Limits, Continuity and Differentiability
Explanation:
$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$
For limit to exist
$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$ ..... (i)
$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$
Put in (i)
${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$
${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$
$ \Rightarrow \alpha = 6$
$\alpha$ = 6, $\beta$ = 6, $\gamma$ = $-$9
$\alpha$ + $\beta$ + $\gamma$ = 3
Explanation:
form : 1$\infty$ $ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$
Now, $\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$ (by L' Hospital Rule)
$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$
So, ${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$
$ = {e^{{3 \over 2} \times 2}} = {e^3}$
$\Rightarrow$ a = 3
$\mathop {\lim }\limits_{h \to 0} {1 \over h}(f(h) - 1)$ is equal to ____________.
Explanation:
Given, $f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$
$\therefore$ $f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$
Now, $f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$
$\therefore$ $f(x) = {({e^3})^x} = {e^{3x}}$
$\therefore$ $f(h) = {e^{3h}}$
Now, $\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$
$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$
Explanation:
$ \Rightarrow $ ${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{x + \sin x} \over 2}} \right)} \over {\left( {{{x + \sin x} \over 2}} \right)}} \times {{\sin \left( {{{x - \sin x} \over 2}} \right)} \over {\left( {{{x - \sin x} \over 2}} \right)}} \times {{{x^2} - {{\sin }^2}x} \over {4{x^4}}}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{x + \sin x} \over x}} \right)\left( {{{x - \sin x} \over {{x^3}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + \cos x} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 - \cos x} \over {3{x^2}}}} \right) \times {1 \over 4}$
= $\mathop {\lim }\limits_{x \to 0} 2 \times 1 \times \left( {{{1 + 1} \over 1}} \right)\left( {{{1 + \sin x} \over {6x}}} \right) \times {1 \over 4}$
= $2 \times 2 \times {1 \over 6} \times {1 \over 4}$ = ${1 \over 6}$
$f(x) = \left\{ {\matrix{ {x + a,} & {x < 0} \cr {|x - 1|,} & {x \ge 0} \cr } } \right.$ and
$g(x) = \left\{ {\matrix{ {x + 1,} & {x < 0} \cr {{{(x - 1)}^2} + b,} & {x \ge 0} \cr } } \right.$,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $\in$ R, then a + b is equal to ____________.
Explanation:
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x + a < 0\& x < 0} \cr {|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr {{{(x + a - 1)}^2} + b} & {x + a \ge 0\& x < 0} \cr {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \ge 0\& x \ge 0} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)\& x \in ( - \infty ,0)} \cr {|x - 1| + 1} & {x \in \phi } \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,\infty )\& x \in [0,\infty )} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in R\& x \in [0,\infty )} \cr } } \right.$
$g[f(x)] = \left[ {\matrix{ {x + a + 1} & {x \in ( - \infty , - a)} \cr {{{(x + a - 1)}^2} + b} & {x \in [ - a,0)} \cr {{{(|x - 1| - 1)}^2} + b} & {x \in [0,\infty )} \cr } } \right.$
g(f(x)) is continuous.
At x = $-$a
-a + a + 1 = (-a + a - 1)2 + b
$ \Rightarrow $ 1 = b + 1
$ \Rightarrow $ b = 0
at x = 0
(a $-$1)2 + b = (|0 - 1| - 1)2 + b
$ \Rightarrow $ (a $-$1)2 + b = b
$ \Rightarrow $ a = 1
$ \Rightarrow $ a + b = 1
Explanation:
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{(a - b + c) + x(a - c) + {x^2}\left( {{a \over 2} + {b \over 2} + {c \over 2}} \right) + ....} \over {{x^2}\left( {1 - {{{x^2}} \over 6}....} \right)}} = 2$
For this limit to exist
a $-$ b + c = 0 & a $-$ c = 0
& ${a \over 2} + {b \over 2} + {c \over 2} = 2$
$ \Rightarrow $ a + b + c = 4
$f(x) = \left\{ {\matrix{ {\min \{ |x|,2 - {x^2}\} ,} & { - 2 \le x \le 2} \cr {[|x|],} & {2 < |x| \le 3} \cr } } \right.$ where [x] denotes the greatest integer $ \le $ x. The number of points, where f is not differentiable in ($-$3, 3) is ___________.
