Question 11 Mark
Find the derivative of function (x + sec x)(x - tan x) (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).
Answer$\text { Here } f ( x )=( x +\sec x )( x -\tan x )$
$\therefore f^{\prime}(x)=\frac{d}{d x}[(x+\sec x)(x-\tan x)]$
$=(x+\sec x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\sec x)$
$=( x +\sec x )\left(1-\sec ^2 x \right)+( x -\tan x )(1+\sec x \tan x )$
View full question & answer→Question 21 Mark
Find the derivative of function $\frac{x}{{1 + \tan x}}$ (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).
AnswerHere $f(x) = \frac{x}{{1 + \tan x}}$
$\therefore \;f'(x) = \frac{d}{{dx}}\left[ {\frac{x}{{1 + \tan x}}} \right]$
$ = \frac{{(1 + \tan x)\frac{d}{{dx}}(x) - x\frac{d}{{dx}}(1 + \tan x)}}{{{{(1 + \tan x)}^2}}}$
$ = \frac{{(1 + \tan x)(1) - x({{\sec }^2}x)}}{{{{(1 + \tan x)}^2}}} = \frac{{1 + \tan x - x{{\sec }^2}x}}{{{{(1 + \tan x)}^2}}}$
View full question & answer→Question 31 Mark
Find the derivative of function $(x + cos x) (x - tan x)$ (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).
AnswerHere $f(x) = (x + co x)(x - tan x)$
$\therefore \;f'(x) = \frac{d}{{dx}}[(x + \cos x) + (x - \tan x)]$
$ = (x + \cos x)\frac{d}{{dx}}(x - \tan x) + (x - \tan x)\frac{d}{{dx}}(x + \cos x)$
$= (x + \cos x)(1 - sec^2x) + (x - \tan x)(1 - \sin x)$
$= -tan^2x(x + \cos x) +(x - \tan x)(1 - \sin x)$
$[\because \;1 + {\tan ^2}x = {\sec ^2}x]$
View full question & answer→Question 41 Mark
Find the derivative of function $(a{x^2} + \sin x)(p + q\;\cos x)$ (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).
AnswerHere $f(x) = (ax^2 + sinx) (p + q cosx)$
$\therefore \;f'(x) = \frac{d}{{dx}}[(a{x^2} + \sin x)(p + q\cos x)]$
$= (a{x^2} + \sin x)\frac{d}{{dx}}(p + q\cos x) + (p + q\cos x)\frac{d}{{dx}}(a{x^2} + \sin x)$
$=\left(a x^2+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x)$
$=-q \sin x\left(a x^2+\sin x\right)+(p+q \cos x)(2 a x+\cos x)$
View full question & answer→Question 51 Mark
Find the derivative of function ($x^2 + 1$) cos x (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).
AnswerHere $f(x) = (x^2 + 1)$ cos x
$\therefore f'(x) = \frac{d}{{dx}}[({x^2} + 1)\cos x]$
$= ({x^2} + 1)\frac{d}{{dx}}(\cos x) + \cos x\frac{d}{{dx}}({x^2} + 1)$
$=\left(x^2+1\right)(-\sin x)+\cos x(2 x)$
$=-x^2 \sin x-\sin x+2 x \cos x$
View full question & answer→Question 61 Mark
Find the derivative of the function $x^4$ (5 sin x - 3 cos x).
AnswerLet f(x) = $x^{4}(5 \sin x-3 \cos x)$
By product rule of differentiation,we have,
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{4} \frac{\mathrm{d}}{\mathrm{dx}}(5 \sin \mathrm{x}-3 \cos \mathrm{x})$ $+(5 \sin x-3 \cos x) \frac{d}{d x}\left(x^{4}\right)$
$=x^{4} \times\left[5 \frac{d}{d x}(\sin x)-3 \frac{d}{d x}(\cos x)\right]$$+(5 \sin x-3 \cos x) \frac{d}{d x}\left(x^{4}\right)$
$=x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)\left(4 x^{3}\right)$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{3}[5 \mathrm{x} \cos \mathrm{x}+3 \mathrm{x} \sin \mathrm{x}+20 \sin \mathrm{x}-12 \cos \mathrm{x}]$
View full question & answer→Question 71 Mark
Find the derivative of function $\frac{{\sin (x + a)}}{{\cos x}}$ (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).
