RD Sharma Solutions Class 12 Mathematics Chapter 9 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 9 FBQ

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 04:14 PM IST

The RD Sharma books are the most prescribed solution books for the Class 12 students. However, when it comes to Mathematics, it is common that students face difficulties in a few chapters, especially Chapter 9, Differentiability. RD Sharma solution In the first place, when sums are hard to solve, objective-type questions like FBQs are even more challenging. In such cases, the RD Sharma Class 12th FBQ solution guide will lend a helping hand.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## RD Sharma Class 12 Solutions Chapter 9 FBQ Differentiability - Other Exercise

Differentiability Excercise:FBQ

Differentiability exercise Fill in the blanks question 1

Answer: $x=-1$
Hint: If f is differentiable at all $x\in R$, we must show $f^{'}(x)$exists at all $x\in R$
Given: The function $f(x)=|x+1|$
Solution:
$f(x)=|x+1|$
We know that
\begin{aligned} &|x+1|=\left\{\begin{array}{l} x+1, x+1 \geq 0 \\ -(x+1), x+1<0 \end{array}\right\} \\\\ &f(x)=\left\{\begin{array}{l} x+1, x \geq-1 \\ -x-1, x<-1 \end{array}\right\} \end{aligned}
\begin{aligned} &\lim _{x \rightarrow-1^{+}}(x+1)=-1+1=0 \\\\ &\lim _{x \rightarrow-1^{-}}(-x-1)=1-1=0 \end{aligned}
Thus, the function is continuous.
\begin{aligned} &x \geq-1, x<-1 \\\\ &f(x)=x+1, f(x)=-x-1 \\\\ &f^{\prime}(x)=1, f^{\prime}(x)=-1 \end{aligned}
\begin{aligned} &f^{\prime}\left(-1^{+}\right)=1, f^{\prime}\left(-1^{-}\right)=-1 \\\\ &f^{\prime}\left(-1^{+}\right)=1 \neq f^{\prime}\left(-1^{-}\right)=-1 \end{aligned}
Therefore, the function $f(x)=|x+1|$ is not differentiable at $x=-1$

Differentiability exercise Fill in the blanks question 2

Answer: $x=\pm 1$
Hint: If f is differentiable at all $x\in R$, we must show$f'(x)$ exists at all $x\in R$
Given: $g(x)=|x-1|+|x+1|$
Solution: we know that
\begin{aligned} g(x) &=\left\{\begin{array}{l} -x-1+1-x, x<-1 \\ x+1+1-x,-1 \leq x<1 \\ x-1+x+1, x \geq 1 \end{array}\right\} \\ &=\left\{\begin{array}{l} -2 x, x<-1 \\ 2,-1 \leq x<1 \\ 2 x, x \geq 1 \end{array}\right\} \end{aligned}
\begin{aligned} &g^{\prime}(x)=\left\{\begin{array}{l} -2, x<-1 \\ 0,-1 \leq x<1 \\ 2, x \geq 1 \end{array}\right\} \\\\ &g^{\prime}\left(-1^{-}\right)=-2, g^{\prime}\left(-1^{+}\right)=0 \end{aligned}
$g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=2$
The above derivative calculations can be done from the first principle also.
Since LHS and RHS derivatives are not equal at the points 1 and -1.
Hence the function is not differentiable at $x=\pm 1$.

Differentiability exercise Fill in the blanks question 3

Answer: $z$
Hint: [r] is not continuous ⇒ not differentiable at every point.
If $\mathrm{f}(\mathrm{x})=\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})$ where $\mathrm{f}_{1}(\mathrm{x})$ is differentiable at $x=C$ and $\mathrm{f}_{2}(\mathrm{x})$ is not differentiable at $x=C$ then $\mathrm{f}(\mathrm{x})=\mathrm{f}_{1}(\mathrm{x})+\mathrm{f}_{2}(\mathrm{x})$ is also not differentiable at $x=c.$
Given: $\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]$
Solution:

