NCERT Solutions for Class 11 Maths Chapter 12 - Limits and Derivatives

$$\textbf{7.}\space\lim\limits_{x\to 2}\space\frac{3x^2-x-10}{x^2-4}.$$

$$\textbf{Sol.}\space\lim\limits_{x\to 2}\frac{3x^2-x-10}{x^2-4}\\=\lim\limits_{x\to 2}\frac{3x^2-6x+5x-10}{x^2-2^2}\\\lbrack\because\space a^2-b^2=(a-b)(a+b)\rbrack\\=\lim\limits_{ x\to 2}\space\frac{3x(x-2)+5(x-2)}{(x-2)(x+2)}\\=\lim\limits_{x\to2}\frac{(x-2)(3x+5)}{(x-2)(x+2)}\\=\frac{3×2+5}{2+2}=\frac{11}{2}.$$

$$\textbf{9.}\space\lim\limits_{x \to 0}\space\frac{ax+b}{cx+1}.\\\textbf{Sol.}\space\lim\limits_{x\to 0}\frac{ax+b}{cx+1}\\=\frac{a×0+b}{c×0+1}=\frac{b}{1}=b.$$

29. Let a₁, a₂, a₃, …, a_n be fixed real numbers and define a function

f(x) = (x – a₁) (x – a₂) … (x – a_n)

$$\text{What is}\space\lim\limits_{x\to a_1}\space\text{f(x)}?\\\text{For some a = a}_1, \text{a}_2, …, \text{a}_\text{n}\space\text{compute}\space\lim\limits_{x\to a}\space\text{f(x)}.$$

Sol. Given, f(x) = (x – a₁) (x – a₂) … (x – a_n)

$$\lim\limits_{x\to a_1}\space\text{f(x)}=\lim_{x\to a_1}\space(x-a_1)(x-a_2)...(x-a_n)$$

= (a₁ – a₁) (a₁ – a₂) … (a₁ – a_n)

= 0 × (a₁ – a₂) … (a₁ – a_n) = 0

$$\text{Again,}\space\lim\limits_{x\to a}\space f(x)\\=\lim\limits_{x\to a}(x-a_1)(x-a_2)...(x-a_n)$$

= (a – a₁) (a – a₂) … (a – a_n)

= (a – a₁) (a – a₂) … (a – a_n).

$$\textbf{32.}\space\text{If f(x) =}\begin{cases}mx^2+n,\space x\lt0\\ nx+m,0\leq x\leq 1\\nx^3+m,\space x\gt1\end{cases}.\\\text{For what integers m and n does both}\\\lim\limits_{x\to0}\space f(x)\space\text{and}\lim\limits_{x\to1}\space\text{f(x) exist?}\\\textbf{Sol.}\space\because\space\text{At}\space x=0,\space\text{limit exists.}$$

Then,  LHL = RHL

$$\Rarr\space\lim\limits_{x\to0^-}\space\text{f(x)}=\lim\limits_{x\to0^+}\space\text{f(x)}\\\Rarr\space\lim\limits_{h\to0}\space f(0-h)=\lim\limits_{h\to0}\space f(0+h)\\\Rarr\space\lim_{h\to0}\space m(0-h)^2+n=\lim\limits_{h\to0}\space n(0+h)+m$$

⇒ m(0 – 0)² + n = n(0 + 0) + m

⇒ n = m …(i)

Again, at x = 1 limit exsts.