AnswerHere $f(x) = \frac{{\sin (x + a)}}{{\cos x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\sin (x + a)}}{{\cos x}}} \right]$
$ = \frac{{\cos x\frac{d}{{dx}}[\sin (x + a)] - \sin (x + a)\frac{d}{{dx}}(\cos x)}}{{{{\cos }^2}x}}$
$= \frac{{\cos x.\cos (x + a) - \sin (x + a)( - \sin x)}}{{{{\cos }^2}x}}$
$= \frac{{\cos x.\cos (x + a) + \sin x\sin (x + a)}}{{{{\cos }^2}x}}$
$= \frac{{\cos (x + a - x)}}{{{{\cos }^2}x}}$ [ $\because$ cos (A-B) = cos A cos B + sin A sin B]
$= \frac{{\cos a}}{{{{\cos }^2}x}}$
View full question & answer→Question 81 Mark
Find the derivative of function $sin^nx$ (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)
AnswerHere $f (x) = sin^nx$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}({\sin ^n}x)$
$= n{\sin ^{n - 1}}x.\frac{d}{{dx}}(\sin x)$
$= nsin^{n-1}x ~cosx$
View full question & answer→Question 91 Mark
Find the derivative of function $f(x) = \cos \left( {x - \frac{\pi }{8}} \right)$ from first principle.
AnswerHere $f(x) = \cos \left( {x - \frac{\pi }{8}} \right)$ $f(x) = \cos \left( {x - \frac{\pi }{8}} \right)$
Then f (x + h) = $\cos \left( {x + h - \frac{\pi }{8}} \right)$
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + h - \frac{\pi }{8}} \right) - \cos \left( {x - \frac{\pi }{8}} \right)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {x - \frac{\pi }{8} + \frac{h}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}$$= \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {x - \frac{\pi }{8} + \frac{h}{2}} \right) \cdot \sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$
$= - \sin \left( {x - \frac{\pi }{8}} \right)$
View full question & answer→Question 101 Mark
Find the derivative of function sin (x + 1) from first principle.
AnswerHere f (x) = sin (x + 1)
Then f (x + h) = sin (x + h + 1)
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin (x + h + 1) - \sin (x + 1)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {\frac{{2x + h + 2}}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {x + 1 + \frac{h}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$ = cos (x + 1)
View full question & answer→Question 111 Mark
Find the derivative of function $(ax + b)^n$ (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).
Answer$\therefore f{\text{'}}(x) = \frac{d}{{dx}}[{(ax + b)^n}]$
$= n{(ax + b)^{n - 1}} \times \frac{d}{{dx}}(ax + b)$
$=n(a x+b)^{n-1} \times a=n a(a x+b)^{n-1}$
View full question & answer→Question 121 Mark
Find the derivative of function $(-x)^{-1}$ from first principle.
AnswerHere $f(x)=(-x)^{-1}=-\frac{1}{x}$
Then $f ( x + h )=-\frac{1}{x+h}$
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{ - \frac{1}{{x + h}} - \left( { - \frac{1}{x}} \right)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - x + x + h}}{{hx(x + h)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{hx(x + h)}} = \frac{1}{{{x^2}}}$
View full question & answer→Question 131 Mark
Find the derivative of function -x from first principle.
AnswerHere f (x) = -x
Then (f(x + h) = -(x + h)
We known that $\mathop {\lim }\limits_{h \to 0} \frac{{ - (x + h) - ( - x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{ - x - h + x}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h} = - 1$
View full question & answer→Question 141 Mark
Find the derivative of $x^{-4}\left(3-4 x^{-5}\right)$
Answer$\text { Here } f(x)=x^{-4}\left(3-4 x^{-5}\right)$
$f^{\prime}(x)=\frac{d}{d x}\left[x^{-4}\left(3-4 x^{-5}\right)\right]$
$=x^{-4} \frac{d}{d x}\left(3-4 x^{-5}\right)+\left(3-4 x^{-5}\right) \frac{d}{d x}\left(x^{-4}\right)$
$=x^{-4}\left(20 x^{-6}\right)+\left(3-4 x^{-5}\right)\left(-4 x^{-5}\right)$
$=20 x^{-10}-12 x^{-5}+16 x^{-10}$