Explanation:
Points of non-differentiability in ($-$3, 3) are at x = $-$2, $-$1, 0, 1, 2.
i.e. 5 points.
Explanation:
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$
This is ${{a - 4} \over 0}$.
limit exist only when $a - 4 = 0$ $ \Rightarrow $ a = 4
Applying L' Hospital Rule
$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$
= ${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$
$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$
f(x) = | 2x + 1 | $-$ 3| x + 2 | + | x2 + x $-$ 2 |, x$\in$R is not differentiable, is __________.
Explanation:
$f(x) = \left\{ {\matrix{ {{x^2} - 7;} & {x > 1} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr {{x^2} + 2x + 3;} & {x < - 2} \cr } } \right.$
$ \therefore $ $f'(x) = \left\{ {\matrix{ {2x;} & {x > 1} \cr {2x - 3;} & { - {1 \over 2} < x < 1} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr {2x + 2;} & {x < - 2} \cr } } \right.$
Check at 1, $-$2 and ${{ - 1} \over 2}$
Non. differentiable at x = 1 and ${{ - 1} \over 2}$
Explanation:
$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$
$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$
$ \therefore $ $\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $
$ = {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)$
$ = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)$
$ = {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)$
$ = {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)$
$\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)$
$ = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)$
$ = \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}$
$ = 1$
$f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right.$
The value of $\lambda $ for which f ''(0) exists, is _______.
Explanation:
and h(x) = x5cos$\left( {{1 \over x}} \right)$
then g''(0) = 0 and h''(0) = 0
So, f''(0+ ) = g''(0+ ) + 10 = 10
and f''(0–) = h''(0–) + 2$\lambda $ = f''(0+)
$ \Rightarrow $ 2$\lambda $ = 10
$ \Rightarrow $ $\lambda $ = 5
Explanation:
$ \Rightarrow $ ${x \over 2} \in ( - 5,5) \to 9$ integers
check continuity at x = 0
$\left. {\matrix{ f & {(0) = } & 0 \cr f & {({0^ + }) = } & 0 \cr f & {({0^ - }) = } & 0 \cr } } \right\}continuous\,at\,x = 0$
function will be discontinuous when
${x \over 2} = \pm 4, \pm 3, \pm 2, \pm 1$
For example checking continuity at x = 4
$\left. {\matrix{ f & {(4) = } & 4 \cr f & {({4^ + }) = } & 4 \cr f & {({4^ - }) = } & 3 \cr } } \right\}discontinuous\,at\,x = 4$
$ \therefore $ 8 points of discontinuity.
f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y.
$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$, then f'(3) is equal to ______.
Explanation:
differentiating partially with respect to x,
f'(x+y) = f'(x) + 0 + y2 + y(2x) [y = constant]
Put x = 0 and y = x
$ \therefore $ f'(x) = f'(0) + x2 ....(2)
putting x = y = 0 at equation (1),
f(0) = 2f(0)
$ \Rightarrow $ f(0) = 0
Given, $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$
This is in $ \frac{0}{0} $ form, so we can apply L' hospital rule.
$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$
$ \Rightarrow f'(0) = 1$
Putting value of f'(0) at equation (2), we get
f'(x) = 1 + x2
$ \therefore $ f'(3) = 1 + 32 = 10
then the value of k is _______ .
Explanation:
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $ = ${2^{ - k}}$
$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$ = 2-k
$ \Rightarrow $ $ {1 \over 8} \times {1 \over {32}} = {2^{ - k}}$
$ \Rightarrow $ 2-8 = 2-k
$ \Rightarrow $ k = 8
(n $ \in $ N) then the value of n is equal to _______.
