Question 12 Marks
Find the derivative of function $\frac{{p{x^2} + qx + r}}{{ax + b}}$ (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 $f(x) = \frac{{p{x^2} + qx + r}}{{ax + b}}$
$\therefore f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{p{x^2} + qx + r}}{{ax + b}}} \right]$
$= \frac{{(ax + b)\frac{d}{{dx}}(p{x^2} + qx + r) - (p{x^2} + qx + r)\frac{d}{{dx}}(ax + b)}}{{{{(ax + b)}^2}}}$
$= \frac{{(ax + b)(2px + q) - (p{x^2} + qx + r)(a)}}{{{{(ax + b)}^2}}}$
$= \frac{{2ap{x^2} + aqx + 2bpx + bq + ap{x^2} - aqx - ar}}{{{{(ax + b)}^2}}}$
$ = \frac{{ap{x^2} + 2bpx + bq - ar}}{{{{(ax + b)}^2}}}$
View full question & answer→Question 22 Marks
Find the derivative of function $\frac{{ax + b}}{{p{x^2} + qx + r}}$ (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$f(x) = \frac{{ax + b}}{{p{x^2} + qx + r}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{ax + b}}{{p{x^2} + qx + r}}} \right]$
$= \frac{{(p{x^2} + qx + r)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}(p{x^2} + qx + r)}}{{{{(p{x^2} + qx + r)}^2}}}$
$ = \frac{{(p{x^2} + qx + r)(a) - (ax + b)(2px + q)}}{{{{(p{x^2} + qx + r)}^2}}}$
$ = \frac{{ap{x^2} + aqx + ar - 2ap{x^2} - aqx - 2bpx - bq}}{{{{(p{x^2} + qx + r)}^2}}}$
$= \frac{{ - ap{x^2} - 2bpx + ar - bq}}{{{{(p{x^2} + qx + r)}^2}}}$
View full question & answer→Question 32 Marks
Find the derivative of function $\frac{1}{{a{x^2} + bx + c}}$(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 $f(x) = \frac{1}{{a{x^2} + bx + c}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left( {\frac{1}{{a{x^2} + bx + c}}} \right)$
$ = \frac{{(a{x^2} + bx + c)\frac{d}{{dx}}(1) - 1.\frac{d}{{dx}}(a{x^2} + bx + c)}}{{{{(a{x^2} + bx + c)}^2}}}$
$ = \frac{{(a{x^2} + bx + c)(0) - 1(2ax + b)}}{{{{(a{x^2} + bx + c)}^2}}}$$ = \frac{{ - (2ax + b)}}{{{{(a{x^2} + bx + c)}^2}}}$
View full question & answer→Question 42 Marks
Find the derivative of function $\frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{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{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}} = \frac{{x + 1}}{{x - 1}}$
$f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{x + 1}}{{x - 1}}} \right]$
$ = \frac{{(x - 1)\frac{d}{{dx}}(x + 1) - (x + 1)\frac{d}{{dx}}(x - 1)}}{{{{(x - 1)}^2}}}$
$= \frac{{(x - 1) \times 1 - (x + 1) \times 1}}{{{{(x - 1)}^2}}}$
$ = \frac{{x - 1 - x - 1}}{{{{(x - 1)}^2}}} = \frac{{ - 2}}{{{{(x - 1)}^2}}},\;x \ne 0,\;1.$
View full question & answer→Question 52 Marks
Find the derivative of function $\frac{{ax + b}}{{cx + d}}$ (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{{ax + b}}{{cx + d}}$
$\therefore \;f'(x) = \frac{d}{{dx}}\left[ {\frac{{ax + b}}{{cx + d}}} \right]$
$= \frac{{(cx + d)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}(cx + d)}}{{{{(cx + d)}^2}}}$
$= \frac{{(cx + d)(a) - (ax + b)(c)}}{{{{(cx + d)}^2}}}$
$ = \frac{{acx + ad - acx - bc}}{{{{(cx + d)}^2}}} = \frac{{ad - bc}}{{{{(cx + d)}^2}}}$
View full question & answer→Question 62 Marks
Find the derivative of function $(ax + b)(cx + d)^2($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 + b) (cx + d)^2$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[(ax + b){(cx + d)^2}]$
$= (ax + b ) \frac{d}{{dx}}{(cx + d)^2} + {(cx + d)^2} \cdot \frac{d}{{dx}}(ax + b)$
$= (ax + b) \times 2(cx + d) \times c + (cx + d)^2 \times a$
$= 2c(ax + b)(cx + d) + a(cx + d)^2$
View full question & answer→Question 72 Marks
Find the derivative of function $\frac{x}{{{{\sin }^n}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}{{{{\sin }^n}x}}$
$\therefore \;f'(x) = \frac{d}{{dx}}\left[ {\frac{x}{{{{\sin }^n}x}}} \right]$
$ = \frac{{{{\sin }^n}x\frac{d}{{dx}}(x) - x\frac{d}{{dx}}({{\sin }^n}x)}}{{{{({{\sin }^n}x)}^2}}}$
$= \frac{{{{\sin }^n}x \times 1 - x.n{{\sin }^{n - 1}}x\frac{d}{{dx}}(\sin x)}}{{{{({{\sin }^n}x)}^2}}}$
$= \frac{{{{\sin }^n}x - nx{{\sin }^{n - 1}}x\cos x}}{{{{({{\sin }^n}x)}^2}}}$
$ = \frac{{{{\sin }^{n - 1}}x[\sin x - nx\cos x]}}{{{{\sin }^{2x}}x}}$
$ = \frac{{\sin x - nx\cos x}}{{{{\sin }^{n + 1}}x}}$
View full question & answer→Question 82 Marks