We have $\mathrm{f}(\mathrm{x})=\mathrm{x}-[\mathrm{x}]$

Let $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{h}(\mathrm{x})$

Where $g(x)=x \text { and } h(x)=-[x]$

We know that [x] is not continuous at every point of of $\mathrm{x} \Rightarrow[\mathrm{x}]$ is not differentiable for all x

So , $h(x) = -[x]$ is also not continuous and this implies

$h(x) = --[x]$ is not differentiable for $x \in z$

Since $h(x)$ is not differentiable $\forall \mathrm{x} \in \mathbb{Z}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{h}(\mathrm{x})$ is also not differentiable for all values of r ie when $x \in z$

Differentiability exercise Fill in the blanks question 4

Answer: $\pm \pi$ hence there are 2 points where f(x) is not differentiable
Hint: : If $f(x)$ is differentiable at all $x\in R$, we must show$f'(x)$exists at all $x\in R$
Given: The function $f(x)=\sin ^{-1}(\sin x)$ is not differentiable is
Solution:
$f(x)=\sin ^{-1}(\sin x)$
The function $f(x)$is continuous everywhere but not differentiable at $(2 n+1) \frac{\pi}{2}, n \in Z$
$f(x)$ is an odd function.
$f^{\prime}(x)=(-1)^{n} \text {, if }(2 n-1) \frac{\pi}{2}
$\therefore x=\pm \pi$ where $f(x)$ is not differentiable
$\therefore$ hence there are 2 points where $f(x)$ is not differentiable

Differentiability exercise Fill in the blanks question 5

Answer: $\pm \pi$
Hint: If $f(x)$ is differentiable at all $x\in R$, we must show$f'(x)$exists at all $x\in R$
Given: $f(x)=\cos ^{-1}(\cos x)$
Solution:
$f(x)=\cos ^{-1}(\cos x)=\left\{\begin{array}{l} 2 \pi+x, x \in(2 \pi,-\pi) \\ -x, x \in(-\pi, 0) \\ x, x \in(0, \pi) \\ 2 \pi-x, x \in(\pi, 2 \pi) \\ 2 \pi+x, x \in(2 \pi, 3 \pi) \end{array}\right\}$
\begin{aligned} &\text { At } x=\pi, \\ &\begin{array}{l} \lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{-}} x=\pi \\\\ \lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{+}}(2 \pi-x)=2 \pi-\pi=\pi \end{array} \end{aligned}
\begin{aligned} &\mathrm{LHS}=\mathrm{RHS} \Rightarrow f(x) \text { is continuous. } \\\\ &\text { At } x=\pi \\\\ &\qquad f^{\prime}(x)=\left\{\begin{array}{l} 1, x \in(0, \pi) \\\\ 0-1, x \in(\pi, 2 \pi) \end{array}\right\} \end{aligned}
\begin{aligned} &\lim _{x \rightarrow \pi^{-}} f^{\prime}(x)=1 \\\\ &\lim _{x \rightarrow \pi^{+}} f^{\prime}(x)=-1 \end{aligned}
$\mathrm{LHS} \neq \mathrm{RHS} \Rightarrow f(x)$ is not differentiable.
$\therefore \quad At , x=\pm \pi f(x)$ is continuous but not differentiable.
At $x=-\pi$
\begin{aligned} &\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi-2 \pi+x}=\lim _{h \rightarrow 0} 2 \pi+(\pi-h) \\\\ &=\lim _{h \rightarrow 0} 2 \pi+\pi-h=3 \pi-0=3 \pi \end{aligned}
$\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{+}}-x=\lim _{h \rightarrow 0}-(\pi+h)$
\begin{aligned} &=\lim _{h \rightarrow 0}-\pi-h \\\\ &=-\pi-0 \end{aligned}
\begin{aligned} &=-\pi \\ &\text { L.H.S } \neq \text { R. H.S } \end{aligned}
$f(x)$ is not continuous at $x=-\pi$
Since we know that not continuous ⇒not differentiable at a point
So $f(x)$ is also not differentiable at $x=-\pi$ as
$f(x)$ is not continuous at $x=-\pi$

Differentiability exercise Fill in the blanks question 6

Hint: If $f(x)$ is differentiable at all $x\in R$, we must show$f'(x)$exists at all $x\in R$
Given: $f(x)=|\sin x|,\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Solution:
We know that$f(x)=|\sin x|$ is continuous everywhere but not differentiable at integer multiples of $\pi$and at $x=0$.
At $x=0$ and $x=k\pi$ ($k$ is an integer),

From the graph,$x = 0$ at this point $\left | \sin x \right |$ is not differentiable.