Explanation:
As it is $\left( {{0 \over 0}} \right)$ form, Apply L'Hospital's Rule.
$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$ = 820
$ \Rightarrow $ 1 + 2 + 3 + .....+ n = 820
$ \Rightarrow $ ${{n\left( {n + 1} \right)} \over 2}$ = 820
$ \Rightarrow $ n2 + n – 1640 = 0
$ \Rightarrow $ (n – 40)(n + 41) = 0
Since n $ \in $ N, so n = 40.
f(x) = $\left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right.$
is continuous, then k is equal to_______.
Explanation:
= $\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$
= $\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$
= 3 + 2 = 5
f(x) is continuous
$ \therefore $ $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ = f(0)
So f(0) = 5 = k
Explanation:
f(x) is non-differentiable at x = 1, 3, 5
$ \therefore $ S is {1, 3, 5}
$\sum\limits_{x \in S} {f(f(x))} $
= f(f(1)) + f(f(3)) + f(f (5))
= f(0) + f(2) + f(0)
= 1 + 1 + 1 = 3
Explanation:
let 3x/2 = t
= $\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$
= $\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} - 9} \right)\left( {{t^2} - 3} \right)} \over {t - 3}}$
= $\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)\left( {{t^2} - 3} \right)$
= 6 $ \times $ 6
= 36
Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$. For a given real number $\alpha$, define
$g(x) = \frac{3x e^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$
for all real numbers $x$.
Then which one of the following statements is TRUE?
For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$
For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$
For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$
For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$
Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f : \mathbb{R} \to \mathbb{R}$ by
$f(x)=\left\{\begin{array}{cc}2-2 x^2-x^2 \sin \frac{1}{x} & \text { if } x \neq 0, \\ 2 & \text { if } x=0 .\end{array}\right.$
Then which one of the following statements is TRUE?
The function $f$ is NOT differentiable at $x = 0$
There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$
For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$
$x = 0$ is a point of local minima of $f$
Let $\mathbb{R}$ denote the set of all real numbers. For a real number $x$, let [ x ] denote the greatest integer less than or equal to $x$. Let $n$ denote a natural number.
Match each entry in List-I to the correct entry in List-II and choose the correct option.
| List–I | List–II |
|---|---|
| (P) The minimum value of $n$ for which the function $ f(x)=\left[\frac{10 x^3-45 x^2+60 x+35}{n}\right] $ is continuous on the interval $[1,2]$, is | (1) 8 |
| (Q) The minimum value of $n$ for which $g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right)$, $x \in \mathbb{R}$, is an increasing function on $\mathbb{R}$, is | (2) 9 |
| (R) The smallest natural number $n$ which is greater than 5 , such that $x=3$ is a point of local minima of $ h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right) $ is | (3) 5 |
| (S) Number of $x_0 \in \mathbb{R}$ such that
$ l(x)=\sum\limits_{k=0}^4\left(\sin |x-k|+\cos \left|x-k+\frac{1}{2}\right|\right) $ $x \in \mathbb{R}$, is NOT differentiable at $x_0$, is |
(4) 6 |
| (5) 10 |
(P) → (1) (Q) → (3) (R) → (2) (S) → (5)
(P) → (2) (Q) → (1) (R) → (4) (S) → (3)
(P) → (5) (Q) → (1) (R) → (4) (S) → (3)
(P) → (2) (Q) → (3) (R) → (1) (S) → (5)
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by
$ f(x)=\left\{\begin{array}{cc} x^2 \sin \left(\frac{\pi}{x^2}\right), & \text { if } x \neq 0, \\ 0, & \text { if } x=0 . \end{array}\right. $
Then which of the following statements is TRUE?