Find the derivative of function $ (px + q) \left( {\frac{r}{x} + s} \right)$ (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) = (px + q)\left( {\frac{r}{x} + s} \right)$
$\therefore f{\text{'}}(x) = \frac{d}{{dx}}[(px + q) \left( {\frac{r}{x} + s} \right) ]$
$ =\frac d{dx}\left(pr+psx +\frac{qr}x+ sq\right)$
$= \frac d{dx}\left(pr\right) +\frac d{dx}\left(psx\right) +\frac d{dx}\left(\frac{qr}x\right) +\frac d{dx}\left(sq\right)$
$= 0 +ps( 1) +qr \left(\frac{-1}{x^2}\right) +0$
$=ps - \frac{qr}{x^2}$
View full question & answer→Question 92 Marks
Find the derivative of the function $\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin 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)
AnswerLet f(x) = $\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$
By quotient rule,we have,
$f^{\prime}(x)=\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\right]$
$=\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \times 2 x-x^{2} \cos x}{\sin ^{2} x}\right]$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x} \cos \frac{\pi}{4}[2 \sin \mathrm{x}-\mathrm{x} \cos \mathrm{x}]}{\sin ^{2} \mathrm{x}}$
View full question & answer→Question 102 Marks
Find the derivative of the function $f(x) = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}$
AnswerHere $f(x) = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}$
$\therefore\;f'(x)=\frac{(3x+7\cos x)\frac d{dx}(4x+5\sin x)-(4x+5\sin x)\frac d{dx}(3x+7\cos x)}{{(3x+7\cos x)}^2}$
$ = \frac{{(3x + 7\cos x)(4+ 5\cos x) - (4x + 5\sin x)(3 - 7\sin x)}}{{{{(3x + 7\cos x)}^2}}}$
$=\;\frac{12x+15x\cos x+28\cos x+35\cos^2x-12x+28x\sin x-15\sin x+35\;\sin^2x}{{(3x+7\cos x)}^2}$
$ = \frac{{15x\cos x + 28\cos x + 28x\sin x - 15\sin x + 35({{\cos }^2}x + {{\sin }^2}x)}}{{{{(3x + 7\cos x)}^2}}}$
$ = \frac{{15x\cos x + 28\cos x + 28x\sin x - 15\sin x + 35}}{{{{(3x + 7\cos x)}^2}}}$
View full question & answer→Question 112 Marks
Find the derivative of function $\frac{{a + b\sin x}}{{c + d\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{{a + b\sin x}}{{c + d\cos x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{a + b\sin x}}{{c + d\cos x}}} \right]$
$= \frac{{(c + d\cos x)\frac{d}{{dx}}(a + b\sin x) - (a + b\sin x)\frac{d}{{dx}}(c + d\cos x)}}{{{{(c + d\cos )}^2}}}$
$= \frac{{(c + d\cos x)(b\cos x) - (a + b\sin x)( - d\sin x)}}{{{{(c + d\cos x)}^2}}}$
$= \frac{{bc\cos x + bd{{\cos }^2}x + ad\sin x + bd{{\sin }^2}x}}{{{{(c + d\cos x)}^2}}}$
$ = \frac{{bc\cos x + ad\sin x + bd({{\cos }^2}x + {{\sin }^2}x)}}{{{{(c + d\cos x)}^2}}}$
$ = \frac{{bc\cos x + ad\sin x + bd}}{{{{(c + d\cos x)}^2}}}$
View full question & answer→Question 122 Marks
Find the derivative of function (x + a) (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 + a
$\therefore {\text{f'}}(x) = \frac{d}{{dx}}(x + a) = 1$
View full question & answer→Question 132 Marks
Find the derivative of function $\frac{{\sec x - 1}}{{\sec x + 1}}$ (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$f(x) = \frac{{\sec x - 1}}{{\sec x + 1}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\sec x - 1}}{{\sec x + 1}}} \right]$
$= \frac{{(\sec + 1)\frac{d}{{dx}}(\sec x - 1) - (\sec x - 1)\frac{d}{{dx}}(\sec x + 1)}}{{{{(\sec x + 1)}^2}}}$
$= \frac{{(\sec x + 1)(\sec x\tan x) - (\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}}}$
View full question & answer→Question 142 Marks
Find the derivative of function $\frac{{\sin x + \cos x}}{{\sin x - \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 + \cos x}}{{\sin x - \cos x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right]$
$= \frac{{(\sin x - \cos x)\frac{d}{{dx}}(\sin x + \cos x) - (\sin x + \cos x)\frac{d}{{dx}}(\sin x - \cos x)}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - {{(\sin x - \cos x)}^2} - {{(\sin x + \cos x)}^2}}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - ({{\sin }^2}x - {{\cos }^2}x + 2\sin x\cos x - {{\sin }^2}x - {{\cos }^2}x - {{\cos }^2}x - 2\sin x\cos x}}{{{{(\sin x - \cos x)}^2}}}$
$= \frac{{ - 2({{\sin }^2}x + {{\cos }^2}x)}}{{{{(\sin x - \cos x)}^2}}}$$= \frac{{ - 2}}{{{{(\sin x - \cos x)}^2}}}$