Differentiability exercise Fill in the blanks question 7

Answer: $\mathrm{a}=\frac{1}{2}$
Hint:
Use differentiability condition
f(x) is said to be differentiable at x=b
Given:
$f(x)= \begin{cases}a x^{2}+3 & x>1 \\ x+\frac{5}{2} & x \leq 1\end{cases}$is differentiable at x=1
Solution:
we have $\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{ax}^{2}+3 & \mathrm{x}>1 \\ \mathrm{x}+\frac{5}{2} & \mathrm{x} \leq 1\end{cases}$
\begin{aligned} \text { L.H.D } &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(1-\mathrm{h})-\mathrm{f}(1)}{-\mathrm{h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h})+5 / 2-(1+5 / 2)}{-\mathrm{h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h}+5 / 2-1-5 / 2}{-\mathrm{h}} \end{aligned}
$=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{-\mathrm{h}}=+1$
\begin{aligned} \mathrm{R} \cdot \mathrm{H} \cdot \mathrm{D}=& \lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(1+\mathrm{h})-\mathrm{f}(1)}{\mathrm{h}} \\ &=\lim _{\mathrm{n} \rightarrow 0} \frac{\mathrm{a}(1+\mathrm{h})^{2}+3-\left(\mathrm{a}(1)^{2}+3\right)}{\mathrm{h}} \end{aligned}
\begin{aligned} &=\lim _{\mathrm{n} \rightarrow 0} \frac{\mathrm{a}\left(1+\mathrm{h}^{2}+2 \mathrm{~h}\right)+3-\mathrm{a}-3}{\mathrm{~h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{a}+\mathrm{ah}^{2}+2 \mathrm{ah}+3-\mathrm{a}-3}{\mathrm{~h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}(\mathrm{ah}+2 \mathrm{a})}{\mathrm{h}} \end{aligned}
\begin{aligned} &\lim _{\mathrm{h} \rightarrow 0} \mathrm{ah}+2 \mathrm{a} \\ &\mathrm{a} \times 0+2 \mathrm{a}=2 \mathrm{a} \end{aligned}
Since $f(x)$ is differentiable at $x=1$
So,$LHD= RHD$
\begin{aligned} &\Rightarrow 1=2 \mathrm{a} \\ &\Rightarrow\mathrm{a}=\frac{1}{2} \end{aligned}

Differentiability exercise Fill in the blanks question 8

Hint: Substitute -1 in the function after differentiating w.r.t x.
Given:
Solution:
\begin{aligned} &f(x)=y=\left\{\begin{array}{l} x^{2}, x \geq 0 \\ -x^{2}, x<0 \end{array}\right\} \\\\ &f^{\prime}(x)=\left\{\begin{array}{l} 2 x, x \geq 0 \\\\ -2 x, x<0 \end{array}\right\} \\\\ &f^{\prime}(-1)=-2(-1)=2 \end{aligned}

Differentiability exercise Fill in the blanks question 9

Hint: Substitute 2 in the function after differentiating w.r.t x.
Given: $f(x)=x|x|$
Solution:
\begin{aligned} &f(x)=\left\{\begin{array}{l} x^{2}, x \geq 0 \\ -x^{2}, x<0 \end{array}\right\} \\\\ &f^{\prime}(x)=\left\{\begin{array}{l} 2 x, x \geq 0 \\\\ -2 x, x<0 \end{array}\right\} \\\\ &f^{\prime}(2)=2(2)=4 \end{aligned}

