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Class 11 NCERT Solutions

Chapter 12: Limits and Derivatives

Master the intuition of approaching values, the first principle of differentiation, and the logic of instantaneous change with our step-by-step logic.

Exercise 12.1
1. $\lim _{\boldsymbol{x} \rightarrow \mathbf{3}} \boldsymbol{x} \boldsymbol{+} \mathbf{3}$.
Solution

We know that the limit of a polynomial function is the value of the function at the prescribed point. i.e., if $f(x)$ is









a polynomial function, then $\lim _{x \rightarrow a} f(x)=f(a)$, obtained by writing $a$ for $x$ in the function.

$$ \therefore \lim _{x \rightarrow 3} x+3=3+3=6 . $$
2. $\lim _{x \rightarrow \pi}\left(x-\frac{\mathbf{2 2}}{\mathbf{7}}\right)$.
Solution

$\lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right)=\left(\pi-\frac{22}{7}\right)$

Remark. $\pi \neq \frac{22}{7}$ since $\pi$ is irrational whereas $\frac{22}{7}$ is rational. However, $\frac{22}{7}$ is an approximate value of $\pi$. $\frac{355}{113}$ is another approximate value of $\pi$.

3. $\lim _{r \rightarrow 1} \pi r^2$.
Solution

$\lim _{r \rightarrow 1} \pi r^2=\pi \times 1^2=\pi$.

4. $\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}$.
Solution

$\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{19}{2}$.

5. $\lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1}$.
Solution

$\lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1}=\frac{(-1)^{10}+(-1)^5+1}{(-1)-1}=\frac{1-1+1}{-2}=-\frac{1}{2}$.

6. $\lim _{x \rightarrow 0} \frac{(x+1)^5-1}{x}$.
Solution

On putting $x=0$, we get $\frac{1^5-1}{0}=\frac{0}{0}$ which is an indeterminate form. Put $x+1=y$, i.e., $\quad x=y-1$ so that $y \rightarrow 1$ as $x \rightarrow 0$.

$$ \begin{aligned} \therefore \lim _{x \rightarrow 0} \frac{(x+1)^5-1}{x}= & \lim _{y \rightarrow 1} \frac{y^5-1^5}{y-1} \\ & {\left[\text { Form } \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}, \text { Here } n=5, a=1\right] } \end{aligned} $$





$$ \begin{array}{ll} =5 \times 1^{5-1} & {\left[n a^{n-1}\right]} \\ =5 & \end{array} $$
7. $\lim _{x \rightarrow 2} \frac{3 x^2-x-10}{x^2-4}$.
Solution

$\lim _{x \rightarrow 2} \frac{3 x^2-x-10}{x^2-4} \quad\left(\right.$ Putting $x=2$, we get the Form $\left.\frac{0}{0}\right)$

Factorising numerator and denominator:

$$ \begin{aligned} {\left[3 x^2-x\right.} & -10=3 x^2-6 x+5 x-10=3 x(x-2)+5(x-2) \\ & =(x-2)(3 x+5)] \\ & =\lim _{x \rightarrow 2} \frac{(x-2)(3 x+5)}{(x-2)(x+2)}[\text { Cancelling }(x-2) \neq 0] \\ & =\lim _{x \rightarrow 2} \frac{3 x+5}{x+2}=\frac{3 \times 2+5}{2+2}=\frac{11}{4} \end{aligned} $$
8. $\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3}$.
Solution

$\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3} \quad\left(\right.$ Putting $x=3$, we get the Form $\left.\frac{0}{0}\right)$

Factorising both numerator and denominator:

$$ \begin{aligned} & {\left[x^4-81=\left(x^2-9\right)\left(x^2+9\right)=(x-3)(x+3)\left(x^2+9\right)\right.} \\ & \text { and } 2 x^2-5 x-3=2 x^2-6 x+x-3=2 x(x-3)+(x-3) \\ & =(x-3)(2 x+1)] \\ & =\lim _{x \rightarrow 3} \frac{(x-3)(x+3)\left(x^2+9\right)}{(x-3)(2 x+1)}[\text { Cancelling }(x-3) \neq 0] \\ & =\lim _{x \rightarrow 3} \frac{(x+3)\left(x^2+9\right)}{2 x+1}=\frac{(3+3)\left(3^2+9\right)}{2(3)+1}=\frac{6 \times 18}{7}=\frac{108}{7} . \end{aligned} $$
9. $\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}$.
Solution

$\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}=b$.

10. $\lim _{z \rightarrow 1} \frac{z^{1 / 3}-1}{z^{1 / 6}-1}$.
Solution

On putting $z=1$, we get the form $\left(\frac{0}{0}\right)$







Put $z^{1 / 6}=y$ so that $y \rightarrow 1$ as $z \rightarrow 1$. Then

$$ \begin{gathered} \begin{array}{l} \lim _{z \rightarrow 1} \frac{z^{1 / 3}-1}{z^{16}-1}=\lim _{y \rightarrow 1} \frac{y^2-1}{y-1} \quad\left[z^{1 / 3}=z^{2 / 6}=\left(z^{1 / 6}\right)^2=y^2\right] \\ \quad\left(\text { This is again } \frac{0}{0} \text { form }\right) \\ =\lim _{y \rightarrow 1} \frac{(y+1)(y-1)}{y-1} \\ \text { Cancelling }(y-1), \quad=\lim _{y \rightarrow 1}(y+1)=1+1=2 \end{array} \text { } \end{gathered} $$

**Alternative approach:**

$$ \lim _{z \rightarrow 1} \frac{z^{1 / 3}-1}{z^{16}-1}=\lim _{z \rightarrow 1} \frac{z^{1 / 3}-1^{1 / 3}}{z^{1 / 6}-1^{1 / 6}} $$

Dividing both numerator and denominator by ( $z-1$ ),

$$ \begin{aligned} & =\lim _{z \rightarrow 1}\left(\frac{\frac{z^{1 / 3}-1^{1 / 3}}{z-1}}{\frac{z^{1 / 6}-1^{1 / 6}}{z-1}}\right)=\frac{\frac{1}{3}(1)^{1 / 3-1}}{\frac{1}{6}(1)^{1 / 6-1}} \quad\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] \\ & =\frac{\frac{1}{3}}{\frac{1}{6}}=\frac{1}{3} \times \frac{6}{1}=2 \end{aligned} $$

Sol. $\lim _{x \rightarrow 1} \frac{a x^2+b x+c}{c x^2+b x+a}=\frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a}=\frac{a+b+c}{c+b+a}=1$.

12. $\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$.
Solution

$\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} \quad\left(\right.$ Putting $x=-2$, we get the Form $\left.\frac{0}{0}\right)$

Using L.C.M. to write as a single fraction:

$$ =\lim _{x \rightarrow-2} \frac{x+2}{2 x(x+2)} $$





$$ \begin{aligned} & =\lim _{x \rightarrow-2} \frac{1}{2 x} \quad[\text { Cancelling }(x+2) \neq 0] \\ & =\frac{1}{2(-2)}=-\frac{1}{4} . \end{aligned} $$
13. $\lim _{x \rightarrow 0} \frac{\sin a x}{b x}$.
Solution

$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\lim _{x \rightarrow 0}\left(\frac{a}{b} \cdot \frac{\sin a x}{a x}\right)=\frac{a}{b} \lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)$

$$ [\because \quad \text { as } x \rightarrow 0, a x \rightarrow 0] $$
$$ =\frac{a}{b} \times 1=\frac{a}{b} . $$
14. $\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$.
Solution

$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0}\left(\frac{a}{b} \cdot \frac{\sin a x}{a x} \cdot \frac{b x}{\sin b x}\right)$

$$ \begin{aligned} & =\frac{a}{b}\left(\lim _{a x \rightarrow 0} \frac{\sin a x}{a x} \div \lim _{b x \rightarrow 0} \frac{\sin b x}{b x}\right)\left(\therefore \frac{c}{d}=\frac{1}{\left(\frac{d}{c}\right)}\right) \\ & \quad[\because \text { as } x \rightarrow 0, a x \rightarrow 0 \text { and } b x \rightarrow 0] \\ & =\frac{a}{b}(1 \div 1)=\frac{a}{b} \end{aligned} $$
15. $\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$.
Solution

Put $\pi-x=t$ so that $t \rightarrow 0$ as $x \rightarrow \pi$.

$$ \therefore \quad \lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}=\frac{1}{\pi} \lim _{t \rightarrow 0} \frac{\sin t}{t}=\frac{1}{\pi} \times 1=\frac{1}{\pi} . $$
16. $\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}$.
Solution

$\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}$.

17. $\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}$.
Solution

$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}=\lim _{x \rightarrow 0} \frac{2 \cos ^2 x-1-1}{\cos x-1}$




$$ =\lim _{x \rightarrow 0} \frac{2\left(\cos ^2 x-1\right)}{(\cos x-1)} $$

Factorising the numerator:

$$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{2(\cos x+1)(\cos x-1)}{\cos x-1} \\ & =\lim _{x \rightarrow 0} 2(\cos x+1) \\ & =2(\cos 0+1)=2(1+1)=4 \end{aligned} $$
18. $\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}$.
Solution

Dividing numerator and denominator by $x$

$$ \begin{aligned} \lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x} & =\lim _{x \rightarrow 0} \frac{a+\cos x}{b\left(\frac{\sin x}{x}\right)}=\frac{\lim _{x \rightarrow 0}(a+\cos x)}{\lim _{x \rightarrow 0} b\left(\frac{\sin x}{x}\right)} \\ & =\frac{a+1}{b(1)}=\frac{a+1}{b} \end{aligned} $$
19. $\lim _{x \rightarrow 0} x \sec x$.
Solution

$\lim _{x \rightarrow 0} x \sec x=\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}=0$.