View full question & answer→Question 152 Marks
Find the derivative of function $f(x) = \frac{{\cos x}}{{1 + \sin x}}$ (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 $f(x) = \frac{{\cos x}}{{1 + \sin x}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$
$= \frac{{(1 + \sin x)\frac{d}{{dx}}(\cos x) - \cos x \cdot \frac{d}{{dx}}(1 + \sin x)}}{{{{(1 + \sin x)}^2}}}$
$= \frac{{(1 + \sin x)( - \sin x) - \cos x(\cos x)}}{{{{(1 + \sin x)}^2}}}$
$ = \frac{{ - \sin x - {{\sin }^2}x - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}$$= \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(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}}$
View full question & answer→Question 162 Marks
Find the derivative of function sin (x + a) (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 (x + a)
$\therefore$ f'(x) = $\frac{d}{{dx}}$ [sin (x + a)]
= cos (x + a). $\frac{d}{{dx}}$(x + a)
= cos (x + a)
View full question & answer→Question 172 Marks
Find the derivative of function $(ax + b)^n (cx + d)^m ($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 + b)^n (cx + d)^m$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[{(ax + b)^n}{(cx + d)^m}]$
$= {(ax + b)^n}\frac{d}{{dx}}{(cx + d)^m}+{(cx + d)^m}\frac{d}{{dx}}{(ax + b)^n} $
$= (ax + b)^n . m (cx + d)^{m-1} . \frac{d}{{dx}}(cx + d) +(cx + d)^m n(ax + b)^{n-1} . \frac{d}{{dx}}(ax + b)$
$= cm (ax + b)^n (cx + d)^{m-1}+ an (cx + d)^m (ax + b)^{n-1}$
$= (ax + b)^{n-1} (cx +d)^{m-1} [cm (ax + b) + an (cx +d)]$
View full question & answer→Question 182 Marks
Find the derivative of function $4\sqrt x - 2$ (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) = 4\sqrt x - 2$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left[ {4\sqrt x - 2} \right]$
$= 4\frac{d}{{dx}}(\sqrt x ) - \frac{d}{{dx}}(2)$
$= 4 \times \frac{1}{{2\sqrt x }} - 0 = \frac{2}{{\sqrt x }}$
View full question & answer→Question 192 Marks
Find the derivative of function $\frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \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{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x = a{x^-4} - b{x^{ - 2}} + \cos x$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}[a{x^{ - 4}} - b{x^2} + \cos x]$$= a\frac{d}{{dx}}({x^{ - 4}}) - b\frac{d}{{dx}}({x^{ - 2}}) + \frac{d}{{dx}}(\cos x)$
$ - a{x^{ - 5}} + 2b{x^{ - 3}} - \sin x = \frac{{ - 4a}}{{{x^5}}} + \frac{{2b}}{{{x^3}}} - \sin x$
$- 4a{x^{ - 5}} + 2b{x^{ - 3}} - \sin x = \frac{{ - 4a}}{{{x^5}}} + \frac{{2b}}{{{x^3}}} - \sin x$
View full question & answer→Question 202 Marks
Find the derivative of $\frac{{{x^n} - {a^n}}}{{x - a}}$ for some constant a.
AnswerHere $f(x)\frac{{{x^n} - {a^n}}}{{x - a}}$
$\therefore \;f(x) = \frac{d}{{dx}}\left[ {\frac{{{x^n} - {a^n}}}{{x - a}}} \right]$
$= \frac{{(x - a)\frac{d}{{dx}}({x^n} - {a^n}) - ({x^n} - {a^n})\frac{d}{{dx}}(x - a)}}{{{{(x - a)}^2}}}$
$ = \frac{{(x - a) \times n{x^{n - 1}} - ({x^n} - {a^n}) \times 1}}{{{{(x - a)}^2}}}$
$= \frac{{n{x^n} - an{x^{n - 1}} - {x^n} + {a^n}}}{{{{(x - a)}^2}}}$
View full question & answer→Question 212 Marks
For some constants, a and b, find the derivative of f(x) $= \frac{{x - a}}{{x - b}}$
AnswerHere f (x) $= \frac{{x - a}}{{x - b}}$
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}\left( {\frac{{x - a}}{{x - b}}} \right)$
$ = \frac{{(x - b)\frac{d}{{dx}}(x - a) - (x - a)\frac{d}{{dx}}(x - b)}}{{{{(x - b)}^2}}}$
$ = \frac{{(x - b) \times 1 - (x - a) \times 1}}{{{{(x - b)}^2}}} = \frac{{x - b - x + a}}{{{{(x - b)}^2}}}$
$= \frac{{a - b}}{{{{(x - b)}^2}}}$
View full question & answer→Question 222 Marks
For some constants a and b, find the derivative of (x - a)(x - b)
AnswerHere f (x) = (x - a)(x - b)
$\therefore \;f{\text{'}}(x) = \frac{d}{{dx}}(x - a)(x - b)$
$ = (x - a)\frac{d}{{dx}}(x - b) + (x - b)\frac{d}{{dx}}(x - a)$
= (x - a) × 1 + (x - b) × 1
= x - a + x - b = 2x - a - b
View full question & answer→Question 232 Marks
Find the derivative of $x^n + ax^{n-1} + a^2x^{n-2} + ...+ a^{n-1} x + a^n $ for some fixed real number a.