$=36 x^{-10}-12 x^{-5}=\frac{36}{x^{10}}-\frac{12}{x^5}$
View full question & answer→Question 151 Mark
Find the derivative of $x^5\left(3-6 x^{-9}\right)$
Answer$\text { Here } f(x)=x^5\left(3-6 x^{-9}\right)$
$=x^5 \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x}\left(x^5\right)$
$=x^5\left(54 x^{-10}\right)+\left(3-6 x^{-9}\right) \times 5 x^4$
$=54 x^{-5}+15 x^4-30 x^{-5}$
$=24 x^{-5}+15 x^4$
$=\frac{24}{x^5}+15 x^4$
View full question & answer→Question 161 Mark
Find the derivative of $x^{-3} (5 + 3x)$
AnswerHere $f(x) = x^{-3}(5 + 3x)$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[{x^{ - 3}}(5 + 3x)]$
$ = {x^{ - 3}}\frac{d}{{dx}}(5 + 3x) + (5 + 3x)\frac{d}{{dx}}({x^{ - 3}})$
$= x^{-3} \times 3 + (5 + 3x) \times (- 3x)^{-4}$
$ = \frac{3}{{{x^3}}} - \frac{3}{{{x^4}}}(5 + 3x)$
$ = \frac{3}{{{x^3}}}\left[ {1 - \frac{{5 + 3x}}{x}} \right]$$ = \frac{3}{{{x^3}}}\left[ {\frac{{x - 5 - 3x}}{x}} \right] = \frac{{ - 3}}{{{x^4}}}(5 + 2x)$
View full question & answer→Question 171 Mark
Find the derivative of
$\left(5 x^{3}+3 x-1\right)(x-1)$
AnswerLet $f(x)=\left(5 x^3+3 x-1\right)(x-1)$
By product rule of differentiation,we have,
$f^{\prime}(x)=\left(5 x^3+3 x-1\right) \frac{d}{d x}(x-1)+(x-1) \frac{d}{d x}\left(5 x^3+3 x+1\right)$
$=\left(5 x^3+3 x-1\right) \times 1+(x-1) \times\left(15 x^2+3\right)$
$=\left(5 x^3+3 x-1\right)+(x-1) \times\left(15 x^2+3\right)$
$=5 x^3+3 x-1+15 x^3+3 x-15 x^2-3$
$=20 x^3-15 x^2+6 x-4$
View full question & answer→Question 181 Mark
Find the derivative of $2x - \frac{3}{4}$
AnswerHere $f(x) = 2x - \frac{3}{4}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left( {2x - \frac{3}{4}} \right)$
$= 2\frac{d}{{dx}}(x) - \frac{d}{{dx}}\left( {\frac{3}{4}} \right)$
= 2 × 1 - 0 = 2
View full question & answer→Question 191 Mark
For some constants a and b, find the derivative of $(ax^2 + b)^2$
AnswerHere $f (x) = (ax^2 + b)^2 = a^2x^4 + b^2 + 2abx^2$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[{a^2}{x^4} + {b^2} + 2ab{x^2}]$
$ = {a^2}\frac{d}{{dx}}({x^4}) + \frac{d}{{dx}}({b^2}) + 2ab\frac{d}{{dx}}({x^2})$
$= a^2 \times 4x^3 + 0 + 2ab \times 2x$
$= 4a^2x^3 + 4abx$
$= 4ax(ax^2 + b)$
View full question & answer→Question 201 Mark
Find the derivative of x at x = 1
AnswerHere $\frac{d}{{dx}}$(x) = 1
$\therefore$ Derivative of x at x = 1
View full question & answer→Question 211 Mark
Find the derivative of function cosec x
AnswerHere f (x) = cosec x
$\therefore \;{\text{f'}}(x) = \frac{d}{{dx}}({\text{cosec}}\,x)$
= - cosec x cot x
View full question & answer→Question 221 Mark
Find the derivative of the function 5sec x + 4cos x
AnswerLet f(x) = 5 sec x + 4 cos x
Therefore, we have
$f^{\prime}(x)=\frac{d}{d x}(5 \sec x+4 \cos x)$
$=5 \frac{d}{d x}(\sec x)+4 \frac{d}{d x}(\cos x)$
= 5 sec x tan x + 4 $\times$ (-sin x)
$\therefore$ f’(x) = 5 sec x tan x – 4 sin x
View full question & answer→Question 231 Mark
Evaluate $\mathop {\lim }\limits_{r \to 1} \;\pi {r^2}$
AnswerHere $\mathop {\lim }\limits_{r \to 1} \;\pi {r^2}$ $ = \pi \times {(1)^2} = \pi$
View full question & answer→Question 241 Mark
Evaluate $\lim \limits_{x \rightarrow \pi}\left(x-\frac{22}{7}\right)$
AnswerWe have, $ \lim \limits_{x \rightarrow \pi}\left(x-\frac{22}{7}\right)=\left(\pi-\frac{22}{7}\right)$
View full question & answer→Question 251 Mark
Evaluate $\mathop {\lim }\limits_{x \to 3} x + 3$
AnswerHere $\mathop {\lim }\limits_{x \to 3} x + 3$ = 3 + 3 = 6
View full question & answer→Question 261 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow4}\frac{4\text{x}+3}{\text{x}-2}$
Answer$\lim\limits_{\text{x}\rightarrow4}\frac{4\text{x}+3}{\text{x}-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}$