Differentiability exercise Fill in the blanks question 10

Answer: $R-\left\{\frac{1}{2}\right\}$
Hint:
Given: $f(x)=|2 x-1|$
Solution:
$f(x)=|2 x-1|$
$f(x)= \begin{cases}(2 x-1) & x \geq 0 \\ -(2 x-1) & x<0\end{cases}$
\begin{aligned} &L_{h \rightarrow 0^{-}} f^{\prime}\left(\frac{1}{2}\right)=\lim _{h} \frac{f\left(\left(\frac{1}{2}\right)+h\right)-f\left(\frac{1}{2}\right)}{h} \\\\ &=\lim _{h \rightarrow 0^{-}} \frac{-\left[2\left(\frac{1}{2}\right)+h-0\right]}{h} \\\\ &=-1 \end{aligned}
\begin{aligned} &\operatorname{Rf}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{\left[\left[2\left(\frac{1}{2}\right)-1\right]-0\right]}{h} \\\\ &=1 \end{aligned}
$L\; f^{\prime}\left(\frac{1}{2}\right) \neq R\; f^{\prime}\left(\frac{1}{2}\right)$
Hence $f(x)$ is differentiable $\forall x \in R-\left\{\frac{1}{2}\right\}$

Differentiability exercise Fill in the blanks question 11

Answer: $x=2$
Hint:
Given: $f(x)=\left\{\begin{array}{l} x+1, x<2 \\ 2 x-1, x \geq 2 \end{array}\right\}$
Solution:
\begin{aligned} &L f(x)=x+1 \\ &f^{\prime}(x)=1 \\ &f(1)=1+1=2 \end{aligned} \begin{aligned} &R f(x)=2 x-1 \\ &f^{\prime}(x)=2 \\ &f(1)=2-1=1 \end{aligned}
$f(2)=2+1=3$ $f(2)=2(2)-1=4-1=3$
$\therefore L f(x)=R f(x)$
Hence, $f(x)=\left\{\begin{array}{l} x+1, x<2 \\ 2 x-1, x \geq 2 \end{array}\right\}$ is not differentiable at $x=2$.

Differentiability exercise Fill in the blanks question 12

Answer:$x=0$ and $x=1$
Given: An example of a function which is continuous everywhere but fails to be differentiable exactly at two points.
Solution:
The function is continuous everywhere but fails to be differentiable exactly at two points $x=0$ and $x=1$
$x+|x-1|$
Hence,$f(x)=x+|x-1|$ is continuous but not differentiable at $x=0,1$.

Differentiability exercise Fill in the blanks question 13

Answer: $f(x)=\cos |x|$ is differentiable at set of R
Given: $f(x)=\cos |x|$
Solution:
The function $f(x)=\cos |x|$ is differentiable everywhere except at points where $x=\left(k+\frac{1}{2}\right) \pi$ where k is an integer.
Hence, $f(x)=\cos |x|$ is differentiable at set of R.

Differentiability exercise Fill in the blanks question 14

Answer: $[0]$
Given: $f(x)=|\sin x|$
Solution: $f(x)=|\sin x|$
$f(x)= \begin{cases}\sin x & x \geq 0 \\ -(\sin x) & x<0\end{cases}$
\begin{aligned} &R f^{\prime}(0)=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{|\sinh |-0}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{\sinh }{h} \\\\ &=1 \end{aligned}
\begin{aligned} &\text { As } h \rightarrow 0^{+},|\sinh |=\sinh \\\\ &\begin{aligned} L f^{\prime}(0) &=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h} \\\\ &=\lim _{h \rightarrow 0^{-}} \frac{|\sinh |-0}{h} \end{aligned} \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0^{-}} \frac{(-\sinh )}{h} \\ &=-1 \end{aligned}
\begin{aligned} &\text { As } h \rightarrow 0^{-},|\sinh |=-\sinh \\\\ &\text { So, } R f^{\prime}(0) \neq L f^{\prime}(0) \end{aligned}
Hence, $f(x)=|\sin x|$ is not differentiable at $x=0$.