AnswerLet $ f(x) = x^n + ax^{n-1} + a^2x^{n-2} + .... + a^{n- 1}x + a^n$
On differentiating both sides, we get
$f'(x) = nx^{n-1} + a(n - 1)x^{n-2} + a^2(n - 2)x^{n-3} + .... + a^{n-1}.1 + 0$
On putting $x = a$ both sides, we get
$f'(a) = na^{n-1} + a(n - 1)a^{n-2} + a^2 (n - 2)a^{n-3} +...+ a^{n-1}$
$= n a^{n-1} + (n - 1) a^{n-1} + (n - 2) a^{n-1} + ... + a^{n-1}$
$= a^{n-1} [n + (n - 1) + (n - 2) + .... + 1]$
$[\because$ sum of n natural numbers $= \frac { n ( n + 1 ) } { 2 }]$
$f'(a)= \frac { n ( n + 1 ) } { 2 }\; a ^ { n - 1 }$
View full question & answer→Question 242 Marks
For the function $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1$ prove that $f'(1) = 100f'(0)$
AnswerHere $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1$
$f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1} \right]$
$= \frac{1}{{100}}\frac{d}{{dx}}({x^{100}}) + \frac{1}{{99}}\frac{d}{{dx}}({x^{99}}) + ... + \frac{1}{2}\frac{d}{{dx}}({x^2}) + \frac{d}{{dx}}(x) + \frac{d}{{dx}}(1)$
$= \frac{1}{{100}} \times 100{x^{99}} + \frac{1}{{99}} \times 99{x^{98}} + ... + \frac{1}{2} \times 2x + 1 + 0$
$= x^{99} + x^{98} + ... + x + 1$
Now $f'(1) = (1)^{99} + (1)^{98} + ...+(1) + 1 = 100$
$f'(0) = (0)^{99} + (0)^{98} + ...+ 0 + 1 = 1$
Which shows that $f'(1) = 100f'(0)$
View full question & answer→Question 252 Marks
Find the derivative of $\left( \frac { x + 1 } { x - 1 } \right)$ from the first principle.
AnswerWe have, $f ( x ) = \frac { x + 1 } { x - 1 }$
By first principle of derivative, we have
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\frac{{(x + h) + 1}}{{(x + h) - 1}} - \frac{{x + 1}}{{x - 1}}} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h + 1)(x - 1) - (x + 1)(x + h - 1)}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + xh - h - 1} \right) - \left( {{x^2} + xh + h - 1} \right)}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2h}}{{h(x + h - 1)(x - 1)}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2}}{{(x + h - 1)(x - 1)}} = \frac{{ - 2}}{{{{(x - 1)}^2}}}$
View full question & answer→Question 262 Marks
Find the derivative of $1/x^2$ from the first principle.
AnswerHere ${\text{f}}(x) = \frac{1}{{{x^2}}}$
Then ${\text{f}}(x + h) = \frac{1}{{{{(x + h)}^2}}}$
We know that ${\text{f'}}(x) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\Rightarrow \;{\text{f'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{(x + h)}^2}}} - \frac{1}{{{x^2}}}}}{h}$$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {{(x + h)}^2}}}{{h{x^2}{{(x + h)}^2}}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} - {x^2} - {h^2} - 2xh}}{{h{x^2}{{(x + h)}^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{h( - h - 2x)}}{{h{x^2}{{(x + h)}^2}}}$
$ = \frac{{ - 2x}}{{{x^2} \times {x^2}}} = \frac{{ - 2}}{{{x^3}}}$
View full question & answer→Question 272 Marks
Find the derivative of $(x -1) (x - 2)$ from first principle.
AnswerWe have, $f(x) = (x - 1)(x - 2)$
$= x^2 - 3x + 2$
By first principle of derivative, we have
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^2} - 3(x + h) + 2} \right] - \left[ {{x^2} - 3x + 2} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left( {{x^2} + {h^2} + 2xh - 3x - 3h + 2} \right] - \left[ {{x^2} - 3x + 2} \right]} \right.}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2} - 3h}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h(2x + h - 3)}}{h}$
$= 2 x - 3$
View full question & answer→Question 282 Marks
Find the derivative of $(x^3 - 27)$ from first principle.