View full question & answer→Question 271 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow-1}\frac{\text{x}^{10}+\text{x}^5+1}{\text{x}+1}$
Answer$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^{10}+(-1)^5+1}{-1-1}=\frac{1-1+1}{-2}=-\frac{1}{2}$
View full question & answer→Question 281 Mark
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\text{x}\sec\text{x}$
Answer$\lim\limits_{\text{x}\rightarrow0}\text{x}\sec\text{x}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\cos\text{x}}=\frac{0}{\cos0}=0$
View full question & answer→Question 291 Mark
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}}{\pi-\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}}{\pi-\text{x}}=\frac{\cos0}{\pi-0}=\frac{1}{\pi}$
View full question & answer→Question 301 Mark
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{b}}{\text{cx}+1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{b}}{\text{cx}+1}=\frac{\text{a(0)+b}}{\text{c(0)+1}}=\text{b}$
View full question & answer→Question 311 Mark
Evaluate the following limits in Exercise:$$$\lim\limits_{\text{x} \rightarrow2}\frac{3\text{x}^2-\text{x}-10}{\text{x}^2-4}$
AnswerAt x = 2, the value of the given rational function takes the form $\frac{0}{0}$.
$\therefore\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-\text{x}-10}{\text{x}^2-4}=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(3\text{x}+5)}{(\text{x}-2)(\text{x}+2)}$
$$$=\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}+5}{\text{x}+2}$
$=\frac{3(2)+5}{2+2}$
$=\frac{11}{4}$
View full question & answer→Question 321 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow1}\pi\text{r}^2$
Answer$\lim\limits_{\text{r}\rightarrow1}\pi\text{r}^2=\pi(1)^2=\pi$
View full question & answer→Question 331 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow-0}\frac{(\text{x}+1)^5-1}{\text{x}}$
Answer$\lim\limits_{\text{x} \rightarrow-0}\frac{(\text{x}+1)^5-1}{\text{x}}$Put x + 1 = y so that y → 1 as x → 0.Put x + 1 = y so that y → 1 as x → 0.
$\text{Accordingly},\lim\limits_{\text{x}\rightarrow0}\frac{(\text{x}+1)^5-1}{\text{x}}=\lim\limits_{\text{y}\rightarrow1}\frac{\text{y}^5-1}{\text{y}-1}$
$=5.1^{5-1}$ $\Big[\lim\limits_{\text{x}\rightarrow\text{a}}\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}\Big]$
$=5$
$\therefore\lim\limits_{\text{x}\rightarrow0}\frac{(\text{x}+5)^5-1}{\text{x}}=5$
View full question & answer→Question 341 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow3} \text{x + 3}$
Answer$\lim\limits_{\text{x} \rightarrow3} \text{x + 3}$ $= 3 +3 = 6$
View full question & answer→Question 351 Mark
Evaluate the following limits in Exercise: $\lim\limits_{\text{x} \rightarrow\pi} \Big(\text{x} - \frac{22}{7}\Big)$
Answer$\lim\limits_{\text{x} \rightarrow\pi} \Big(\text{x} - \frac{22}{7}\Big)=\Big(\pi-\frac{22}{7}\Big)=\Big(\frac{22}{7}-\frac{22}{7}\Big)=0$
View full question & answer→Question 361 Mark
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) $\lim\limits_{\text{x}\rightarrow 0} \frac{3{^{2+\text{x}}-9}}{\text{x}}$ is equal to $9\log2$ Reason (R) $\lim\limits_{\text{x}\rightarrow 0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$ is equal to $\log\text{a}.$
- A is true, R is true; R is acorrect explanation of A.
- A is true, R is true; R is not a correct explanation of A.
- A is true; R is false
- A is false; R is true.
AnswerDirections: 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) $\lim\limits_{\text{x}\rightarrow 0} \frac{3{^{2+\text{x}}-9}}{\text{x}}$ is equal to $9\log2$ Reason (R) $\lim\limits_{\text{x}\rightarrow 0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$ is equal to $\log\text{a}.$
- A is true, R is true; R is acorrect explanation of A.