20. $\lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x}$.
Solution

Dividing numerator and denominator by $x$

$$ \begin{aligned} \lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x} & =\lim _{x \rightarrow 0} \frac{\frac{\sin a x}{x}+b}{a+\frac{\sin b x}{x}} \\ & =\lim _{x \rightarrow 0} \frac{a\left(\frac{\sin a x}{a x}\right)+b}{a+b\left(\frac{\sin b x}{b x}\right)} \\ & =\frac{\lim _{a x \rightarrow 0}\left[a\left(\frac{\sin a x}{a x}\right)+b\right]}{\lim _{b x \rightarrow 0}\left[a+b\left(\frac{\sin b x}{b x}\right)\right]} \\ & =\frac{a(1)+b}{a+b(1)}=\frac{a+b}{a+b}=1 . \end{aligned} $$
21. $\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)$.
Solution

$\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)=\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)$

(on changing all T-ratios in terms of $\sin x$ and $\cos x$ )

$$ \begin{aligned} & \text { Taking L.C.M; } \quad=\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin x} \\ & \text { Rationalising the numerator, }=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{\sin x} \times \frac{1+\cos x}{1+\cos x}\right) \\ & =\lim _{x \rightarrow 0} \frac{1-\cos ^2 x}{\sin x(1+\cos x)}=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sin x(1+\cos x)} \\ & \text { Cancelling } \sin x,=\lim _{x \rightarrow 0} \frac{\sin x}{(1+\cos x)}=\frac{\sin 0}{(1+\cos 0)}=\frac{0}{(1+1)} \\ & =\frac{0}{2}=0 \end{aligned} $$
22. $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$.
Solution

Put $x=\frac{\pi}{2}+t$ so that $t=x-\frac{\pi}{2}$ and as $x \rightarrow \frac{\pi}{2}, t \rightarrow 0$

$$ \begin{aligned} & \therefore \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}=\lim _{t \rightarrow 0} \frac{\tan (\pi+2 t)}{t} \\ & \quad=\lim _{t \rightarrow 0} \frac{\tan 2 t}{t}=\lim _{t \rightarrow 0} 2 \cdot \frac{\tan 2 t}{2 t} \\ & {[\because \tan (\pi+\theta)=\tan \theta]} \\ & \quad=2 \lim _{2 t \rightarrow 0} \frac{\tan 2 t}{2 t}=2 \times 1=2 \end{aligned} $$
23. Find $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$, where $f(x)$ $ =\left\{\begin{array}{cc} 2 x+3, & x \leq 0 \\ 3(x+1), & x>0 \end{array} .\right. $
Solution

In the neighbourhood of $0, f(x)$ is defined differently.

Therefore, we shall find both left hand limit and right hand limit.

$$ \begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x+3) \\ & {\left[\because \text { when } x \rightarrow 0^{-}, \quad x<0 \text { and } f(x)=2 x+3 \text { (given) }\right]} \\ & =2 \times 0+3=3 \end{aligned} $$






$$ \begin{aligned} & \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 3(x+1) \\ & {[\because \text { when } x}\left.\rightarrow 0^{+}, \quad x>0 \text { and } f(x)=3(x+1) \text { (given) }\right] \\ &=3(0+1)=3 \\ & \Rightarrow \quad \quad \lim _{x \rightarrow 0^{-}} f(x)=3=\lim _{x \rightarrow 0^{+}} f(x) \\ & \therefore \quad \quad \lim _{x \rightarrow 0} f(x) \text { exists and }=3 \end{aligned} $$

In the neighbourhood of $1, f(x)=3(x+1)$

$$ \begin{aligned} & {\left[\because x \rightarrow 1^{-} \text {or } x \rightarrow 1^{+} \Rightarrow x>0\right] } \\ & \therefore \quad \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} 3(x+1)=3(1+1)=6 . \end{aligned} $$
24. Find $\lim _{x \rightarrow 1} f(x)$, where $f(x)=\left\{\begin{array}{cc}x^2-1, & x \leq 1 \\ -x^2-1, & x>1\end{array}\right.$.
Solution

Here $f(x)$ is defined differently in the neighbourhood of 1 . Therefore, we shall find both left hand limit and right hand limit.

$$ \begin{aligned} & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^2-1\right) \\ & \quad\left[\because \text { when } x \rightarrow 1^{-}, \quad x<1 \text { and } f(x)=x^2-1\right] \\ & \quad=1^2-1=0 \\ & \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(-x^2-1\right) \end{aligned} $$

$\left[\because \quad\right.$ when $x \rightarrow 1^{+}, \quad x>1$ and $\left.f(x)=-x^2-1\right]$

$$ =-1^2-1=-2 $$

Since $\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$, therefore, $\lim _{x \rightarrow 1} f(x)$ does not exist.

25. Evaluate $\lim _{x \rightarrow 0} f(x)$, where $f(x)=\left\{\begin{array}{ll}\frac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$.
Solution

We know that $|x|=\left\{\begin{array}{rll}x, & \text { if } & x \geq 0 \\ -x, & \text { if } & x<0\end{array}\right.$

We shall find both left hand limit and right hand limit.





$$ \begin{aligned} & \text { Now } \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{-}} \frac{-x}{x}=\lim _{x \rightarrow 0^{-}}(-1)=-1 \\ & {\left[x \rightarrow 0^{-} \Rightarrow x<0 \Rightarrow|x|=-x\right]} \end{aligned} $$

Again $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{+}} \frac{x}{x}=\lim _{x \rightarrow 0^{+}}(1)=1$

$$ \left[x \rightarrow 0^{+} \Rightarrow x>0 \Rightarrow|x|=x\right] $$

Since $\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$, therefore, $\lim _{x \rightarrow 0} f(x)$ does not exist.

26. Find $\lim _{x \rightarrow 0} f(x)$, where $f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{array}\right.$.
Solution

L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|}$

$$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{x}{-x}\left[\because x \rightarrow 0^{-} \Rightarrow x<0 \therefore|x|=-x\right] \\ & =\lim _{x \rightarrow 0^{-}}(-1)=-1 \end{aligned} $$
$$ \begin{aligned} \text { R.H.L. } & =\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{x}{|x|} \\ & =\lim _{x \rightarrow 0^{+}} \frac{x}{x} \quad\left[\because x \rightarrow 0^{+} \Rightarrow x>0 \quad \therefore|x|=x\right] \\ & =\lim _{x \rightarrow 0^{+}} 1=1 \end{aligned} $$

Since L.H.L. $\neq$ R.H.L., therefore, $\lim _{x \rightarrow 0} f(x)$ does not exist.

27. Find $\lim _{x \rightarrow 5} f(x)$, where $f(x)=|x|-5$.
Solution$ \begin{aligned} & \text { Sol. L.H.L. }=\lim _{x \rightarrow 5^{-}}(|x|-5) \\ & \quad=\lim _{x \rightarrow 5^{-}}(x-5)\left[\because x \rightarrow 5^{-} x \text { is slightly less than } 5\right. \\ & \quad=5-5=0 \quad \Rightarrow x>0 \quad \therefore|x|=x] \end{aligned} $ $ \begin{aligned} \text { R.H.L. } & =\lim _{x \rightarrow 5^{+}}(|x|-5) \\ & =\lim _{x \rightarrow 5^{+}}(x-5) \quad\left[\because x \rightarrow 5^{+} \Rightarrow x>0 \therefore|x|=x\right] \\ & =5-5=0 \end{aligned} $ Since L.H.L. $=0=$ R.H.L. $\therefore \quad \lim _{x \rightarrow 5} f(x)$ exists and $=0$.
28. Suppose $f(x)=\left\{\begin{array}{cc}a+b x, & x<1 \\ 4, & x=1 \\ b-a x, & x>1\end{array}\right.$ and if $\lim _{x \rightarrow 1} f(x)=f(1)$ what are possible values of $a$ and $\boldsymbol{b}$ ?
Solution

Here $f(x)$ is defined differently in the neighbourhood of 1 . Therefore, we shall find both left hand limit and right hand limit.

$$ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(a+b x)=a+b \times 1=a+b $$

Again $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(b-a x)=b-a \times 1=b-a$

Also $\quad f(1)=4$

$$ \begin{array}{ll} \because & \lim _{x \rightarrow 1} f(x)=f(1) \text { (given), } \\ \therefore & \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)=4 \end{array} $$
$$ \therefore \quad \lim _{x \rightarrow 1^{-}} f(x)=4 \text { and } \lim _{x \rightarrow 1^{+}} f(x)=4 $$

$\Rightarrow \quad a+b=4$ and $b-a=4$

Adding, $\quad 2 b=8 \quad \therefore \quad b=4$

Putting $b=4$ in $a+b=4$, we have $a+4=4$ or $a=0$.

29. Let $a_1, a_2, \ldots, a_n$ be fixed real numbers and define a function $ f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right) . $ What is $\lim _{x \rightarrow a_i} f(x)$ ? For some $a \neq a_1, a_2, \ldots a_n$, compute $\lim _{x \rightarrow a} f(x)$.
Solution

$f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$


Let us find $\lim _{x \rightarrow a_i} f(x)$ for $i=1$

i.e., $\lim _{x \rightarrow a_1} f(x)=\lim _{x \rightarrow a_1}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$

Putting $x=a_1$

$$ =\left(a_1-a_1\right)\left(a_1-a_2\right) \ldots\left(a_1-a_n\right) $$
$$ =0\left(a_1-a_2\right) \ldots\left(a_1-a_{\mathrm{n}}\right)=0 $$

Again $\lim _{x \rightarrow a_i} f(x)=\lim _{x \rightarrow a_i}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_i\right)$

$$ \ldots\left(x-a_n\right) $$

Putting $x=a_i$

$$ \begin{aligned} & =\left(a_i-a_1\right)\left(a_i-a_2\right) \ldots\left(a_i-a_i\right) \ldots\left(a_i-a_n\right) \\ & =\left(a_i-a_1\right)\left(a_i-a_2\right) \ldots 0 \ldots\left(a_i-a_n\right)=0 \end{aligned} $$

for all $i=1,2,3, \ldots, n$.