AnswerWe have, $f(x) = x^3 - 27$
By using first principle of derivative,
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$\therefore \quad {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^3} - 27} \right] - \left[ {{x^3} - 27} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3{h^2}x + {h^3} - 27 - {x^3} + 27}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{3{x^2}h + 3x{h^2} + {h^3}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0}$ $( 3x^2 + 3xh + h^2) = 3x^2$
View full question & answer→Question 292 Marks
Find the derivative of 99x at x = 100
AnswerHere $\frac{d}{{dx}}(99x) = 99$
$\therefore $ Derivative of 99x at x = 100 = 99
View full question & answer→Question 302 Marks
Find the derivative of the function $5 \sin x-6 \cos x+7$
View full question & answer→Question 312 Marks
Find the derivative of function $3 \cot x + 5\ cosec\ x$
AnswerHere $f (x) = 3 \cot x + 5\ cosec\ x$
$\therefore \;{\text{f'}}(x) = \frac{d}{{dx}}[3\cot x + 5\ {\text{cosec }}\ x]$
$= 3\frac{d}{{dx}}(\cot x) + 5\frac{d}{{dx}}(cosec\ x)$
$= - 3\ cosec^2\ x - 5\ cosec\ x \cot x$
View full question & answer→Question 322 Marks
Find the derivative of function sec x.
AnswerHere f (x) = sec x
$\therefore \;{\text{f'}}(x) = \frac{d}{{dx}}(\sec x)$
= sec x tan x
View full question & answer→Question 332 Marks
Find the derivative of function $\sin x \cos x.$
AnswerHere $f (x) = \sin x \cos x$
$\therefore \;f{\text{}}(x) = \frac{d}{{dx}}(\sin x\cos x)$
$= \sin x\frac{d}{{dx}}(\cos x) + \cos x\frac{d}{{dx}}(\sin x)$
$= \sin x (- \sin x) + \cos x(\cos x) = \cos^2x - \sin^2x = \cos 2 x$
View full question & answer→Question 342 Marks
Find the derivative of cos x from first principle.
AnswerHere f (x) = cos x
Then f (x + h) = cos (x + h)
We know that $f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h) - f(x)}}{h}$
$\Rightarrow f{\text{'}}(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\cos (x + h) - \cos x}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} - \sin \left( {\frac{{2x + h}}{2}} \right).\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}$
= -sin x
View full question & answer→Question 352 Marks
Find the derivative of $x^2 - 2$ at $x = 10$
AnswerHere $\frac{d}{{dx}}({x^2} - 2) = 2x - 0 = 2x$
$\therefore$ Derivative of $x^2 - 2$ at $x = 10 = 2 \times 10= 20$
View full question & answer→Question 362 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}} = \frac{{a \times 0 + b}}{{c \times 0 + 1}} = \frac{b}{1} = b$
View full question & answer→Question 372 Marks
Evaluate $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}$
AnswerHere $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}\left[ {\frac{0}{0}{\text{from}}} \right]$
$= \mathop {\lim }\limits_{x \to 3} \frac{{({x^2} + 9)(x + 3)(x - 3)}}{{(x - 3)(2x + 1)}}$
$= \mathop {\lim }\limits_{x \to 3} \frac{{({x^2} + 9)(x + 3)}}{{(2x + 1)}} = \frac{{({3^2} + 9)(3 + 3)}}{{(2 \times 3 + 1)}} = \frac{{108}}{7}$
View full question & answer→Question 382 Marks
Evaluate $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}$
AnswerHere $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}\left[ {\frac{0}{0}{\text{from}}} \right]$
$= \mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)(3x + 5)}}{{(x + 2)(x - 2)}}$
$= \mathop {\lim }\limits_{x \to 2} \frac{{3x - 5}}{{x + 2}} = \frac{{6 + 5}}{{2 + 2}} = \frac{{11}}{4}$
View full question & answer→Question 392 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{x}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{x}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{{(X + 1)}^5} - 1}}{{(x + 1) - 1}}$
Putting x + 1 = y, as $x \to 0,\;y \to 1$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{{y^5} - 1}}{{y - 1}} = 5.{(1)^{5 - 1}}$
$= 5 \times 1 = 5\left[ {\because \;\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n \cdot {a^{n - 1}}} \right]$
View full question & answer→Question 402 Marks
Evaluate $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}$
$ = \frac{{{{( - 1)}^{10}} + {{( - 1)}^5} + 1}}{{ - 1 - 1}} = \frac{{1 - 1 + 1}}{{ - 2}} = \frac{{ - 1}}{2}$
View full question & answer→Question 412 Marks
Evaluate $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}}$
AnswerHere $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}} = \frac{{4 \times 4 + 3}}{{4 - 2}} = \frac{{19}}{2}$
View full question & answer→Question 422 Marks
If $f(x) = \left\{ {\begin{array}{*{20}{c}} {m{x^2} + n,} \\ {nx + m,} \\ {n{x^3} + m,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {0 \le x \le 1} \\ {x > 1} \end{array}$ .For what integers m and n does both $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ exist?
AnswerIt is given that
$\mathop {\lim }\limits_{x \to 0} f(x) and \mathop {\lim }\limits_{x \to 1} f(x) $both exist.