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be functions defined by
$ f(x)=\left\{\begin{array}{ll} x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0, \end{array} \quad \text { and } g(x)= \begin{cases}1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise } .\end{cases}\right. $
Let $a, b, c, d \in \mathbb{R}$. Define the function $h: \mathbb{R} \rightarrow \mathbb{R}$ by
$ h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in \mathbb{R} . $
Match each entry in List-I to the correct entry in List-II.
| List-I | List-II |
|---|---|
| (P) If $a = 0$, $b = 1$, $c = 0$, and $d = 0$, then | (1) $h$ is one-one. |
| (Q) If $a = 1$, $b = 0$, $c = 0$, and $d = 0$, then | (2) $h$ is onto. |
| (R) If $a = 0$, $b = 0$, $c = 1$, and $d = 0$, then | (3) $h$ is differentiable on $\mathbb{R}$. |
| (S) If $a = 0$, $b = 0$, $c = 0$, and $d = 1$, then | (4) the range of $h$ is $[0, 1]$. |
| (5) the range of $h$ is $\{0, 1\}$. |
The correct option is
$ f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} . $
Then, the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$ is equal to :
(i) ${f_1}(x) = \sin (\sqrt {1 - {e^{ - {x^2}}}} )$,
(ii) ${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {\tan { - ^1}x}}if\,x \ne 0,\,where \hfill \cr 1\,if\,x = 0 \hfill \cr} \right.$
the inverse trigonometric function tan$-$1x assumes values in $\left( { - {\pi \over 2},{\pi \over 2}} \right)$,
(iii) ${f_3}(x) = [\sin ({\log _e}(x + 2))]$, where for $t \in R,\,[t]$ denotes the greatest integer less than or equal to t,
(iv) ${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right)\,if\,x \ne 0 \hfill \cr 0\,if\,x = 0 \hfill \cr} \right.$
| LIST-I | LIST-II |
|---|---|
| P. The function $ f_1 $ is | 1. NOT continuous at $ x = 0 $ |
| Q. The function $ f_2 $ is | 2. continuous at $ x = 0 $ and NOT differentiable at $ x = 0 $ |
| R. The function $ f_3 $ is | 3. differentiable at $ x = 0 $ and its derivative is NOT continuous at $ x = 0 $ |
| S. The function $ f_4 $ is | 4. differentiable at $ x = 0 $ and its derivative is continuous at $ x = 0 $ |
If $\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + x + 1} \over {x + 1}} - ax - b} \right) = 4$, then
Let $f(x) = \left\{ {\matrix{ {{x^2}\left| {\cos {\pi \over x}} \right|,} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$
x$\in$R, then f is
If $\mathop {\lim }\limits_{x \to 0} {[1 + x\ln (1 + {b^2})]^{1/x}} = 2b{\sin ^2}\theta $, $b > 0$ and $\theta \in ( - \pi ,\pi ]$, then the value of $\theta$ is
Which of the following is true?
$g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}.$ Then, $g$ is
Let $g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}};0 < x < 2,m$ and $n$ are integers, $m \ne 0,n > 0$, and let $p$ be the left hand derivative of $|x - 1|$ at $x = 1$. If $\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p$, then
Let $f(x)=2+\cos x$ for all real $x$.
STATEMENT - 1 : For each real $t$, there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(C)=0$.
STATEMENT - 2 : $f(t)=f(t+2 \pi)$ for each real $t$.
The line $y=x$ meets $y=k e^{\mathrm{x}}$ for $k \leq 0$ at
The positive value of $k$ for which $k e^{x}-x=0$ has only one root is
For $k > 0$, the set of all values of $k$ for which $k e^{x}-x=0$ has two distinct roots is
Let $f(x) = {{{x^2} - 6x + 5} \over {{x^2} - 5x + 6}}$.
Match the conditions/expressions in Column I with statements in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | If $ - 1 < x < 1$, then $f(x)$ satisfies | (P) | $0 < f(x) < 1$ |
| (B) | If $1 < x < 2$, then $f(x)$ satisfies | (Q) | $f(x) < 0$ |
| (C) | If $3 < x < 5$, then $f(x)$ satisfies | (R) | $f(x) > 0$ |
| (D) | If $x > 5$, then $f(x)$ satisfies | (S) | $f(x) < 1$ |
In the following [x] denotes the greatest integer less than or equal to x.