⇒ LHL = RHL

$$\Rarr\space\lim_{x\to1^+}\space f(x)=\lim_{x\to1^-}\space f(x)$$

$$\Rarr\space\lim\limits_{h\to0}\space f(1+h)=\lim\limits_{h\to0}\space f(1-h)\\\Rarr\space\lim\limits_{h\to0}\space n(1+h)^3+m\\=\lim\limits_{h\to0}\space n(1-h)+m$$

⇒ n(1 + 0)² + m = n(1 – 0) + m

⇒ n + m = m + n …(II)

$$\text{Hence, from equations (i) and (ii) for}\space\lim\limits_{x\to0}\space\text{f(x)}\\\text{to be existed, we need n = m and}\lim\limits_{x\to1}\space\text{f(x)}\\\text{exists for any integral value of m and n.}$$

4. Find the derivative of the following functions from first principle:

(i) x³ – 27

(ii) (x – 1) (x – 2)

$$\text{(iii)}\space\frac{1}{x^3}\\\text{(iv)}\space\frac{x+1}{x-1}$$

Sol. (i) Let f(x) = x³ – 27

We have

$$\text{f'(x)}=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\Rarr\space\text{f'(x)}=\lim\limits_{h\to0}\space\frac{[(x+h)^3-27]-(x^3-27)}{h}\\=\lim\limits_{h\to0}\space\frac{(x+h)^3-27-x^3+27}{h}\\=\lim\limits_{h\to0}\space\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}$$

∵ (a + b)³ = a³ + b³ + 3a²b + 3ab²

$$=\lim\limits_{h\to0}\space\frac{h^3+3x^2h+3xh^2}{h}\\=\lim\limits_{h\to0}\space\frac{h(h^2+3x^2-3xh)}{h}$$

= 0 + 3x² – 0

= 3x².

(ii) Let f(x) = (x – 1) (x – 2) = x² – 2x – x + 2

⇒ f(x) = x² – 3x + 2

We have,

$$f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}$$

[(x+h)²-3(x+h)+2]

$$\Rarr\space f'(x)=\lim\limits_{h\to0}\frac{-(x^2-3x+2)}{h}$$

(∵ f(x) = x² – 3x + 2)

$$=\lim\limits_{h\to0}\\\space\frac{x^2+xh^2+2xh-3x-3h+2-x^2+3x-2}{h}$$

$$=\lim\limits_{h\to0}\space\frac{h^2+2xh-3h}{h}\\=\lim\limits_{h\to0}\space\frac{h(h+2x-3)}{h}\\=\lim\limits_{h\to0}\space h+2x-3=0+2x-3=2x-3\\\therefore\space\text{f'(x)=2x-3.}$$

$$\text{(iii)\space}\text{f(x)}=\frac{1}{x^2}\\f'(x)=\lim\limits_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\=\lim\limits_{h\to0}\space\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\\=\lim\limits_{h\to0}\space\frac{x^2-(x+h)^2}{hx^2(x+h)^2}\\=\lim\limits_{h\to0}\space\frac{x^2-(x+h)^2}{hx^2(x+h)^2}\\\lim\limits_{x\to0}\space\frac{x^2-x^2-h^2-2xh}{hx^2(x+h)^2}\\=\lim\limits_{h\to0}\space\frac{-h(h+2x)}{hx^2(x+h)^2}$$

$$=\frac{-2x}{x^2×x^2}=\frac{-2}{x^3}\\=\lim\limits_{h\to0}|5+h|-5\\=\lim\limits_{h\to0}(5+h)-5\\\lim\limits_{h\to0}h$$

= 0

Then LHL = RHL

$$\text{Hence}\space\lim\limits_{x\to5}\space f(x)=0.$$

$$\text{(iv)\space Let}\space f(x)=\frac{x+1}{x-1}\\\text{we have,}\\\text{f'(x)}=\lim_{h\to0}\space\frac{f(x+h)-f(x)}{h}\\\Rarr\space f'(x)=\lim\limits_{h\to0}\space\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}\\\bigg(\because\space \text{f(x)}=\frac{x+1}{x-1}\bigg)\\=\lim\limits_{h\to0}\\\space\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x+h-1)(x-1)h}\\(x^2-x+hx-h+x-1)$$

$$=\lim\limits_{h\to0}\space\frac{-(x^2+hx-x+x+h-1)}{(x+h-1)(x-1)h}\\=\lim\limits_{h\to0}\space\frac{x^2+hx-x-1-x^2-hx-h+1}{(x+h-1)(x-1)h}\\=\lim\limits_{h\to0}\space\frac{-2h}{(x+h-1)(x-1)h}\\=\frac{-2}{(x+0-1)(x-1)}=\frac{2}{(x-1)^2}$$