$\Rightarrow \;\mathop {\lim }\limits_{x \to {0^ - }} f(x) and = \mathop {\lim }\limits_{x \to {0^ +}}f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)$
Now $\mathop {\lim }\limits_{x \to {0^- }} f(x) = \mathop {\lim }\limits_{x \to {0 }} (mx^2+ n) = n$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0 }} (nx + m) = m$
Now$ \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) \Rightarrow n = m ...(i)$
For $\mathop {\lim }\limits_{x \to 0} f(x)$ to exist we need $m = n$
Also, $\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1}} (nx + m) = n + m$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1}} (nx^3 + m) = n + m$
Now $\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x) \Rightarrow n + m = n + m$
Thus $\mathop {\lim }\limits_{x \to 0} f(x)$ exists for any integral value of $m$ and $n.$
View full question & answer→Question 432 Marks
If the function f(x) satisfies $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi ,$ then evaluate $\mathop {\lim }\limits_{x \to 1} f(x)$.
AnswerGiven, $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi \Rightarrow \frac{{\mathop {\lim }\limits_{x \to 1} [f(x) - 2]}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)}} = \pi $
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} [f(x) - 2] = \pi \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \left( {{1^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \times 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f(x) - 2 = 0$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) = 2$
View full question & answer→Question 442 Marks
If f(x) = $ \left\{ {\begin{array}{*{20}{c}} {|x| + 1} \\ 0 \\ {|x| - 1} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$ for what values of a does $\mathop {\lim }\limits_{x \to a}$ f(x) exists?
AnswerHere f(x) = $ \left\{ {\begin{array}{*{20}{c}} {|x| + 1} \\ 0 \\ {|x| - 1} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$
$\Rightarrow$ f(x) = $\left\{ {\begin{array}{*{20}{c}} { - x + 1,} \\ {0,} \\ {x - 1,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 0} \\ {x = 0} \\ {x > 0} \end{array}$
$\therefore \;\mathop {\lim }\limits_{x \to a}$ f(x) exists for all a $\ne$ 0
Now we see that $\mathop {\lim }\limits_{x \to 0}$ f(x) exist or not
L.H.L.= $\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ - }}$ (-x + 1) = 1
R.H.L =$\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0 }}$ (x - 1) = -1
$\therefore$ L.H.L at x = 0 $\ne$ R.H.L as x 0
Thus $\therefore \;\mathop {\lim }\limits_{x \to 0}$ f(x) does not exist.
View full question & answer→Question 452 Marks
Let $a_1, a_2, ....,a_n$ be fixed real numbers and define a function $f(x) = (x - a_1)(x - a_2)... (x - a_n). $ What is & $\mathop {\lim }\limits_{x \to {a_1}} f(x)$? For some $a \ne {a_1},\;{a_2}...{a_n},$ compute $\mathop {\lim }\limits_{x \to {a}} f(x)$
AnswerHere $f(x) = (x – a_1) (x – a_2)… (x – a_n)$
Now $\mathop {\lim }\limits_{x \to {a_1}} f(x) = \mathop {\lim }\limits_{x \to {a_1}} (x - {a_1})(x - {a_2})...(x - {a_n})$
$= (a_1 – a_1) (a_1 – a_2) … (a_1 – a_n)$
$= 0 \times (a_1 – a_2) … (a_1 – a_n) = 0.$
Also $\mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} (x - {a_1})(x - {a_2})...(x - {a_n})$
$= (a – a_1) (a – a_2) … (a – a_n)$
View full question & answer→Question 462 Marks
Suppose f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$ and if $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1) what are possible values of a and b?
AnswerHere f(x) = $ \left\{ {\begin{array}{*{20}{c}} {a + bx,} \\ {4,} \\ {b - ax,} \end{array}} \right.\begin{array}{*{20}{c}} {x < 1} \\ {x = 1} \\ {x > 1} \end{array}$
Also $\mathop {\lim }\limits_{x \to 1}$ f(x) = f(1)
$\Rightarrow \mathop {\lim }\limits_{x \to 1}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = f(1) = 4
$\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = 4 and $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = 4
Now $\mathop {\lim }\limits_{x \to {1^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ - }}$ (a + bx)
Put x = 1 - h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a + b(1 - h)] = $\mathop {\lim }\limits_{h \to 0}$ [a + b - bh] = a + b
Also $\mathop {\lim }\limits_{x \to {1^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {1^ + }}$ (b - ax)
Put x = 1 + h as x $\to$ 1, h $\to$ 0
$\therefore \;\mathop {\lim }\limits_{h \to 0}$ [a - b(1 + h)] = $\mathop {\lim }\limits_{h \to 0}$ [b - a - ah] = b - a
Putting values from (ii) and (iii) in (i)
$\therefore $ a + b = 4 and -a + b = 4
Solving these equations, we have
a = 0 and b = 4
View full question & answer→Question 472 Marks
Find $\mathop {\lim }\limits_{x \to 5} f(x)$, where f(x) = |x| - 5
AnswerHere f(x) = |x| - 5
L.H.L. $\mathop {\lim }\limits_{x \to 5} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} |x| - 5$
Put x = 5 – h as x → 5, h → 0
$\therefore \;\mathop {\lim }\limits_{h \to 0} \left| {5 - h} \right| - 5 = \mathop {\lim }\limits_{h \to 0} 5 - h - 5 = \mathop {\lim }\limits_{h \to 0} ( - h) = 0$
R.H.L. $= \mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} |x| - 5$
Put x = 5 + h as x → 5, h →0
$\therefore \;\mathop {\lim }\limits_{h \to 0} \left| {5 + h} \right| - 5 = \mathop {\lim }\limits_{h \to 0} 5 + h - 5 = \mathop {\lim }\limits_{h \to 0} h = 0$
Now L.H.L. = R.H.L
Thus limit exists at x = 5 and $\mathop {\lim }\limits_{x \to 0} f(x) = 0$
View full question & answer→Question 482 Marks
Find $\mathop {\lim }\limits_{x \to 0} f(x)$ where $f ( x ) = \left\{ \begin{array} { c c } { \frac { x } { | x | } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
AnswerHere $f ( x ) = \left\{ \begin{array} { c c } { \frac { x } { | x | } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
L.H.L. $ = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{|x|}}$
Put x = 0 – h as x → 0, h → 0
$\therefore \mathop {\lim }\limits_{h \to 0} \frac{{0 - h}}{{|0 - h|}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{| - h|}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h} = - 1$
R.H.L. $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{|x|}}$
Put x = 0 + h as $\mathrm { x } \rightarrow 0 , \mathrm { h } \rightarrow 0$
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{0 + h}}{{|0 + h|}} = \mathop {\lim }\limits_{n \to 0} \frac{h}{{|h|}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = 1$
Now L.H.L. $\neq$ R.H.L.