Again $\lim _{x \rightarrow a} f(x)$ for some $a \neq a_1, a_2, \ldots, a_n$

$$ \lim _{x \rightarrow a}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right) $$

Putting $x=a$

$$ =\left(a-a_1\right)\left(a-a_2\right) \ldots\left(a-a_n\right) . $$
30. If $\boldsymbol{f}(\boldsymbol{x})=\left\{\begin{array}{cc}|\boldsymbol{x}|+1, & x<0 \\ 0, & x=0 \\ |\boldsymbol{x}|-1, & x>0\end{array}\right.$ For what value(s) of $a$ does $\lim _{x \rightarrow a} f(x)$ exist?
Solution

We know that $|x|=\left\{\begin{array}{cc}x, & x>0 \\ -x, & x<0\end{array}\right.$

$$ \therefore \quad f(x)=\left\{\begin{array}{cc} -x+1, & x<0 \\ 0, & x=0 \\ x-1, & x>0 \end{array}\right. $$

Since $a \in \mathrm{R}$, three cases arise.

Case 1. When $a<0, f(x)=-x+1$

$$ \therefore \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}(-x+1)=-a+1 $$




$$ \Rightarrow \quad \lim _{x \rightarrow a} f(x) \text { exists for all } a<0 . $$

Case 2. When $a>0, f(x)=x-1$

$$ \begin{aligned} & \therefore \quad \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}(x-1)=a-1 \\ & \Rightarrow \quad \lim _{x \rightarrow a} f(x) \text { exists for all } a>0 \end{aligned} $$

Case 3. When $a=0$.

$$ \begin{gathered} \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-x+1)=-0+1=1 \\ \lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(x-1)=0-1=-1 \end{gathered} $$

Since $\lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x \rightarrow a^{+}} f(x)$ for $a=0, \lim _{x \rightarrow a} f(x)$ does not exist.

Hence $\lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$.

31. If the function $\boldsymbol{f}(\boldsymbol{x})$ satisfies $\lim _{\boldsymbol{x} \rightarrow 1} \frac{\boldsymbol{f}(\boldsymbol{x})-2}{\boldsymbol{x}^2-1}=\pi$, evaluate $\lim _{x \rightarrow 1} f(x)$.
Solution

Given: $\quad \lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$

$$ \therefore \quad \frac{\lim _{x \rightarrow 1}(f(x)-2)}{\lim _{x \rightarrow 1}\left(x^2-1\right)}=\pi $$

But $\lim _{x \rightarrow 1}\left(x^2-1\right)=1^2-1=1-1=0$, therefore we must have $\lim _{x \rightarrow 1}(f(x)-2)=0$,

because if $\lim _{x \rightarrow 1}(f(x)-2)$ is non-zero, then the given limit becomes $\frac{\text { non-zero }}{0}=\infty$ which does not exist and hence can’t be $\pi$ (given).

$$ \begin{aligned} & \therefore \quad \lim _{x \rightarrow 1} f(x)-\lim _{x \rightarrow 1} 2=0 \\ & \Rightarrow \quad \lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} 2=2 . \quad \therefore \lim _{x \rightarrow 1} f(x)=2 . \end{aligned} $$





$\left[\begin{array}{r}\text { Remark : If } \lim _{x \rightarrow a} \frac{g(x)}{h(x)}=a \text { real number } l \text { and } \lim _{x \rightarrow a} h(x)=h(a)=0, \\ \text { then } \lim _{x \rightarrow a} g(x) \text { must be } 0 .\end{array}\right]$

32. If $f(x)=\left\{\begin{array}{cc}m x^2+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^3+m, & x>1\end{array}\right.$. For what integers $m$ and $n$ does both $\lim _{x \rightarrow 0} f(x)$ and $\lim _{\boldsymbol{x} \rightarrow \mathbf{1}} \boldsymbol{f}(\boldsymbol{x})$ exist?
Solution

Here $f(x)$ is defined differently in the neighbourhood of 0 as well as 1 .

Therefore, we shall find both left hand limit and right hand limit both for $x=0$ and $x=1$.

Now $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(m x^2+n\right)=m \times 0^2+n=n$

Again $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(n x+m)=n \times 0+m=m$

$\because \quad \lim _{x \rightarrow 0} f(x)$ exist (given),

$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \quad \therefore \quad n=m$


Also $\quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(n x+m)=n \times 1+m=n+m$

Again $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(n x^3+m\right)=n \times 1^3+m=n+m$

Here $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=n+m$, therefore, $\lim _{x \rightarrow 1} f(x)$ exists for all values of $m$ and $n$ and $\lim _{x \rightarrow 1} f(x)=m+n$

From (i), we conclude that $m$ and $n$ must be equal integers.

Exercise 12.2
1. Find the derivative of $\boldsymbol{x}^{\mathbf{2}}-\mathbf{2}$ at $\boldsymbol{x}=\mathbf{1 0}$.
Solution

Here $f(x)=x^2-2,(x=) a=10$.

$$ \begin{aligned} & \because f(a+h)=f(10+h)=(10+h)^2-2 \\ & =100+h^2+20 h-2=h^2+20 h+98 \\ & \text { and } f(a)=f(10)=(10)^2-2=100-2=98 \end{aligned} $$





We know that $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

$$ \begin{aligned} & \therefore f^{\prime}(10)=\lim _{h \rightarrow 0} \frac{f(10+h)-f(10)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h^2+20 h+98-98}{h}=\lim _{h \rightarrow 0} \frac{h^2+20 h}{h} \\ & =\lim _{h \rightarrow 0} \frac{h(h+20)}{h} \\ & \text { cancellling } h,=\lim _{h \rightarrow 0}(h+20)=0+20=20 \end{aligned} $$
2. Find the derivative of $99 x$ at $x=100$.
Solution

Here, $\quad f(x)=99 x, \quad(x=) a=100$

$$ \begin{aligned} & f(a+h)=f(100+h)=99(100+h), f(a)=f(100)=99(100) \\ & \begin{aligned} f^{\prime}(100) & =\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h} \\ & =\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h} \\ & =\lim _{h \rightarrow 0} \frac{9900+99 h-9900}{h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \end{aligned} \\ & \text { cancelling } h,=\lim _{h \rightarrow 0} 99=99 \end{aligned} $$

Hence the derivative of $99 x$ at $x=100$ is 99 .

3. Find the derivative of $\boldsymbol{x}$ at $\boldsymbol{x}=\mathbf{1}$.
Solution

Here $f(x)=x, a=1$.

$$ \begin{aligned} f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(1+h)-1}{h}=\lim _{h \rightarrow 0} \frac{h}{h} \\ & =\lim _{h \rightarrow 0} 1=1 \end{aligned} $$
4. Find the derivative of the following functions from first principle. (i) $x^3-27$ (ii) $(x-1)(x-2)$ (iii) $\frac{1}{x^2}$ (iv) $\frac{x+1}{x-1}$.
Solution

(i) Let $f(x)=x^3-27$.

Changing $x$ to $x+h, f(x+h)=(x+h)^3-27$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{\left[(x+h)^3-27\right]-\left(x^3-27\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(x+h)^3-27-x^3+27}{h} \\ & =\lim _{h \rightarrow 0} \frac{(x+h)^3-x^3}{h} \\ & =\lim _{h \rightarrow 0} \frac{x^3+h^3+3 x h(x+h)-x^3}{h} \\ & =\lim _{h \rightarrow 0} \frac{h^3+3 x^2 h+3 x h^2}{h}=\lim _{h \rightarrow 0} \frac{h\left(h^2+3 x^2+3 x h\right)}{h} \\ & =\lim _{h \rightarrow 0}\left(h^2+3 x^2+3 x h\right)=0+3 x^2+0=3 x^2 \\ & \Rightarrow \frac{d}{d x}(f(x))=3 x^2 \\ & \text { Hence } \frac{d}{d x}\left(x^3-27\right)=3 x^2 \end{aligned} $$

(ii) Let $f(x)=(x-1)(x-2)=x^2-3 x+2$.


Changing $x$ to $x+h, f(x+h)=(x+h)^2-3(x+h)+2$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{\left[(x+h)^2-3(x+h)+2\right]-\left(x^2-3 x+2\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(x^2+h^2+2 h x-3 x-3 h+2-x^2+3 x-2\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h^2+2 h x-3 h}{h}=\lim _{h \rightarrow 0} \frac{h(h+2 x-3)}{h} \\ & =\lim _{h \rightarrow 0}(h+2 x-3)=0+2 x-3=2 x-3 \\ & \Rightarrow \frac{d}{d x}(f(x))=2 x-3 . \text { Hence } \frac{d}{d x}((x-1)(x-2))=2 x-3 . \end{aligned} $$



(iii) Let $f(x)=\frac{1}{x^2}$.


Changing $x$ to $x+h, f(x+h)=\frac{1}{(x+h)^2}$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$

$$ =\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} $$

Taking L.C.M.