$$\textbf{8.}\space\text{Find the derivative of}\space\frac{x^n-a^n}{x-a}\\\space\text{for some constant a.}$$

$$\textbf{Sol.}\space\text{Let}\space y=\frac{x^n-a^n}{x-a}$$

Differentiating y w.r.t. x,

$$\frac{dy}{dx}=\frac{(x-a)\frac{d}{dx}(x^n-a^n)-(x^n-a^n)\frac{d}{dx}(x-a)}{(x-a)^2}\\=\frac{(x-a)[nx^{n-1}-0]-[x^n-a^n](1-0)}{(x-a)^2}$$

$$=\frac{(x-a)nx^{n-1}-x^n+a^n}{(x-a)^2}\\=\frac{x×nx^{n-1}-anx^{n-1}-x^n+a^n}{(x-a)^2}\\=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$$

$$\textbf{3.}\space(px+q)\bigg(\frac{r}{x}+s\bigg).\\\textbf{Sol.}\space\text{Let}\space y=\text{(px+q)}\bigg(\frac{r}{x}+s\bigg)$$

Differentiating y w.r.t. x,

$$\frac{dy}{dx}=(px+q)\frac{d}{dx}\bigg(\frac{r}{x}+s\bigg)+\\\bigg(\frac{r}{x}+s\bigg)\frac{d}{dx}(px+q)$$

(By product rule)

$$=(px+q)\bigg[-\frac{r}{x^2}+0\bigg]+\\\bigg(\frac{r}{x}+s\bigg)(p×1+0)\\=(px+q)\bigg(-\frac{r}{x^2}\bigg)+\bigg(\frac{r}{x}+s\bigg)p\\=-\frac{pxr}{x^2}-\frac{qr}{x^2}+\frac{pr}{x}+ps\\=-\frac{pr}{x}-\frac{qr}{x^2}+\frac{pr}{x}+ps\\\Rarr\space\frac{dy}{dx}=ps-\frac{qr}{x^2}$$

$$\textbf{20.}\space\frac{\text{a+b sin x}}{\text{c+d cos x}}.\\\textbf{Sol.}\space\text{Let}\space y=\frac{\text{a+b sin x}}{\text{c+d cos x}}$$

Differentaiting w.r.t. x, we get

$$\text{(c + d cos x)}\frac{d}{dx}(\text{a+b sin x})\\\frac{dy}{dx}=\frac{-(a+b \space \text{sin x})\frac{d}{dx}(\text{c+d cos x})}{(\text{c+ d cos x})^2}\\=\\\frac{(\text{c+d cos x})[0+b \space cos x]-(a+b \space sin x)(0-d \space sin x)}{(\text{c+d cos x})^2}\\=\\\frac{(\text{c+d cos x})(b \space cos x)-(a+b\space sin x)(-d \space sin x)}{(\text{c+d cos x})^2}\\=\frac{\text{bc cos x + bd cos}^2x +\text{ad\space sin x + bd\space sin}^ 2 x}{(c+d\space cos x)^2}\\=\frac{\text{bc cos x + ad sin x + bd}(\text{cos}^2x + \text{sin}^2 x)}{(\text{c+d cos x})^2}\\\Rarr\space\frac{dy}{dx}=\frac{bc \space cos x + ad \space sin x + bd}{\text{(c+d cos x)}^2}$$

23. (x² + 1) cos x.

Sol. Let y = (x² + 1) cos x

Differentiating y w.r.t. x, we get

$$\frac{dy}{dx}=(x^2+1)\frac{d}{dx}(\text{cos x}) \space +\\\text{cos x}\frac{d}{dx}(x^2+1)$$

(By product formula)

= (x² + 1) (– sin x) + cos x (2x)

= – x² sin x – sin x + 2x cos x.