Thus limit does not exist at x = 0.
View full question & answer→Question 492 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0}$ f(x), where f(x) = $\left\{ \begin{array} { l l } { \frac { | x | } { x } , } & { x \neq 0 } \\ { 0 , } & { x = 0 } \end{array} \right.$
AnswerWe have, f(x) = $\left\{ {\begin{array}{*{20}{l}} {\frac{{|x|}}{x},}&{x \ne 0} \\ {0,}&{x = 0} \end{array}} \right.$
LHL = $\mathop {\lim }\limits_{x \to {0^ - }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 - h|}}{{(0 - h)}}$ [putting x = 0 - h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
= $\mathop {\lim }\limits_{h \to 0} \frac{{| - h|}}{{ - h}}$
= $\mathop {\lim }\limits_{h \to 0} \frac{{ h}}{{ - h}}$ [$\because$ |-x| = x]
= - 1
RHL = $\mathop {\lim }\limits_{x \to {0^ + }}$ f(x) = $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{|x|}}{x} = \mathop {\lim }\limits_{h \to 0} \frac{{|0 + h|}}{{(0 + h)}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h}$ = 1 [putting x = 0 + h as x $\rightarrow$ 0, then h $\rightarrow$ 0]
$\mathop {\lim }\limits_{h \to {0^ - }}$ f(x) $\ne \mathop {\lim }\limits_{x \to {0^ + }}$ f(x)
Limit does not exist at x = 0
View full question & answer→Question 502 Marks
Find $\mathop {\lim }\limits_{x \to 1} f(x)$ where f(x) = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$
AnswerHere $\mathop {\lim }\limits_{x \to 1} f(x)$ = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$
L.H.L. $\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} ({x^2} - 1)$
Put x = 1 - h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [{(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [1 + {h^2} - 1]$
R.H.L. $= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} ( - {x^2} - 1)$
Put x = 1 + h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [ - {(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [ - 1 - {h^2} - 2h - 1]$
$= - {(0)^2} - 2 \times 0 - 2 = - 2$
$\therefore$ LHL $\neq$ RHL
Therefore limit of given function does not exist
View full question & answer→Question 512 Marks
Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x)$ where $f(x) = \left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$
AnswerHere f(x) = $\left\{ {\begin{array}{*{20}{c}} {2x + 3}&{x \le 0} \\ {3(x + 1)}&{x > 0} \end{array}} \right.$
Now $\mathop {\lim }\limits_{x \to 0} f(x)$ = $\mathop {\lim }\limits_{x \to 0} 2x + 3$ = 2 $\times$ 0 + 3 = 3
$\mathop {\lim }\limits_{x \to 1} f(x)$ = $\mathop {\lim }\limits_{x \to 1} 3(x + 1) $ = 3(1 + 1) = 3 $\times$ 2 = 6
View full question & answer→Question 522 Marks
Evaluate $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$
AnswerHere $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \frac{\pi }{2} + y$ as $x \to \frac{\pi }{2},\;y \to 0$
$\therefore \;\mathop {\lim }\limits_{y \to 0} \frac{{\tan 2\left( {\frac{\pi }{2} + y} \right)}}{{\frac{\pi }{2} + y - \frac{\pi }{2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan (\pi + 2y)}}{y}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{\tan 2y}}{{2y}} \times 2 = 1 \times 2 = 2$
View full question & answer→Question 532 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$
AnswerHere $\mathop {\lim }\limits_{x \to 0} ({\text{cosec }}x - \cot x)$
$= \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right)$
$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x/2}}{{2\sin x/2\cos x/2}} = \mathop {\lim }\limits_{x \to 0} \tan x/2 = 0$
View full question & answer→Question 542 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}};$ a, b, a + b $\neq$ 0.