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{x^2-(x+h)^2}{x^2(x+h)^2}\right]=\lim _{h \rightarrow 0} \frac{x^2-\left(x^2+h^2+2 x h\right)}{h x^2(x+h)^2} \\ & =\lim _{h \rightarrow 0} \frac{-2 x h-h^2}{h x^2(x+h)^2}=\lim _{h \rightarrow 0} \frac{-h(2 x+h)}{h x^2(x+h)^2} \\ & =\lim _{h \rightarrow 0} \frac{-(2 x+h)}{x^2(x+h)^2}=\frac{-(2 x+0)}{x^2(x+0)^2}=\frac{-2 x}{x^2(x)^2}=\frac{-2}{x^3} \\ & \Rightarrow \frac{d}{d x}(f(x))=\frac{-2}{x^3} \quad \text { Hence } \frac{d}{d x}\left(\frac{1}{x^2}\right)=\frac{-2}{x^3} \end{aligned} $$

(iv) Here $f(x)=\frac{x+1}{x-1}$


Changing $x$ to $x+h, f(x+h)=\frac{x+h+1}{x+h-1}$

We know that $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$$ =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{(x+h)+1}{(x+h)-1}-\frac{x+1}{x-1}\right] $$

Using L.C.M.,

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{(x+1+h)(x-1)-(x+1)(x-1+h)}{(x+h-1)(x-1)}\right] \\ = & \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\left[\left(x^2-x+x-1+h x-h-x^2+x-h x-x+1-h\right.\right.}{(x+h-1)(x-1)}\right] \end{aligned} $$


$$ =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-2 h}{(x+h-1)(x-1)}\right] $$
$$ \text { cancelling } h,=\lim _{h \rightarrow 0} \frac{-2}{(x+h-1)(x-1)} $$
$$ \therefore \quad \frac{d}{d x}\left(\frac{x+1}{x-1}\right)=-\frac{2}{(x-1)^2} . $$
5. For the function $ f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^2}{2}+x+1, $ prove that $\boldsymbol{f}^{\prime}(\mathbf{1})=\mathbf{1 0 0} \boldsymbol{f}^{\prime}(\mathbf{0})$.
Solution

Given $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^2}{2}+x+1$

$$ \begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots+\frac{2 x}{2}+1+0 \\ & {\left[\because \frac{d}{d x} x^n=n x^{n-1}, \frac{d}{d x}(c)=0, \frac{d}{d x}\left(c f(x)=c \frac{d}{d x} f(x)\right]\right.} \\ & \text { or } \quad f^{\prime}(x)=x^{99}+x^{98}+\ldots+x+1 \end{aligned} $$

Putting $x=0$ and $x=1$ in ( $i$ ), we have

$$ \begin{aligned} & f^{\prime}(0) \\ \text { and } & =0+0+\ldots+0+1=1 \\ f^{\prime}(1) & =1^{99}+1^{98}+\ldots+1+1 \\ & =1+1+\ldots+1+1=100 \\ \therefore \quad & \\ \therefore \quad f^{\prime}(1) & =100 \times 1 \\ & =100 f^{\prime}(0) \end{aligned} $$

[By (ii)]

6. Find the derivatives of $x^n+a x^{n-1}+a^2 x^{n-2}+\ldots \boldsymbol{a}^{\boldsymbol{n}-1} \boldsymbol{x}+\boldsymbol{a}^{\boldsymbol{n}}$ for some fixed real number $\boldsymbol{a}$.
Solution

$f(x)=x^n+a x^{n-1}+a^2 x^{n-2}+\ldots+a^{n-1} x+a^n$

$$ \Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^n+a x^{n-1}+a^2 x^{n-2}+\ldots+a^{n-1} x+a^n\right) $$





$$ \begin{aligned} = & \frac{d}{d x}\left(x^n\right)+a \frac{d}{d x}\left(x^{n-1}\right)+ \\ & +a^2 \frac{d}{d x}\left(x^{n-2}\right)+\ldots \\ & +a^{n-1} \frac{d}{d x}(x)+\frac{d}{d x}\left(a^n\right) \\ = & n x^{n-1}+a(n-1) x^{n-2}+a^2(n-2) x^{n-3}+\ldots \\ & +a^{n-1}(1)+0 \\ = & n x^{n-1}+a(n-1) x^{n-2}+a^2(n-2) x^{n-3}+\ldots \\ & +a^{n-1} \end{aligned} $$
7. For some constants $\boldsymbol{a}$ and $\boldsymbol{b}$, find the derivative of (i) $(\boldsymbol{x}-\boldsymbol{a})(\boldsymbol{x}-\boldsymbol{b})$ (ii) $\left(a x^2+b\right)^2$ (iii) $\frac{x-a}{x-b}$.
Solution

(i) Let $f(x)=(x-a)(x-b)$; | uv form By the product rule:

$$ \text { then } \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(x-a) \times(x-b)+(x-a) \times \frac{d}{d x}(x-b) \\ & =1 \times(x-b)+(x-a) \times 1=x-b+x-a \\ & =2 x-(a+b) \end{aligned} $$

(ii) Let $f(x)=\left(a x^2+b\right)^2=a^2 x^4+2 a b x^2+b^2$, then

$$ \begin{aligned} f^{\prime}(x) & =a^2 \frac{d}{d x}\left(x^4\right)+2 a b \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left(b^2\right) \\ & =a^2\left(4 x^3\right)+2 a b(2 x)+0 \end{aligned} $$

$\left[\because b\right.$ is constant (given) $\Rightarrow b^2$ is also constant. For example $3^2=9=$ constant]

$$ =4 a^2 x^3+4 a b x=4 a x\left(a x^2+b\right) $$

(iii) Here $f(x)=\frac{x-a}{x-b}$


Using the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(x-a) \cdot(x-b)-(x-a) \cdot \frac{d}{d x}(x-b)}{(x-b)^2} \\ & =\frac{1 \cdot(x-b)-(x-a) \cdot 1}{(x-b)^2}=\frac{(x-b)-(x-a)}{(x-a)^2} \\ & =\frac{x-b-x+a}{(x-b)^2}=\frac{a-b}{(x-b)^2} \end{aligned} $$
8. Find the derivative of $\frac{x^{\boldsymbol{n}}-a^{\boldsymbol{n}}}{x-a}$ for some constant $a$.
Solution

Let $f(x)=\frac{x^n-a^n}{x-a}$, then using the quotient rule



$$ \begin{aligned} & \boldsymbol{f}^{\prime}(\boldsymbol{x})=\frac{\frac{d}{d x}\left(x^n-a^n\right) \cdot(x-a)-\left(x^n-a^n\right) \cdot \frac{d}{d x}(x-a)}{(x-a)^2} \\ &=\frac{n x^{n-1} \cdot(x-a)-\left(x^n-a^n\right) \cdot 1}{(x-a)^2} \\ & {\left[\because a^n \text { is a constant, } \therefore \frac{d}{d x} a^n=0\right] } \\ &=\frac{n x^n-n a x^{n-1}-x^n+a^n}{(x-a)^2} \\ &=\frac{\left.(\because-1) x^{n-1} \cdot x=x^{n-1} \cdot x^1=x^{n-1+1}=x^n\right)}{(x-a)^{n-1}+a^n} \end{aligned} $$
9. Find the derivative of (i) $\mathbf{2 x}-\frac{\mathbf{3}}{\mathbf{4}}$ (ii) $\left(5 x^3+3 x-1\right)(x-1)$ (iii) $\boldsymbol{x}^{-\mathbf{3}} \boldsymbol{(} \mathbf{5} \boldsymbol{+} \mathbf{3} \boldsymbol{x} \boldsymbol{)}$ (iv) $x^5\left(3-6 x^{-9}\right)$ (v) $\boldsymbol{x}^{-\mathbf{4}}\left(\mathbf{3}-\mathbf{4} \boldsymbol{x}^{-\mathbf{5}}\right)$ (vi) $\frac{2}{x+1}-\frac{x^2}{3 x-1}$.
Solution

(i) Let $f(x)=2 x-\frac{3}{4}$, then

$$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left(2 x-\frac{3}{4}\right) \\ & =\frac{d}{d x}(2 x)-\frac{d}{d x}\left(\frac{3}{4}\right) \\ & =2 \frac{d}{d x}(x)-0=2(1)=2 \end{aligned} $$

(ii) Here $f(x)=\left(5 x^3+3 x-1\right)(x-1)$


Using the product rule

$$ \begin{aligned} f^{\prime}(x)= & \frac{d}{d x}\left(5 x^3+3 x-1\right) \times(x-1)+\left(5 x^3+3 x-1\right) \\ & \times \frac{d}{d x}(x-1) \\ = & \left(15 x^2+3\right)(x-1)+\left(5 x^3+3 x-1\right) \times 1 \end{aligned} $$



$$ \begin{aligned} & =15 x^3-15 x^2+3 x-3+5 x^3+3 x-1 \\ & =20 x^3-15 x^2+6 x-4 . \end{aligned} $$

(iii) Here $f(x)=x^{-3}(5+3 x)$

$$ \begin{aligned} & =5 x^{-3}+3 x^{-2}, \text { then } \\ f^{\prime}(x) & =\frac{d}{d x}\left(5 x^{-3}+3 x^{-2}\right)=5 \frac{d\left(x^{-3}\right)}{d x}+3 \frac{d\left(x^{-2}\right)}{d x} \\ & =5\left(-3 x^{-4}\right)+3\left(-2 x^{-3}\right) \\ & =-\frac{15}{x^4}-\frac{6}{x^3}=-\frac{15+6 x}{x^4} \\ & =-\frac{3}{x^4}(5+2 x) \end{aligned} $$