Match the functions in Column I with the properties Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $x|x|$ | (P) | continuous in ($-1,1$). |
| (B) | $\sqrt{|x|}$ | (Q) | differentiable in ($-1,1$) |
| (C) | $x+[x]$ | (R) | strictly increasing in ($-1,1$) |
| (D) | $|x-1|+|x+1|$ | (S) | not differentiable at least at one point in ($-1,1$) |
For $x>0, \mathop {\lim }\limits_{x \to 0}\left((\sin x)^{1 / x}+(1 / x)^{\sin x}\right)$ is :
0
-1
1
2
If $f(x-y)=f(x) \circ g(y)-f(y) \circ g(x)$ And $g(x-y) =g(x) \circ g(y)+f(x) \circ f(y)$ for all $x, y \in \mathrm{R}$. If right-hand derivative at $x=0$ exists for $f(x)$, find the derivative of $g(x)$ at $x=0$
Let α and β be the real numbers such that
$ \lim\limits_{x \to 0} \frac{1}{x^3} \left( \frac{\alpha}{2} \int\limits_0^x \frac{1}{1-t^2} \, dt + \beta x \cos x \right) = 2. $
Then the value of α + β is ___________.
$ \beta=\lim \limits_{x \to 0} \frac{e^{x^{3}}-\left(1-x^{3}\right)^{\frac{1}{3}}+\left(\left(1-x^{2}\right)^{\frac{1}{2}}-1\right) \sin x}{x \sin ^{2} x}, $
then the value of $6 \beta$ is ___________.
Explanation:
Given,
$\beta = \mathop {\lim }\limits_{x \to 0} {{{e^{{x^3}}} - {{(1 - {x^3})}^{{1 \over 3}}} + ({{(1 - {x^2})}^{{1 \over 2}}} - 1)\sin x} \over {x{{\sin }^2}x}}$
$ = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^3}\, + \,...) - \left( {1 - {{{x^3}} \over 3}\, + \,...} \right) + \left( {\left( {1 - {1 \over 2}{x^2}\, + \,...} \right) - 1} \right)x} \over {x\,.\,{{{{\sin }^2}x} \over {{x^2}}}\,.\,{x^2}}}$
$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{x^3} + {{{x^3}} \over 3} - {{{x^3}} \over 2}} \right) + {x^4}(......)} \over {{x^3}}}$
$ = \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {1 + {1 \over 3} - {1 \over 2}} \right)} \over {{x^3}}}$ [Neglecting higher power of x]
$ = 1 + {1 \over 3} - {1 \over 2} = {5 \over 6}$
$\therefore$ $6\beta = 6 \times {5 \over 6} = 5$
$ f(x)=\sin \left(\frac{\pi x}{12}\right) \quad \text { and } \quad g(x)=\frac{2 \log _{\mathrm{e}}(\sqrt{x}-\sqrt{\alpha})}{\log _{\mathrm{e}}\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)} . $
Then the value of $\lim \limits_{x \rightarrow \alpha^{+}} f(g(x))$ is
Explanation:
$\mathop {\lim }\limits_{x \to {\alpha ^ + }} f(g(x)) = f\left( {\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)} \right)$ [As $f(x)$ is continuous function so we can write this]
Now,
$\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left( {{e^{\sqrt x }} - {e^{\sqrt \alpha }}} \right)}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left( {{e^{\sqrt \alpha }}\left( {{{{e^{\sqrt x }}} \over {{e^{\sqrt \alpha }}}} - 1} \right)} \right)}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}\left[ {{e^{\sqrt \alpha }}({e^{\sqrt x - \sqrt \alpha }} - 1)} \right]}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {{{\log }_e}{e^{\sqrt \alpha }} + {{\log }_e}({e^{\sqrt x - \sqrt \alpha }} - 1)}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}\left[ {{{{e^{\sqrt x - \sqrt \alpha }} - 1} \over {(\sqrt x - \sqrt \alpha )}} \times (\sqrt x - \sqrt \alpha )} \right]}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}\left[ {\mathop {\lim }\limits_{x \to {\alpha ^ + }} \left( {{{{e^{\sqrt x - \sqrt \alpha }} - 1} \over {(\sqrt x - \sqrt \alpha )}} \times (\sqrt x - \sqrt \alpha )} \right)} \right]}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}(1 \times \sqrt x - \sqrt \alpha )}}$ [using $\mathop {\lim }\limits_{x \to 0} {{{e^2} - 1} \over x} = 1$]