AnswerWe have,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{x} + \frac{{bx}}{x}}}{{\frac{{ax}}{x} + \frac{{\sin bx}}{x}}}$
[dividing both numerator and denominator by x]
$ = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin (ax)}}{{ax}} \times a + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + \mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}} \times b}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}{\text{ and }}} {\mathop {\lim }\limits_{x \to a} f(x) + g(x) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)} \right]$
$ = \frac{{a\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + b\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}}$
$= \frac { a ( 1 ) + b } { a + b ( 1 ) }$ $\left[ {\because \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]$
$= \frac { a + b } { a + b } = 1$
View full question & answer→Question 552 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} x\sec x$
AnswerHere $\mathop {\lim }\limits_{x \to 0} x\sec x$
$ = \mathop {\lim }\limits_{x \to 0} x \times \frac{1}{{\cos x}}\rightarrow\mathop {\lim }\limits_{x \to 0} \frac{x}{{\cos x}} = \frac{0}{1} = 0$
View full question & answer→Question 562 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}.$
AnswerWe have, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ax}}{{b\sin x}} + \frac{{x\cos x}}{{b\sin x}}} \right)$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sin x}} + \frac{1}{b}\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x}}{{\sin x}}$
$ = \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\mathop {\lim }\limits_{bx \to 0} \frac{{\cos x}}{{\left( {\frac{{\sin x}}{x}} \right)}}$
$ = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}} + \frac{1}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \cos x}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)}}$$\left[\because {\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$
$ = \frac{a}{b} \times \frac{1}{1} + \frac{1}{b} \times \frac{1}{1}$$\left[\because {\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
$= \frac { a + 1 } { b }$
View full question & answer→Question 572 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$$= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{1 - \cos x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{2{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{(2\sin x/2\cos x/2)}^2}}}{{{{\sin }^2}x/2}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}x/2{{\cos }^2}x/2}}{{{{\sin }^2}x/2}} = \mathop {\lim }\limits_{x \to 0} 4{\cos ^2}x/2$ = 4/2 = 2
View full question & answer→Question 582 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}} = \frac{1}{\pi }$
View full question & answer→Question 592 Marks
Evaluate $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}$
AnswerLet y= $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}\left[ {\frac{0}{0}{\text{from}}} \right]$
Put $x = \pi + y$, as $x \to \pi ,\;y \to 0$
$\therefore y=\;\mathop {\lim }\limits_{y \to 0} \frac{{\sin [\pi - \pi - y]}}{{\pi [\pi - \pi - y]}}=\mathop {\lim }\limits_{y \to 0} \frac{{\sin ( - y)}}{{ - \pi y}}$
$ = \mathop {\lim }\limits_{y \to 0} \frac{{ - \sin y}}{{ - \pi y}} = \frac{1}{\pi }\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = \frac{1}{\pi } \times1 = \frac{1}{\pi }$
View full question & answer→Question 602 Marks
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},\;a,\;b \ne 0$
AnswerHere $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}}$$= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times ax \times \frac{1}{{\frac{{\sin bx}}{{bx}} \times bx}}} \right]$
$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}} \times \frac{1}{{\frac{{sinbx}}{{bx}}}} \times \frac{{ax}}{{bx}}} \right] = \frac{a}{b}$$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin ax}}{{ax}}\frac{1}{{\frac{{\sin bx}}{{bx}}}}} \right]$
$ = \frac{a}{b} \times 1 \times 1 = \frac{a}{b}$
View full question & answer→Question 612 Marks
Evaluate $\lim \limits_{x \rightarrow 0} \frac{\sin a x}{b x}$
Answer Given, $\lim _{x \rightarrow 0} \frac{\sin a x}{b x}$
$$Applying the limits in the given expression we get,$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{0}{0}$
Multiplying and dividing the given expression by a we get,
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{b x} \times \frac{a}{a}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin a x}{a x} \times \frac{a}{b}$
We know that: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$= \frac{a}{b} \lim _{a x \rightarrow 0} \frac{\sin a x}{a x}=\frac{a}{b} \times 1=\frac{a}{b}$
View full question & answer→Question 622 Marks
Evaluate $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$
AnswerHere $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{\frac{{x + 2}}{{2x}}}}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{2x}} \times \frac{1}{{x + 2}}$
$ = \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2x}} = \frac{1}{{2 \times - 2}} = \frac{{ - 1}}{4}$
View full question & answer→Question 632 Marks
Evaluate $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}},\;a + b + c \ne 0$
AnswerHere $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}}$
$ = \frac{{a \times {{(1)}^2} + b \times 1 + c}}{{c \times {{(1)}^2} + b \times 1 + a}} = \frac{{a + b + c}}{{c + b + a}} = 1$
View full question & answer→Question 642 Marks
Evaluate $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}$
AnswerHere $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{1/3}} - 1}}{{{z^{1/6}} - 1}}\left[ {\frac{0}{0}{\text{form}}} \right]$
$= \mathop {\lim }\limits_{z \to 1} \frac{{{{({z^{1/6}} - 1)}^2}}}{{{z^{1/6}} - 1}}$
$ = \mathop {\lim }\limits_{z \to 1} \frac{{({z^{1/6}} + 1)({z^{1/6}} - 1)}}{{({z^{1/6}} - 1)}} = \mathop {\lim }\limits_{z \to 1}$
$= z^{1/6} + 1$
$= (1)^{1/6} + 1 = 1 + 1 = 2$
View full question & answer→