(iv) Let $f(x)=x^5\left(3-6 x^{-9}\right)=3 x^5-6 x^{-4}$, then

$$ \begin{aligned} f^{\prime}(x) & =3 \frac{d}{d x}\left(x^5\right)-6 \frac{d}{d x}\left(x^{-4}\right) \\ & =3\left(5 x^4\right)-6\left(-4 x^{-5}\right)=15 x^4+\frac{24}{x^5} \end{aligned} $$

(v) Let $f(x)=x^{-4}\left(3-4 x^{-5}\right)=3 x^{-4}-4 x^{-9}$, then

$$ \begin{aligned} f^{\prime}(x) & =3 \frac{d}{d x}\left(x^{-4}\right)-4 \frac{d}{d x}\left(x^{-9}\right) \\ & =3\left(-4 x^{-5}\right)-4\left(-9 x^{-10}\right)=-\frac{12}{x^5}+\frac{36}{x^{10}} \\ & =\frac{12}{x^5}\left(\frac{3}{x^5}-1\right) \end{aligned} $$

(vi) Let $f(x)=\frac{2}{x+1}-\frac{x^2}{3 x-1}$

$$ \therefore f^{\prime}(x)=2 \frac{d}{d x}\left(\frac{1}{x+1}\right)-\frac{d}{d x}\left(\frac{x^2}{3 x-1}\right) $$

Using the quotient rule:

$$ \boldsymbol{f}^{\prime}(\boldsymbol{x})=2\left[\frac{\frac{d}{d x}(1) \cdot(x+1)-1 \frac{d}{d x}(x+1)}{(x+1)^2}\right] $$


$$ \begin{aligned} & -\left[\frac{\frac{d}{d x}\left(x^2\right) \cdot(3 x-1)-x^2 \frac{d}{d x}(3 x-1)}{(3 x-1)^2}\right] \\ = & 2\left[\frac{0(x+1)-1 \cdot 1}{(x+1)^2}\right]-\left[\frac{2 x(3 x-1)-x^2 \cdot 3}{(3 x-1)^2}\right] \\ = & 2\left(\frac{-1}{(x+1)^2}\right)-\left[\frac{6 x^2-2 x-3 x^2}{(3 x-1)^2}\right] \\ = & \frac{-2}{(x+1)^2}-\frac{\left(3 x^2-2 x\right)}{(3 x-1)^2} . \end{aligned} $$
10. Find the derivative of $\cos \boldsymbol{x}$ from first principle.
Solution

Let $f(x)=\cos x$.

Changing $x$ to $x+h, f(x+h)=\cos (x+h)$

We know that

$$ \begin{aligned} \boldsymbol{f}^{\prime}(\boldsymbol{x}) & =\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}} \\ & =\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}}{2 \cdot \frac{h}{2}} \\ & {\left[\because \cos \mathrm{C}-\cos \mathrm{D}=-2 \sin \frac{\mathrm{C}+\mathrm{D}}{2} \sin \frac{\mathrm{C}-\mathrm{D}}{2}\right] } \\ & =-\lim _{h \rightarrow 0}-\sin \left(x+\frac{h}{2}\right) \cdot \frac{\sin h / 2}{h / 2} \\ & =-\lim _{h \rightarrow 0} \sin \left(x+\frac{h}{2}\right) \cdot \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin h / 2}{h / 2} \\ & =-\sin (x+0) \cdot 1=-\sin x \\ \Rightarrow \frac{d}{d x}(f(x)) & =-\sin x \quad \text { Hence } \frac{d}{d x}(\cos x)=-\sin x . \end{aligned} $$
11. Find the derivative of the following functions: (i) $\sin x \cos x$ (ii) $\sec x$ (iii) $5 \sec x+4 \cos x$ (iv) $\operatorname{cosec} x$ (v) $3 \cot x+5 \operatorname{cosec} x$ (vi) $5 \sin x-6 \cos x+7$ (vii) $2 \tan x-7 \sec x$.
Solution

(i) Let $\quad f(x)=\sin x \cos x$; | uv form

Applying Product rule,

$$ \text { then } \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(\sin x \cos x) \\ & =\frac{d}{d x}(\sin x) \times \cos x+\sin x \times \frac{d}{d x}(\cos x) \\ & =\cos x \cdot \cos x+\sin x(-\sin x) \\ & =\cos ^2 x-\sin ^2 x=\cos 2 x \end{aligned} $$

(ii) Here $f(x)=\sec x=\frac{1}{\cos x}$


Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(1) \times \cos x-1 \times \frac{d}{d x}(\cos x)}{\cos ^2 x} \\ & =\frac{0 \times \cos x-(-\sin x)}{\cos ^2 x}=\frac{\sin x}{\cos ^2 x} \\ & =\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}=\sec x \tan x \end{aligned} $$

(iii) Here $f(x)=5 \sec x+4 \cos x$

$$ \begin{aligned} \therefore \quad f^{\prime}(x) & =5 \frac{d}{d x}(\sec x)+4 \frac{d}{d x}(\cos x) \\ & =5(\sec x \tan x)+4(-\sin x) \\ & =5 \sec x \tan x-4 \sin x \end{aligned} $$

(iv) Here $f(x)=\operatorname{cosec} x=\frac{1}{\sin x}$


Applying the quotient rule:

$$ f^{\prime}(x)=\frac{\frac{d}{d x}(1) \times \sin x-1 \times \frac{d}{d x}(\sin x)}{\sin ^2 x} $$




$$ \begin{aligned} & =\frac{0 \times \sin x-\cos x}{\sin ^2 x}=-\frac{\cos x}{\sin ^2 x} \\ & =-\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}=-\operatorname{cosec} x \cot x . \end{aligned} $$

(v) Here $f(x)=3 \cot x+5 \operatorname{cosec} x$

$$ \begin{aligned} \therefore \quad f^{\prime}(x) & =3\left(-\operatorname{cosec}^2 x\right)+5(-\operatorname{cosec} x \cot x) \\ & =-3 \operatorname{cosec}^2 x-5 \operatorname{cosec} x \cot x \end{aligned} $$

(vi) Here $f(x)=5 \sin x-6 \cos x+7$

$$ \begin{aligned} \therefore \quad f^{\prime}(x) & =5(\cos x)-6(-\sin x)+0 \\ & =5 \cos x+6 \sin x \end{aligned} $$

(vii) Here $f(x)=2 \tan x-7 \sec x$

$$ \therefore \quad f^{\prime}(x)=2 \sec ^2 x-7 \sec x \tan x . $$
Miscellaneous Exercise
1. Find the derivative of the following functions from first principle:
(i) $-x$
(ii) $(-x)^{-1}$
(iii) $\sin (x+1)$
(iv) $\cos \left(x-\frac{\pi}{8}\right)$.
Solution

(i) Here $f(x)=-x$

Changing $x$ to $x+h, f(x+h)=-(x+h)$

$$ \begin{aligned} & \therefore \begin{aligned} f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}=\lim _{h \rightarrow 0} \frac{(-x-h+x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-h}{h} \end{aligned} \\ & \text { cancelling } h, \quad=\lim _{h \rightarrow 0}(-1)=-1 \end{aligned} $$

(ii) Let $f(x)=(-x)^{-1}=\frac{1}{-x}=-\frac{1}{x}$.


Changing $x$ to $x+h, f(x+h)=\frac{-1}{x+h}$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$





$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{-\frac{1}{x+h}-\left(-\frac{1}{x}\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-\frac{1}{x+h}+\frac{1}{x}\right] \end{aligned} $$

Taking L.C.M.

$$ \begin{aligned} & \text { or } \begin{aligned} f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{-x+x+h}{h(x+h)(x)} \\ = & \lim _{h \rightarrow 0} \frac{h}{h(x+h)(x)} \\ \text { cancelling } h & =\lim _{h \rightarrow 0} \frac{1}{(x+h)(x)}=\frac{1}{(x+0) x}=\frac{1}{(x)(x)}=\frac{1}{x^2} \\ \Rightarrow \frac{d}{d x}(f(x)) & =\frac{1}{x^2} \text { Hence } \frac{d}{d x}\left((-x)^{-1}\right)=\frac{1}{x^2} \end{aligned} \end{aligned} $$

(iii) Here $f(x)=\sin (x+1)$

changing $x$ to $x+h, f(x+h)=\sin (x+h+1)$

$$ \begin{aligned} & \therefore \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\sin (x+h+1)-\sin (x+1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 \cos \frac{(x+h+1)+(x+1)}{2} \sin \frac{(x+h+1)-(x+1)}{2}}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+2+h}{2}\right) \sin \left(\frac{x+h+1-x-1}{2}\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2 \cos \left(x+1+\frac{h}{2}\right) \sin \frac{h}{2}}{2 \cdot \frac{h}{2}} \\ & =\lim _{h \rightarrow 0} \cos \left(x+1+\frac{h}{2}\right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}} \end{aligned} $$


$$ \begin{aligned} & =\lim _{h \rightarrow 0} \cos \left(x+1+\frac{h}{2}\right) \times \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \overline{2}}{\frac{h}{2}} \\ & =\cos (x+1+0) \times 1=\cos (x+1) \end{aligned} $$

