 3 \in (1,\,5)$, 

For $x = 3, $   $ f(x) = 4,\;\;f'(x) = 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 5. continuity and differentiation→5. continuity and differentiation — all question types→ગણિત ધોરણ 12 સાયન્સ question papers→Gujarat Board ધોરણ 12 સાયન્સ question papers→Gujarat Board question paper generator→

Similar questions

$\tan ^{-1}\left(\tan \frac{31 \pi}{6}\right)=$ ________.
→
જો બિંદુ $R (4, y, z)$ એ બિંદુઓ $P (2, -3, 4)$ અને $Q (8, 0, 10)$ ને જોડતી રેખા પર આવેલ હોય તો $R$ નું ઉગમબિંદુથી અંતર મેળવો.
→
વક્ર y = x|x|, X-અક્ષ અને રેખાઓ x = - 1 તથા x = 1 વડે આવૃત્ત પ્રદેશનું ક્ષેત્રફળ _________ છે.
→
વિકલ સમીકરણ $y = x\frac{{dy}}{{dx}} + \sqrt {{a^2}{{\left( {\frac{{dy}}{{dx}}} \right)}^2} + {b^2}} $ ના કક્ષા અને પરિમાણ મેળવો.
→
Consider two events $A$ and $B$ such that $P(A) = \frac{1}{4},\,\,P\left( {\frac{B}{A}} \right) = \frac{1}{2},\,\,P\left( {\frac{A}{B}} \right) = \frac{1}{4}.$ For each of the following statements, which is true

$I.$    $P\,({A^c}/{B^c}) = \frac{3}{4}$

$II.$   The events $A$ and $B$ are mutually exclusive

$III.$  $P(A/B) + P(A/{B^c}) = 1$

→
જો $A = (k , 1, -1) ; B = (2k, 0, 2)$ અને $C = (2 + 2k, k, 1)$ જો $AB \perp  BC$, હોય, તો $k$ નું મૂલ્ય....
→
$\int_{}^{} {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}\;dx = } $
→
$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $
→
જો $u = {{x + y} \over {x - y}}$, તો ${{\partial u} \over {\partial x}} + {{\partial u} \over {\partial y}} = $
→
જો$f(x)\begin{vmatrix}{c1}&{x}&{x^2}\\{x}&{x^2}&{1} \\{x^2}&{1}&x\end{vmatrix}$ હોય તો $f\left( {\sqrt[3]{3}} \right) = .........$
→