- A is true, R is true; R is not a correct explanation of A.
- A is true; R is false
- A is false; R is true.
View full question & answer→Question 371 Mark
Find the derivative of f(x) = 10x.
Answer f′ ( x) = $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{10(x+h)-10(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{10 h}{h}=\lim _{h \rightarrow 0}(10)=10$
View full question & answer→Question 381 Mark
Find the derivative of f(x) = 3 at x = 0 and at x = 3.
AnswerLet f(x)=3
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{3-3}{h}=\lim _{h \rightarrow 0} \frac{0}{h}=0$
Similarly $f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h \rightarrow 0} \frac{3-3}{h}=0$
View full question & answer→Question 391 Mark
Find the derivative of sin x at x = 0.
AnswerLet f(x) = sin x. Therefore,
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (0+h)-\sin (0)}{h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=1$
View full question & answer→Question 401 Mark
Find the derivative of the function $f(x) = 2x^2 + 3x - 5 at x = - 1$. Also, prove that f'(0) + 3f'(-1) = 0.
AnswerFirst, we find the derivatives of f(x ) at x = -1 and x = 0. We have,
${f^\prime }( - 1) = \mathop {\lim }\limits_{h \to 0} \frac{{f( - 1 + h) - f( - 1)}}{h}$
$\left[ {\because {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{( - 1 + h)}^2} + 3( - 1 + h) - 5} \right] - \left[ {2{{( - 1)}^2} + 3( - 1) - 5} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2\left( {1 + {h^2} - 2h} \right) - 3 + 3h - 5} \right] - [2 - 3 - 5\} }}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} - h}}{h} = \mathop {\lim }\limits_{h \to 0} (2h - 1)$= 2(0) - 1 = -1
and ${f^\prime }(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h}$
$\left[ {\because {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {2{{(0 + h)}^2} + 3(0 + h) - 5} \right] - \left[ {2{{(0)}^2} + 3(0) - 5} \right]}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{2{h^2} + 3h}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} (2h + 3)$
= 2(0) + 3 = 3
Now, f'(0) + 3f'(-1) = 3 - 3 = 0.
Hence proved.
View full question & answer→Question 411 Mark
Find the derivative at x = 2 of the function f(x) = 3x.
AnswerWe have $f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{3(2+h)-3(2)}{h}$ $=\lim _{h \rightarrow 0} \frac{6+3 h-6}{h}=\lim _{h \rightarrow 0} \frac{3 h}{h}=\lim _{h \rightarrow 0} 3=3$
Therefore, derivative of the function 3x at x = 2 is 3.
View full question & answer→Question 421 Mark
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x}$
Answer$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} \cdot \frac{1}{{\cos x}}$
$ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right) \cdot \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\cos x}}} \right)$
$\left[\because {\mathop {\lim }\limits_{x \to a} f(x)g(x) = \mathop {\lim }\limits_{x \to a} f(x)\mathop {\lim }\limits_{x \to a} g(x)} \right]$
$= 1 \times \frac { 1 } { \cos 0 } = 1 \times \frac { 1 } { 1 } = 1$
View full question & answer→Question 431 Mark
Evaluate $\lim _\limits{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$
AnswerWe have, $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$ $=\lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x} \cdot \frac{2 x}{\sin 2 x} \cdot 2\right]$
$=2 \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div\left[\frac{\sin 2 x}{2 x}\right]$
$=2 \cdot \lim _{4 x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div \lim _{2 x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]$
$=2.1 ÷1=2(\text { as } x \rightarrow 0,4 x \rightarrow 0 \text { and } 2 x \rightarrow 0)$
View full question & answer→Question 441 Mark
Find the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - 1}}{x}.$
AnswerPut y = 1 + x, then y $\rightarrow$ 1 as x $\rightarrow$ 0.