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {{2{{\log }_e}(\sqrt x - \sqrt \alpha )} \over {\sqrt \alpha + {{\log }_e}(\sqrt x - \sqrt \alpha )}}$
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {2 \over {{{\sqrt \alpha } \over {{{\log }_e}(\sqrt x - \sqrt \alpha )}} + 1}}$
From graph you can see $\log _e^{{0^ + }} \to \, - \alpha $
$\therefore$ $\mathop {\lim }\limits_{x \to {\alpha ^ + }} {\log _e}(\sqrt x - \sqrt \alpha ) = \log _e^{{0^ + }} = - \alpha $
$ = \mathop {\lim }\limits_{x \to {\alpha ^ + }} {2 \over {{{\sqrt \alpha } \over { - \alpha }} + 1}}$
$ = {2 \over {0 + 1}}$
$ = 2$
$\therefore$ $f\left( {\mathop {\lim }\limits_{x \to {\alpha ^ + }} g(x)} \right) = f(2) = \sin \left( {{{\pi \times 2} \over {12}}} \right) = \sin {\pi \over 6} = {1 \over 2}$
Explanation:
As, we know the composite function (fog) (x) is discontinuous at the points, where g(x) is discontinuous for given domain. And, since g(x) is discontinuous at x = 0 lies in interval ($-$1, 1), so value of c = 1.
And, since (fog) (x) is not differentiable at the point where g(x) is not differentiable as well as at those points also where g(x) attains the values so that f(g(x)) is non-differentiable.
Since g(x) is not continuous at x = 0$ \in $ ($-$1, 1) so fog(x) is not differentiable and as $f(x) = |2x - 1| + |2x + 1|$ is not differentiable at x = $-$1/2 and 1/2, so (fog) (x) is not differentiable for those x, for which g(x) = $-$1/2 or 1/2.
But g(x) $ \ge $ 0, so g(x) can be 1/2 only and for x = $-$1/2 and 1/2, g(x) = ${1 \over 2}$.
So, (fog) (x) is not differentiable at x = $-$1/2, 0, 1/2, therefore value of d = 3
$ \therefore $ c + d = 1 + 3 = 4.
$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}$
is ...........
Explanation:
$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (2\sin 2x\cos x)} \over {2\sin 2x\sin {{3x} \over 2} + \left( {\cos {{5x} \over 2} - \cos {{3x} \over 2}} \right) - \sqrt 2 (1 + cos2x)}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{8\sqrt 2 \sin 2x\cos x} \over {2\sin 2x\sin {{3x} \over 2} - 2\sin 2x\sin {x \over 2} - \sqrt 2 (2{{\cos }^2}x)}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x\cos x} \over {\sin 2x\left( {\sin {{3x} \over 2} - \sin {x \over 2}} \right) - \sqrt 2 {{\cos }^2}x}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x\cos x} \over {2\sin 2x\cos x\sin {x \over 2} - \sqrt 2 {{\cos }^2}x}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 \sin 2x} \over {2\sin 2x\sin {x \over 2} - \sqrt 2 \cos x}}$
$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{8\sqrt 2 \sin x} \over {4\sin x\sin {x \over 2} - \sqrt 2 }}$
$ = {{8\sqrt 2 } \over {{4 \over {\sqrt 2 }} - \sqrt 2 }} = {{16} \over {4 - 2}} = 8$
$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$
is equal to a non-zero real number, is .............