(iv) Let $f(x)=\cos \left(x-\frac{\pi}{8}\right)$


Changing $x$ to $x+h ; f(x+h)=\cos \left(x+h-\frac{\pi}{8}\right)$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\boldsymbol{\operatorname { l i m }}_{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos \left(x+h-\frac{\pi}{8}\right)-\cos \left(x-\frac{\pi}{8}\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{x+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}\right) \sin \left(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2}\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{2 x+h-\frac{\pi}{4}}{2}\right) \sin \frac{h}{2}}{2 \cdot \frac{h}{2}} \\ & =\lim _{h \rightarrow 0}\left[-\sin \left(x+\frac{h}{2}-\frac{\pi}{8}\right) \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}\right] \\ & \left.=-\lim _{h \rightarrow 0} \sin \left(x+\frac{h}{2}-\frac{\pi}{8}\right) \cdot \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \sin \frac{\mathrm{C}-\mathrm{D}}{2}\right] \\ & =-\sin \left(x+0-\frac{\pi}{8}\right) \cdot 1=-\sin \left(x-\frac{\pi}{8}\right)\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\ & \Rightarrow \frac{d}{d x}(f(x))=-\sin \left(x-\frac{\pi}{8}\right) \end{aligned} $$


$$ \text { Hence } \quad \frac{d}{d x}\left[\cos \left(x-\frac{\pi}{8}\right)\right]=-\sin \left(x-\frac{\pi}{8}\right) . $$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers):

2. $(\boldsymbol{x}+\boldsymbol{a})$.
Solution

Here $f(x)=x+a$

$$ \Rightarrow \quad f^{\prime}(x)=1+0=1 $$
3. $(p x+q)\left(\frac{r}{x}+s\right)$.
Solution

Let $f(x)=(p x+q)\left(\frac{r}{x}+s\right)=p r+p s x+\frac{q r}{x}+q s$, then

$$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(p r)+p s \frac{d}{d x}(x)+q r \frac{d}{d x}\left(x^{-1}\right)+\frac{d}{d x}(q s) \\ & =0+p s(1)+q r\left(-1 x^{-2}\right)+0 \\ & =p s-\frac{q r}{x^2} \end{aligned} $$
4. $(a x+b)(c x+d)^{\mathbf{2}}$.
Solution

Here $f(x)=(a x+b)(c x+d)^2$.

Using the product rule

$$ \begin{array}{r} f^{\prime}(x)=\frac{d}{d x}(a x+b) \times(c x+d)^2+(a x+b) \times \frac{d}{d x}(c x+d)^2 \\ =(a \times 1+0) \times(c x+d)^2+(a x+b) \times 2(c x+d) \\ \times \frac{d}{d x}(c x+d) \end{array} $$

[By the chain rule: here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ ]

$$ \begin{aligned} & =a(c x+d)^2+2(a x+b)(c x+d) \times c \left\lvert\, \because \frac{d}{d x}(c x+d)=c \times 1+0=c\right. \\ & =a(c x+d)^2+2 c(a x+b)(c x+d) \end{aligned} $$
5. $\frac{a x+b}{c x+d}$.
Solution

Here $f(x)=\frac{a x+b}{c x+d}$.

Applying the quotient rule:




$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(a x+b) \times(c x+d)-(a x+b) \times \frac{d}{d x}(c x+d)}{(c x+d)^2} \\ & =\frac{a(c x+d)-(a x+b) c}{(c x+d)^2}=\frac{a c x+a d-a c x-b c}{(c x+d)^2} \\ & =\frac{a d-b c}{(c x+d)^2} . \end{aligned} $$
6. $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$.
Solution

Let $f(x)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x-1}{x}}=\frac{x+1}{x-1}$

Applying the quotient rule:

$$ \begin{aligned} \therefore \quad f^{\prime}(x) & =\frac{\frac{d}{d x}(x+1) \cdot(x-1)-(x+1) \frac{d}{d x}(x-1)}{(x-1)^2} \\ & =\frac{1 \cdot(x-1)-(x+1) \cdot 1}{(x-1)^2}=\frac{-2}{(x-1)^2} \end{aligned} $$
7. $\frac{1}{a x^2+b x+c}$.
Solution

Let $f(x)=\frac{1}{a x^2+b x+c}$

By the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(1) \cdot\left(a x^2+b x+c\right)-1 \cdot \frac{d}{d x}\left(a x^2+b x+c\right)}{\left(a x^2+b x+c\right)^2} \\ & =\frac{(0)\left(a x^2+b x+c\right)-(2 a x+b)}{\left(a x^2+b x+c\right)^2} \\ & =-\frac{2 a x+b}{\left(a x^2+b x+c\right)^2} \end{aligned} $$
8. $\frac{a x+b}{p x^2+q x+r}$.
Solution

Here $f(x)=\frac{a x+b}{p x^2+q x+r}$.




Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(a x+b) \times\left(p x^2+q x+r\right)-(a x+b) \times \frac{d}{d x}\left(p x^2+q x+r\right)}{\left(p x^2+q x+r\right)^2} \\ & =\frac{a\left(p x^2+q x+r\right)-(a x+b)(2 p x+q)}{\left(p x^2+q x+r\right)^2} \\ & =\frac{a p x^2+a q x+a r-2 a p x^2-a q x-2 b p x-b q}{\left(p x^2+q x+r\right)^2} \\ & =\frac{-a p x^2-2 b p x+a r-b q}{\left(p x^2+q x+r\right)^2} \end{aligned} $$
9. $\frac{p x^2+q x+r}{a x+b}$.
Solution

Here $f(x)=\frac{p x^2+q x+r}{a x+b}$.

Using the quotient rule

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}\left(p x^2+q x+r\right) \times(a x+b)-\left(p x^2+q x+r\right) \times \frac{d}{d x}(a x+b)}{(a x+b)^2} \\ & =\frac{(2 p x+q)(a x+b)-\left(p x^2+q x+r\right)(a)}{(a x+b)^2} \\ & =\frac{2 a p x^2+2 b p x+a q x+b q-a p x^2-a q x-a r}{(a x+b)^2} \\ & =\frac{a p x^2+2 b p x+b q-a r}{(a x+b)^2} \end{aligned} $$
10. $\frac{a}{x^4}-\frac{b}{x^2}+\cos x$.
Solution

Let $f(x)=\frac{a}{x^4}-\frac{b}{x^2}+\cos x=a x^{-4}-b x^{-2}+\cos x$, then

$$ f^{\prime}(x)=a \frac{d}{d x}\left(x^{-4}\right)-b \frac{d}{d x}\left(x^{-2}\right)+\frac{d}{d x}(\cos x) $$



$$ \begin{aligned} & =a\left(-4 x^{-5}\right)-b\left(-2 x^{-3}\right)+(-\sin x) \\ & =-\frac{4 a}{x^5}+\frac{2 b}{x^3}-\sin x . \end{aligned} $$
11. $4 \sqrt{x}-2$.
Solution

Let $f(x)=4 \sqrt{x}-2$, then

$$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(4 \sqrt{x})-\frac{d}{d x}(2) \\ & =4 \frac{d}{d x}\left(x^{1 / 2}\right)-0=4\left(\frac{1}{2} x^{1 / 2-1}\right)\left[\because \frac{d}{d x} x^n=n x^{n-1}\right] \\ & =2 x^{-\frac{1}{2}}=\frac{2}{x^{\frac{1}{2}}}=\frac{2}{\sqrt{x}} . \end{aligned} $$
12. $(a x+b)^n$.
Solution

Let $f(x)=(a x+b)^n$

Differentiating both sides w.r.t $x$,

$$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}(a x+b)^n \\ & =n(a x+b)^{\mathrm{n}-1} \frac{d}{d x}(a x+b) \end{aligned} $$

[By the chain rule: Here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ where $u$ is a function of $x$ ]

$$ \begin{aligned} & \Rightarrow f^{\prime}(x)=n(a x+b)^{n-1}[a(1)+0]=n(a x+b)^{n-1} a \\ & =n a(a x+b)^{n-1} \end{aligned} $$
13. $(a x+b)^n(c x+d)^m$.
Solution

Let $f(x)=(a x+b)^n(c x+d)^m$, then

$$ f^{\prime}(x)=\frac{d}{d x}\left[(a x+b)^n(c x+d)^m\right] $$

By the product rule:

$$ \begin{aligned} & \quad=\frac{d}{d x}(a x+b)^n \cdot(c x+d)^m+(a x+b)^n \cdot \frac{d}{d x}(c x+d)^m \\ & =n(a x+b)^{n-1} \frac{d}{d x}(a x+b) \cdot(c x+d)^m+(a x+b)^n m(c x+d)^{m-1} \\ & \frac{d}{d x}(c x+d) \end{aligned} $$





[By the chain rule: here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ ]

$$ \begin{aligned} & =n a(a x+b)^{n-1}(c x+d)^m+(a x+b)^n \cdot m c(c x+d)^{m-1} \\ & =n a(a x+b)^{n-1}(c x+d)^{m-1}(c x+d)+(a x+b)^{n-1} \\ & (a x+b) m c(c x+d)^{m-1} \\ & {\left[\because(c x+d)^m=(c x+d)^{m-1+1}=(c x+d)^{m-1}(c x+d) .\right.} \\ & \text { similarly } \left.(a x+b)^n=(a x+b)^{n-1}(a x+b)\right] \\ & \text { Taking }(a x+b)^{n-1}(c x+d)^{m-1} \text { common } \\ & =(a x+b)^{n-1}(c x+d)^{m-1}[n a(c x+d)+m c(a x+b)] \\ & =(a x+b)^{n-1}(c x+d)^{m-1}[n a c x+n a d+m a c x+m b c] \\ & =(a x+b)^{n-1}(c x+d)^{m-1}[a c(n+m) x+n a d+m b c] . \end{aligned} $$
14. $\sin (x+a)$.
Solution

Here $f(x)=\sin (x+a)=\sin x \cos a+\cos x \sin a$.

$$ \begin{aligned} \therefore f^{\prime}(x) & =\frac{d}{d x}(\sin x \cos a)+\frac{d}{d x}(\cos x \sin a) \\ \quad & =\cos a \frac{d}{d x}(\sin x)+\sin a \frac{d}{d x}(\cos x) \\ & =\cos a \cos x+\sin a(-\sin x) \\ & =\cos x \cos a-\sin x \sin a \\ & =\cos (x+a) \end{aligned} $$

Alternatively:

$$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}[\sin (x+a)]=\cos (x+a) \cdot \frac{d}{d x}(x+a) \\ & =\cos (x+a) \times(1+0)=\cos (x+a) \times 1=\cos (x+a) \end{aligned} $$
15. $\operatorname{cosec} x \cot x$.
Solution

Let $\quad f(x)=\operatorname{cosec} x \cot x$; I uv form

By the product rule:

$$ \text { then } \begin{aligned} f^{\prime}(x)= & \frac{d}{d x}(\operatorname{cosec} x) \times \cot x+\operatorname{cosec} x \times \frac{d}{d x}(\cot x) \\ & =(-\operatorname{cosec} x \cot x) \cot x+\operatorname{cosec} x\left(-\operatorname{cosec}^2 x\right) \\ = & -\operatorname{cosec} x \cot ^2 x-\operatorname{cosec}^3 x \\ = & -\operatorname{cosec} x\left(\cot ^2 x+\operatorname{cosec}^2 x\right) \end{aligned} $$
16. $\frac{\cos x}{1+\sin x}$.
Solution

Here $f(x)=\frac{\cos x}{1+\sin x}$.




Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(\cos x) \times(1+\sin x)-\cos x \times \frac{d}{d x}(1+\sin x)}{(1+\sin x)^2} \\ & =\frac{-\sin x \times(1+\sin x)-\cos x(\cos x)}{(1+\sin x)^2} \\ & =\frac{-\sin x-\sin ^2 x-\cos ^2 x}{(1+\sin x)^2} \\ & =\frac{-\sin x-\left(\sin ^2 x+\cos ^2 x\right)}{(1+\sin x)^2} \\ & =\frac{-\sin x-1}{(1+\sin x)^2}=\frac{-(1+\sin x)}{(1+\sin x)^2} \\ & =\frac{-1}{1+\sin x} \end{aligned} $$
17. $\frac{\sin x+\cos x}{\sin x-\cos x}$.
Solution

Let $f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}$, using the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(\sin x+\cos x) \cdot(\sin x-\cos x)-(\sin x+\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^2} \\ & =\frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2} \\ & =\frac{-\sin x \cos x-\sin ^2 x-\cos ^2 x-\sin x \cos x}{(\sin x-\cos x)^2} \\ & =\frac{-2 \sin ^2 x-2 \cos ^2 x}{(\sin x-\cos x)^2}=\frac{-2\left(\sin ^2 x+\cos ^2 x\right)}{(\sin x-\cos x)^2} \\ & =\frac{-2}{(\sin x-\cos x)^2}\left[\because \sin ^2 x+\cos ^2 x=1\right] \end{aligned} $$
18. $\frac{\sec x-1}{\sec x+1}$.
Solution

Here $f(x)=\frac{\sec x-1}{\sec x+1}$

Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(\sec x-1) \times(\sec x+1)-(\sec x-1) \times \frac{d}{d x}(\sec x+1)}{(\sec x+1)^2} \\ & =\frac{\sec x \tan x(\sec x+1)-(\sec x-1)(\sec x \tan x)}{(\sec x+1)^2} \\ & =\frac{\sec ^2 x \tan x+\sec x \tan x-\sec ^2 x \tan x+\sec x \tan x}{(\sec x+1)^2} \\ & =\frac{2 \sec x \tan x}{(\sec x+1)^2} \end{aligned} $$
19. $\sin ^n x$.
Solution

Let $f(x)=\sin ^n x=(\sin x)^n$

Applying the chain rule:

$$ \begin{aligned} & {\left[\text { Here } \frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u \text { where } u=\sin x \text { is a function of } x\right]} \\ & f^{\prime}(x)=n(\sin x)^{n-1} \frac{d}{d x}(\sin x) \\ & \quad=n \sin ^{n-1} x \cos x \end{aligned} $$
20. $\frac{a+b \sin x}{c+d \cos x}$.
Solution

Here $f(x)=\frac{a+b \sin x}{c+d \cos x}$

Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(a+b \sin x) \times(c+d \cos x)-(a+b \sin x) \times \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^2} \\ & =\frac{b \cos x(c+d \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^2} \end{aligned} $$



$$ \begin{aligned} & =\frac{b c \cos x+b d \cos ^2 x+a d \sin x+b d \sin ^2 x}{(c+d \cos x)^2} \\ & =\frac{b c \cos x+a d \sin x+b d\left(\cos ^2 x+\sin ^2 x\right)}{(c+d \cos x)^2} \\ & =\frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^2} \end{aligned} $$
21. $\frac{\sin (x+a)}{\cos x}$.
Solution

Here $f(x)=\frac{\sin (x+a)}{\cos x}$.

Applying the quotient rule:

$$ \begin{aligned} & f^{\prime}(x)=\frac{\frac{d}{d x}[\sin (x+a)] \times \cos x-\sin (x+a) \times \frac{d}{d x}(\cos x)}{\cos ^2 x} \\ & =\frac{\cos (x+a) \frac{d}{d x}(x+a)(\text { By chain rule }) \times \cos x-\sin (x+a)(-\sin x)}{\cos ^2 x} \\ & \quad=\frac{\cos (x+a)(1) \cos x+\sin (x+a) \sin x}{\cos ^2 x} \end{aligned} $$

Using $\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})$,

$$ =\frac{\cos [(x+a)-x]}{\cos ^2 x}=\frac{\cos a}{\cos ^2 x} . $$
22. $x^4(5 \sin x-3 \cos x)$.
Solution

Here $f(x)=x^4(5 \sin x-3 \cos x)$

Applying the product rule:

$$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left(x^4\right) \times(5 \sin x-3 \cos x)+x^4 \times \frac{d}{d x}(5 \sin x- \\ &3 \cos x) \\ &=4 x^3(5 \sin x-3 \cos x)+x^4(5 \cos x+3 \sin x) \\ &=x^3(20 \sin x-12 \cos x+5 x \cos x+3 x \sin x) \end{aligned} $$
23. $\left(x^2+1\right) \cos x$.
Solution

Let $f(x)=\left(x^2+1\right) \cos x$,

Applying the product rule:





$$ \text { then } \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left(x^2+1\right) \times \cos x+\left(x^2+1\right) \times \frac{d}{d x}(\cos x) \\ & =2 x \cos x+\left(x^2+1\right)(-\sin x) \\ & =2 x \cos x-\left(x^2+1\right) \sin x \end{aligned} $$
24. $\left(a x^2+\sin x\right)(p+q \cos x)$.
Solution

Here $f(x)=\left(a x^2+\sin x\right)(p+q \cos x)$

Applying the product rule:

$$ \begin{array}{r} f^{\prime}(x)=\frac{d}{d x}\left(a x^2+\sin x\right) \times(p+q \cos x)+\left(a x^2+\sin x\right) \\ \times \frac{d}{d x}(p+q \cos x) \\ =(2 a x+\cos x)(p+q \cos x)+\left(a x^2+\sin x\right)(-q \sin x) \\ =(2 a x+\cos x)(p+q \cos x)-q \sin x\left(a x^2+\sin x\right) \end{array} $$
25. $(x+\cos x)(x-\tan x)$.
Solution

Let $\quad f(x)=(x+\cos x)(x-\tan x) ; \quad$ I u form Applying the product rule:

$$ \text { then } \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}(x+\cos x)(x-\tan x)+( x+\cos x) \\ & \times \frac{d}{d x}(x-\tan x) \\ &=(1-\sin x)(x-\tan x)+(x+\cos x)\left(1-\sec ^2 x\right) \end{aligned} $$
26. $\frac{4 x+5 \sin x}{3 x+7 \cos x}$.
Solution

Let $f(x)=\frac{4 x+5 \sin x}{3 x+7 \cos x}$, using the quotient rule:

$$ \begin{aligned} f^{\prime}(x)= & \frac{\frac{d}{d x}(4 x+5 \sin x) \cdot(3 x+7 \cos x)-(4 x+5 \sin x) \frac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^2} \\ = & \frac{(4+5 \cos x)(3 x+7 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^2} \\ = & \frac{\left(12 x+28 \cos x+15 x \cos x+35 \cos ^2 x\right)}{(3 x+7 \cos x)^2} \end{aligned} $$



$$ \begin{aligned} & =\frac{12 x+28 \cos x+15 x \cos x+35 \cos ^2 x-12 x+28 x \sin x-15 \sin x+35 \sin ^2 x}{(3 x+7 \cos x)^2} \\ & \quad=\frac{28(\cos x+x \sin x)+15(x \cos x-\sin x)+35\left(\cos ^2 x+\sin ^2 x\right)}{(3 x+7 \cos x)^2} \\ & \quad=\frac{28(\cos x+x \sin x)+15(x \cos x-\sin x)+35}{(3 x+7 \cos x)^2} \\ & \quad \boldsymbol{x}^{\mathbf{2}} \boldsymbol{\operatorname { c o s }}\left(\frac{\pi}{\mathbf{4}}\right) \end{aligned} $$

Sol. Let

$$ \begin{aligned} f(x) & =\frac{x^2 \cos \frac{\pi}{4}}{\sin x}=\frac{1}{\sqrt{2}} \frac{x^2}{\sin x} \\ f^{\prime}(x) & =\frac{d}{d x}\left(\frac{1}{\sqrt{2}} \frac{x^2}{\sin x}\right) \\ & =\frac{1}{\sqrt{2}} \frac{d}{d x} \frac{x^2}{\sin x} \end{aligned} $$

Using the quotient rule:

$$ \begin{aligned} & =\frac{1}{\sqrt{2}}\left[\frac{\frac{d}{d x}\left(x^2\right) \cdot \sin x-x^2 \cdot \frac{d}{d x} \sin x}{\sin ^2 x}\right] \\ & =\frac{2 x \sin x-x^2 \cos x}{\sqrt{2} \sin ^2 x} \\ & =\frac{x(2 \sin x-x \cos x)}{\sqrt{2} \sin ^2 x} \end{aligned} $$
28. $\frac{x}{1+\tan x}$.
Solution

Here $f(x)=\frac{x}{1+\tan x}$.