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - 1}}{x} = \mathop {\lim }\limits_{y \to 1} \frac{{\sqrt y - 1}}{{y - 1}}$
$ = \mathop {\lim }\limits_{y \to 1} \frac{{{y^{\frac{1}{2}}} - {1^{\frac{1}{2}}}}}{{y - 1}}$
$ = \frac{1}{2}{(1)^{\frac{1}{2} - 1}}\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$
$= \frac { 1 } { 2 }$
View full question & answer→Question 451 Mark
Evaluate: $\mathop {\lim }\limits_{x \to 1} \frac{{{x^{15}} - 1}}{{{x^{10}} - 1}}$
AnswerWe have,
$\mathop {\lim }\limits_{x \to 1} \frac{{{x^{15}} - 1}}{{{x^{10}} - 1}}$ = $\mathop {\lim }\limits_{x \to 1} \frac{{{x^{15}} - 1}}{{{x^{10}} - 1}} \times \frac{{(x - 1)}}{{(x - 1)}}$
= $\mathop {\lim }\limits_{x \to 1} \frac{{{x^{15}} - 1}}{{x - 1}} \div \mathop {\lim }\limits_{x \to 1} \frac{{{x^{10}} - 1}}{{x - 1}}$
$\mathop {\lim }\limits_{x \to 1} \frac{{{x^{15}} - {{(1)}^{15}}}}{{x - 1}} \div \mathop {\lim }\limits_{x \to 1} \frac{{{x^{10}} - {{(1)}^{10}}}}{{x - 1}}$
= $15(1)^{14} \div 10(1)^9 [\because {\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}}}$ = $na^{n - 1}$]
= $\frac { 15 } { 10 } = \frac { 3 } { 2 }$
View full question & answer→Question 461 Mark
Find the limit: $\lim \limits_{x \rightarrow 1}\left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]$
AnswerFirst, we rewrite the function as a rational function.
$\left[\frac{x-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]=\left[\frac{x-2}{x(x-1)}-\frac{1}{x\left(x^{2}-3 x+2\right)}\right]$
$=\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x-1)(x-2)}\right]$
$=\left[\frac{x^{2}-4 x+4-1}{x(x-1)(x-2)}\right]$
$=\frac{x^{2}-4 x+3}{x(x-1)(x-2)}$
Evaluating the function at 1, we get it of the form $\frac{0}{0}$
Hence $\lim _{x \rightarrow 1}\left[\frac{x^{2}-2}{x^{2}-x}-\frac{1}{x^{3}-3 x^{2}+2 x}\right]=\lim _{x \rightarrow 1} \frac{x^{2}-4 x+3}{x(x-1)(x-2)}$
$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{x(x-1)(x-2)}$
$=\lim _{x \rightarrow 1} \frac{x-3}{x(x-2)}=\frac{1-3}{1(1-2)}=2$ $$
View full question & answer→Question 471 Mark
Find the limit: $\lim \limits_{x \rightarrow 2}\left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$
AnswerEvaluating the function at 2, we get it of the form $\frac{0}{0}$
Therefore, we have,
$\lim _{x \rightarrow 2} \frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}=\lim _{x \rightarrow 2} \frac{x^{2}(x-2)}{(x-2)(x-3)}$
$=\lim _{x \rightarrow 2} \frac{x^{2}}{(x-3)}=\frac{(2)^{2}}{2-3}=\frac{4}{-1}=-4$
View full question & answer→Question 481 Mark
Find the limit: $\lim \limits_{x \rightarrow 2}\left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$
AnswerEvaluating the function at 2, we get it of the form $\frac{0}{0}$
Therefore, we have,
$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^{2}}$
$=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0}$ = $\infty$
View full question & answer→Question 491 Mark
Find the derivative of $\frac { x + \cos x } { \tan x }.$
AnswerLet $y = \frac { x + \cos x } { \tan x }$
On differentiating both sides w.r.t. x, we get
$\frac { d y } { d x } = \frac { \frac { d } { d x } \tan x ( x + \cos x ) - ( x + \cos x ) \frac { d } { d x } ( \tan x ) } { ( \tan x ) ^ { 2 } }$ [using quotient rule of derivate]
$= \frac { \tan x ( 1 - \sin x ) - ( x + \cos x ) \sec ^ { 2 } x } { ( \tan x ) ^ { 2 } }$
View full question & answer→Question 501 Mark
Find the derivative of $\frac{x^{5}-\cos x}{\sin x}$
AnswerLet $h(x)=\frac{x^{5}-\cos x}{\sin x}$ We use the quotient rule on this function wherever it is defined.Then,we have,
$h^{\prime}(x)=\frac{\left(x^{5}-\cos x\right)^{\prime} \sin x-\left(x^{5}-\cos x\right)(\sin x)^{\prime}}{(\sin x)^{2}}$
$=\frac{\left(5 x^{4}+\sin x\right) \sin x-\left(x^{5}-\cos x\right) \cos x}{\sin ^{2} x}$
$=\frac{-x^{5} \cos x+5 x^{4} \sin x+1}{(\sin x)^{2}}$
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