Explanation:
$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$
$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left\{ {{1 \over x}{{\log }_e}(1 - x)} \right\}}} - {e^{ - 1}}} \over {{x^a}}}$
$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{{1 \over x}\left( { - x - {{{x^2}} \over 2} - {{{x^3}} \over 3} - ...} \right)}} - {e^{ - 1}}} \over {{x^a}}}$
$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ - 1}}.{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - {e^{ - 1}}} \over {{x^a}}}$
$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{ \left( {-{x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ....} \right)}} - 1} \over {{x^a}}}$
$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{{e^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}} - 1} \over {^{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)}}} \times {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$
$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} {{\left( { - {x \over 2} - {{{x^2}} \over 3} - {{{x^3}} \over 4} - ...} \right)} \over {{x^a}}}$
$ = {e^{ - 1}}\mathop {\lim }\limits_{x \to {0^ + }} \left( { - {1 \over 2}{x^{1 - a}} - {1 \over 3}{x^{2 - a}} - ...} \right)$
The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is
$ = {e^{ - 1}}\left( { - {1 \over 2}} \right) = - {1 \over {2e}}$
Explanation:
$ = {({\log _2}9)^{{{2.{{\log }_2}} \over {{{\log }_2}9}}}} \times {7^{{1 \over 2}.{{\log }_7}4}}$
$ = {({\log _2}9)^{{{\log }_{\log {2^{{9^{{2^2}}}}}}}}} \times {7^{{{\log }_7}2}}$
$ = {2^2} \times 2 = 8$
If $g(x) = \int\limits_x^{\pi /2} {[f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt} $
for $x \in \left( {0,\,{\pi \over 2}} \right]$, then $\mathop {\lim }\limits_{x \to 0} g(x)$ =
Explanation:
Let $g(x) = \int\limits_x^{\pi /2} {[f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt} $
$ = \int\limits_x^{\pi /2} {{d \over {dt}}(f(t)\cos ect))} $
So, $g(x) = f(\pi /2)\cos ec{\pi \over 2} - f(x)\cos ecx$
$ = 3 - f(x)\cos ecx$
$\therefore$ $g(x) = 3 - {{f(x)} \over {\sin x}}$
$\mathop {\lim }\limits_{x \to 0} g(x) = 3 - \mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}$
As the above is a 0/0 form, use L'Hospital's rule to get
$\mathop {\lim }\limits_{x \to 0} g(x) = 3 - \mathop {\lim }\limits_{x \to 0} {{f'(x)} \over {\cos x}} = 3 - f'(0) = 3 - 1 = 2$
Let $\alpha$, $\beta$ $\in$ R be such that $\mathop {\lim }\limits_{x \to 0} {{{x^2}\sin (\beta x)} \over {\alpha x - \sin x}} = 1$. Then 6($\alpha$ + $\beta$) equals _________.
Explanation:
Here, $\mathop {\lim }\limits_{x \to 0} {{{x^2}\sin (\beta x)} \over {\alpha x - \sin x}} = 1$
$\mathop {\lim }\limits_{x \to 0} {{{x^2}\left( {\beta x - {{{{(\beta x)}^3}} \over {3!}} + {{{{(\beta x)}^5}} \over {5!}} - ....} \right)} \over {\alpha x - \left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}} = 1$
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {(\alpha - 1)x + {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ...}} = 1$
Limit exists only, when $\alpha$ $-$ 1 = 0
$\Rightarrow$ $\alpha$ = 1 ...... (i)
$\therefore$ $ \Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {{x^3}\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - ...} \right)}} = 1$
$\Rightarrow$ 6$\beta$ = 1 ....... (ii)
From Eqs. (i) and (ii), we get
6($\alpha$ + $\beta$) = 6$\alpha$ + 6$\beta$
= 6 + 1
= 7