Applying the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(x) \times(1+\tan x)-x \times \frac{d}{d x}(1+\tan x)}{(1+\tan x)^2} \\ & =\frac{1(1+\tan x)-x \sec ^2 x}{(1+\tan x)^2}=\frac{1+\tan x-x \sec ^2 x}{(1+\tan x)^2} . \end{aligned} $$
29. $(x+\sec x)(x-\tan x)$.
Solution

Here $f(x)=(x+\sec x)(x-\tan x)$

Applying the product rule:

$$ \begin{array}{r} f^{\prime}(x)=\frac{d}{d x}(x+\sec x) \times(x-\tan x)+(x+\sec x) \\ \times \frac{d}{d x}(x-\tan x) \\ =(1+\sec x \tan x)(x-\tan x)+(x+\sec x)\left(1-\sec ^2 x\right) \end{array} $$
30. $\frac{x}{\sin ^n x}$.
Solution

Let $f(x)=\frac{x}{\sin ^n x}$, using the quotient rule:

$$ \begin{aligned} f^{\prime}(x) & =\frac{\frac{d}{d x}(x) \cdot \sin ^n x-x \cdot \frac{d}{d x}\left(\sin ^n x\right)}{\left(\sin ^n x\right)^2} \\ & =\frac{1 \cdot \sin ^n x-x \cdot n \sin ^{n-1} x \cos x}{\sin ^{2 n} x} \end{aligned} $$
$$ \begin{aligned} & \because \text { by chain rule, } \frac{d}{d x} \sin ^n x=\frac{d}{d x}(\sin x)^n=\frac{d}{d x} u^n(\text { where } u=\sin x)=n u^{n-1} \frac{d}{d x} \\ & \left.n \sin ^{n-1} x \frac{d}{d x} \sin x=n \sin ^{n-1} x \cos x\right] \\ & =\frac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x} \\ & {\left[\because \sin ^n x=(\sin x)^n=(\sin x)^{n-1+1}=(\sin x)^{n-1} \sin x=\sin ^{n-1} x \sin x\right]} \\ & =\frac{\sin x-n x \cos x}{\sin ^{n+1} x} . \\ & (\because 2 n-(n-1)=2 n-n+1=n+1) \end{aligned} $$

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Mental Ability Test

General Instruction:

1. There are 35 MCQ's in the Test.

2. Passing %age is 50.

3. After time gets over, test will be submitted itself.

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1 / 35

Find the letter to be placed in place of ‘?’ in the figure given.

 

 

 

 

 

 

 

 

2 / 35

A pattern is given below. You have to identify which among the following pieces will not be required to complete the pattern.

 

 

 

 

 

 

 

 

 

 

3 / 35

Which symbol replaces the ‘?’. Figure below represent a balance.

 

 

 

 

 

 

 

 

 

 

 

4 / 35

Find the missing number in the series 2, 10, 26, _____, 242.

5 / 35

Find the number in the position of ‘?’.

 

 

 

 

 

 

6 / 35

In each of the following questions, you are given a figure (X) followed by four alternative figures (1), (2), (3) and (4) such that figure (X) is embedded in one of them. Trace out the alternative figure which contains fig. (X) as its part.

Find out the alternative figure which contains figure (X) as its part.

 

 

 

 

 

7 / 35

In this multiplication question the five letters represent five different digits. What are the actual figures ? There is no zero.

SEAM
        T 

MEATS

8 / 35

Find the next number in the sequence 6, 24, 60, 120 ______

9 / 35

In a dairy, there are 60 cows and buffalos. The number of cows is twice that of buffalos. Buffalo X ranked seventeenth in terms of milk delivered. If there are 9 cows ahead of Buffalo. X, how many buffalos are after in rank in terms of milk delivered ?

10 / 35

The sheet of paper shown in the figure (X) given on the left hand side, in each problem, is folded to form a box. Choose from amongst the alternatives
(1), (2), (3) and (4), the boxes that are similar to the box that will be formed.

How many dots lie opposite to the face having three dots, when the given figure is folded to form a cube?

 

 

 

 

 

 

 

11 / 35

If A is the brother of B; B is the sister of C; and C is the father of D, how D is related to A?

12 / 35

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

Kumar is at 40 m to the right of Ankur.
Dev is are 60 m in the south of Kumar.
Nilesh is at a distance of 25 m in the west of Ankur.
Pintu is at a distance of 90 m in the North of Dev.

Which one is in the North-East of the person who is to the left of Kumar?

13 / 35

In the following letter sequence, some of the letters are missing. These are given in order as one of the alternatives below. Choose the correct alternative.

ab__aa__bbb__aaaa__bb

14 / 35

In each of the following questions two statements are given. Which are followed by four conclusions (1), (2), (3) and (4). Choose the conclusions which logically follow from the given statements.

Statements: 

Some tables are T.V.

Some T.V. are radios.

Conclusions:

(1) Some tables are radios.
(2) Some radios are tables.
(3) All the radios are T.V.
(4) All the T.V. are tables.

15 / 35

In each problem, out of the five figures marked (1), (2), (3), (4) and (5), four are similar in a certain manner. However, one figure is not like the other four.

Choose the figure which is different from the rest.

 

 

 

 

16 / 35

Mohan started going for regular morning walks for controlling his blood sugar level. He did so for a month and also started taking Yoga lessons, without going for any pathological examination. He underwent pathological test after two months and found that the blood sugar level has come down. Presuming that he had not changed his food habits during these two months, which statement among the alternatives given below follows most logically ?

17 / 35

Find the odd man out

 

 

 

 

 

 

18 / 35

 

19 / 35

Akbar used to board the train from Metro Station A for going to her office. Since Station A is a terminus. she had no problem in getting a seat. Ever since she shifted to Locality B she finds it difficult to get a seat, as by the time the train reaches Locality B it becomes crowded. Find the statement among the alternatives which must be true as per the given information.

20 / 35

Question given below has a problem and two statements I & II. Decide if the information given in the statement is sufficient for answering the problem:
K, R, S and T are four players in Indian Cricket team. Who is the oldest among them? I : The total age of K & T together is more than that of S II : The total age of R & K together is less than that of S.

21 / 35

Here are some words translated from an artificial language

mie pie is blue light
mie tie is blue berry
aie tie is rasp berry

Which words could possibly mean “light fly”?

22 / 35

In each of the following problems, a square transparent sheet (X) with a pattern is given. Figure out from amongst the four alternatives as to how the patter would appear when the transparent sheet is folded at the dotted line.

Find out from amongst the four alternatives as to how the pattern would appear when the transparent sheet is folded at the dotted line.

 

 

 

 

 

23 / 35

Directions: The following series are based on a specific pattern. In these series one number is missing, find that odd one.

4, 11, 19, 41, ?, 161

24 / 35

Find the next number in the sequence 0, 2, 24, 252 . ______

25 / 35

If in certain code, STUDENT is written as RSTEDMS, then how would TEACHER be written in the same code ?

26 / 35

Each of these questions given below contains three elements. These elements may or may not have some inter linkage. Each group of elements may fit into one of these diagrams at (A), (B), (C), (D) and/or (E). You have to indicate the group of elements which correctly fits into the diagrams. 

Which of the following diagrams indicates the best relation between Class, Blackboard and School ?

27 / 35

In each of the following questions, you are given a combination of alphabets and/or numbers followed by four alternatives (1), (2), (3) and (4). Choose the alternative which is closely resembles the water-image of the given combination.

Choose the alternative which is closely resembles the water-image of the given combination.

 

 

 

 

28 / 35

In each of the following questions you are given a combination of alphabets and/or numbers followed by four alternatives (1), (2), (3) and (4). Choose the alternative which is closely resembles the mirror image of the given combination.

Choose the alternative which is closely resembles the mirror image of the given  combination.

 

29 / 35

Identify the number in the position of ‘?’

 

 

 

 

 

 

 

30 / 35

Identify which among the pieces given below will not be required to complete the triangular pattern shown below.

 

 

 

 

 

 

 

 

31 / 35

Which group of letters is different from others?

32 / 35

Each of the following questions consists of two sets of figures. Figures A, B, C and D constitute the Problem Set while figures 1, 2, 3, 4 and 5 constitute the Answer Set. There is a definite relationship between figures A and B. Establish a similar relationship between figures C and D by selecting a suitable figure from the Answer Set that would replace the question mark (?) in fig. (D).

 

 

 

 

 

33 / 35

Each of the following questions consists of five figures marked A, B, C, D and E called the Problem Figures followed by five other figures marked 1, 2, 3, 4 and 5 called the Answer Figures. Select a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.

Select a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.

 

 

 

 

 

34 / 35

Which of the following diagram/sets indicate the relation between women, mothers and parents?

35 / 35

Fill in the missing number

 

 

 

 

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Pos.NameScoreDuration
1dfd50 %11 seconds
2dfdg50 %22 seconds
3dssd29 %11 hours 46 minutes 8 seconds
4jhh0 %17 minutes 25 seconds

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