Differentiability exercise Fill in the blanks question 15

Answer:$[-1,0,1]$
Given:$f(x)=\frac{1}{\log |x|}$
Solution:
The function $\log \left | x \right |$ is not defined at $x=0$ . So, $x=0$ is a point of discontinuity.
Also, for $f(x)$ to be defined,$\log \left | x \right |=0$, that is $x\neq I$(Integer)
Hence, 1 and -1 are also points of discontinuity.
Clearly$f(x), x \in R-\{0,1,-1\}$ where the function is not differentiable

Differentiability exercise Fill in the blanks question 16

Answer: $x=1$
Given: The greatest integer function, $f(x)=[x], 0
Solution:
RHD:
$f(x)=x$
\begin{aligned} R f^{\prime}(0) &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\\\ &=\lim _{h \rightarrow 0} \frac{[1+h]-[1]}{h} \\\\ &=\lim _{h \rightarrow 0} \frac{(1-1)}{h} \\\\ &=0 \end{aligned}
LHD
\begin{aligned} &L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\\\ \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0^{+}} \frac{[1-h]-[1]}{-h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{0-1}{-h} \\\\ &=\lim _{h \rightarrow 0} \frac{1}{h} \\\\ &=\infty \end{aligned}
Since RHD $\neq$ LHD
$\therefore f(x)$ is not differentiable at $x=1$

Class 12 RD Sharma Chapter 9 FBQ Solution is one of the difficult chapters that the students face. The FBQs are tricky and make students spend more time on them. This chapter consists of two exercises, ex 9.1 and 9.2. Next comes the FBQ section, with 16 questions to be answered. Some common concepts are Differentiability functions, continuous function, discontinuous function, and a set of differentiable points. The RD Sharma Class 12 Chapter 9 FBQ helps you solve it.

All the solutions in RD Sharma Class 12th FBQ book are given by experts who possess wide knowledge in this field. The RD Sharma books follow the NCERT pattern, which helps the students of the CBSE board. The presence of numerous practice questions has become a one-stop solution for clarifying homework doubts and exam preparation.

If you find difficulties filling the blanks in the Differentiability concept, get the help of the solutions present in the RD Sharma Class 12 Solutions Chapter 9 FBQ Solution book. Answers are provided for every FBQ question present in the book. This RD Sharma solution book will help you get accurate answers and save time.

The RD Sharma Class 12 Solutions Differentiability FBQ can be downloaded from the Career360 website for free of cost. It takes only a few swipes and clicks to visit the Career360 website and download the best of RD Sharma solution books.

The RD Sharma Class 12th FBQ book is used for preparing question papers by teachers for class tests, and even for public examinations, it leads the students on the right path. Once the students practice sums from the RD Sharma books regularly, they can witness themselves crossing their benchmark scores. Therefore, RD Sharma solution books are the best guidance for the students at their homes.

Therefore, if you have any nagging doubts regarding the FBQs in the Differentiability chapter, you can very well download the RD Sharma Class 12 Solutions Chapter 9 FBQ book for reference. An abundance of students has benefitted from this set of books. All you have to do it, look for the book you need, download them, and start practising regularly.

## RD Sharma Chapter wise Solutions

1. Which solution book can the students refer to clear their doubts regarding the mathematics chapter 9 FBQ?

The syllabus of Differentiability is a bit hard for the students; they can get help from the RD Sharma Class 12th FBQ solution books.

2. Do RD Sharma books contain solutions only for exercise problems?

The RD Sharma books provide solutions for all exercise problems and MCQ, FBQ, and VSA that are given in the textbook.

3. Do the RD Sharma solution books follow the NCERT syllabus?

Yes, the RD Sharma solution books follow the NCERT syllabus. Hence, the CBSE board students can use it as their guide.

4. How can I access the RD Sharma solution books from the Career 360 website?

It is an easy-peasy task to access the RD Sharma books. First, visit the Career 360 website and search for the name of the solutions book that you are looking for. For example, searching for RD Sharma Class 12th FBQ will give you the FBQ solution book of this chapter.

5. How much should I pay to download the RD Sharma solution books?

The RD Sharma books are available completely free of cost. So be it, students or teachers, no one is required to pay monetary charges. You can download and use it whenever required.

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