Download App

STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
Function $f\left( x \right) = \left( {x - 2} \right)\left| {x - 3} \right|$ is monotonically increasing in
 
  • ✓
    $\left( { - \infty ,\frac{5}{2}} \right) \cup \left( {3,\infty } \right)$
  • B
    $\left( {\frac{5}{2},\infty } \right)$
  • C
    $\left( {2,\infty } \right)$
  • D
    $\left( { - \infty ,3} \right)$
✓

Answer

Correct option: A.
$\left( { - \infty ,\frac{5}{2}} \right) \cup \left( {3,\infty } \right)$
a
Given : $f(x)=(x-1)|(x-2)(x-3)|$

at $x < 0 \quad f(x)=(x-1)(-x-2)(-x-3)$

$=(x-1)(x+2)(x+3)$

$(x-1)\left(x^2+3 x+2 x+6\right)$

$=(x-1)\left(x^2+5 x+6\right)$

$=x^3+5 x^2-1 x^2-6$

$f(x)=x^3+4 x^2-6$

Differentiating w.r.t $x$

$f^1(x)=3 x^2-12 x=11$

Evaluate $f ^1( x )=0$

Given $f(x)=(-x)$

Given: $f(x)=(x-1)|(x-2)(x-3)|$

Finding critical points for $f(x)$

$f(x)=(-x+2)(-x+3)(x-1)$

$f^1(x)=3 x^2-12 x-11$

Evaluate $f ^{ l }( x )=0$

$3 x^2-12 x=11=0$

Therefore for $x < 2$

$x =\frac{6-\sqrt{3}}{3}$

$x=\alpha-\frac{1}{\sqrt{3}}$

Therefore the intervals where the function is increasing is

Therefore $\left(2-\frac{1}{\sqrt{3}}, 2\right)$ is the interval where the function is decreasing.

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 "M.C.Q (1 Marks)" from STD 12 - 6. Application of derivatives→STD 12 - 6. Application of derivatives — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be two non-constant differentiable functions. If $f^{\prime}(x)=\left(e^{(f(x)-g(x))}\right) g^{\prime}(x)$ for all $x \in R$, and $f(1)=g(2)=1$, then which of the following statement (s) is (are) $TRUE$? 

$(A)$ $f (2)<1-\log _{ e } 2$  $(B)$ $f (2)>1-\log _{ e } 2$  $(C)$ $g(1)>1-\log _e 2$  $(D)$ $g(1)<1-\log _e 2$

→
$\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}=$
→
Find the equation of a line passing through $(1,2,-3)$ and parallel to the line $\frac{x-2}{1}=\frac{y+1}{3}=\frac{z-1}{4}$.
→
The value of $\frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} d x}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} d x}$ is $........$
→
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion $ (A) : \frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)^4=(\text{x}^2+\text{x}+1)^3(2\text{x}+1)$
Reason $ (R) : (\text{fog}\ '=\text{f '}[\text{g(x)}].\text{g '(x)}$
→
Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $\text{aRb}$ if a is congruent to $b \ \forall \ a, b \in T.$ Then $R$ is:
→
If $I_n =$ $\int\limits_0^1 {\,\,\frac{{dx}}{{{{\left( {1\,\, + \,\,{x^2}} \right)}^n}}}} $; $n \in N$, then which of the following statements hold good ?
→
Let $'a'$ $(a < 0, a \notin  I)$ be a constant and $'t'$ be a parameter, then set of values of $'t'$ for which the function $f(x)=\left( \frac{\left| \left[ t \right]+1 \right|+a}{\left| \left[ t \right]+1 \right|+1-a} \right)x$ ,is a decreasing function of $x$ ([.] denote greatest integer function) is
→
Let $f(1) = -2$ and $f'(x)  \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval
→
If $a=\sin ^{-1}(\sin (5))$ and $b=\cos ^{-1}(\cos (5))$, then $a^2+b^2$ is equal to
→