Limits, Continuity and Differentiability

$ \mathrm{e}^{\mathrm{x}_0}+\mathrm{x}_0=0 $

$ \begin{aligned} & g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\ & g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \end{aligned} $

$ g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) $

$\Rightarrow$ As very clear $\mathrm{g}\left(\mathrm{x}_0\right)+\mathrm{x}_0 \Rightarrow 0$

$ \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim \limits_{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right| $

$ \mathrm{g}^{\prime}(\mathrm{x})=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+1\right)-\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right) \mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}+1\right)^2}\right) $

$ \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}{\left(\mathrm{e}^{\mathrm{x}_0}+1\right)^2}\right)=1-\frac{\alpha}{3} $

$ \Rightarrow \lim \limits_{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|1-\frac{\alpha}{3}\right| $

The function $ f : \mathbb{R} \to \mathbb{R} $ is defined by:

$ f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x}, & \text{if } x \neq 0, \\ 2, & \text{if } x = 0. \end{cases} $

Explanation:

(A) To determine differentiability at $ x = 0 $, consider the right-hand derivative (RHD):

$ \lim\limits_{h \rightarrow 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = 0 $

Similarly, for the left-hand derivative (LHD) at $ x = 0 $, we also find it equals 0. Therefore, $ f $ is differentiable at $ x = 0 $.

(B) For $ x \in (0, \delta) $, the derivative of $ f $ is:

$ f^{\prime}(x) = -\left(4x + 2x \sin \frac{1}{x}\right) + \cos \frac{1}{x} $

Thus:

$ f^{\prime}(x) = -\left(2x\left(4 - \sin \frac{1}{x}\right)\right) + \cos \frac{1}{x} $

Since $\cos \frac{1}{x}$ oscillates, we cannot assert that $ f(x) $ is strictly decreasing on $ (0, \delta) $.

(C) For $ x \in (-\delta, 0) $ and any $ \delta > 0 $, $ f(x) $ is not increasing since $\cos \frac{1}{x}$ oscillates between -1 and 1.

(D) Evaluating at $ x = 0 $:

$ f(0) = 2 $

For small $ h $:

$ f(0 + h) < 2 \quad \text{and} \quad f(0 - h) < 2 $

Therefore, $ x = 0 $ is a local maximum, not a local minimum.

$\begin{aligned} & \text { (P) Let } g(x)=10 x^3-45 x^2+60 x+35 \\\\ & g^{\prime}(x)=30\left(x^2-3 x+2\right)=30(x-1)(x-2)\end{aligned}$

$\Rightarrow g(x)$ is decreasing in $[1,2]$

Now, $f(1)=\left[\frac{g(1)}{n}\right]=\left[\frac{60}{n}\right], f(2)=\left[\frac{55}{n}\right]$

For $f(x)$ to be continuous in $[1,2]$ its integral value should remain same in whole interval for $n=9, f(1)=6, f(2)=6$

$(P) \rightarrow(2)$

(Q)

$ \begin{aligned} & g^{\prime}(x)=\left(2 n^2-13 n-15\right)\left(3 x^2+3\right) \\ & =(2 n-15)(n+1)\left(3 x^2+3\right)>0 \end{aligned} $

for $g(x)$ to be increasing $\min n=8$

$\Rightarrow Q \rightarrow 1$ matching

(R) $h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right)$

$h(3)=0$ for $n>5$

at $n=6 \Rightarrow n(3+\delta)>h(3)$

$ n(3-\delta)>n(3) $

$\Rightarrow h(x)$ has local minima at $x=3$ for $n=6$

$R \rightarrow 4$ matching

(S) $\quad I(x)=\sin |x|+\cos \left|x+\frac{1}{2}\right|+\sin |x-1|+\cos \left|x-\frac{1}{2}\right|$ $+\ldots \ldots+\sin |x-4|+\cos \left|x-\frac{7}{2}\right|$

as $\sin |x-a|$ is non differentiable at $x=a$

but $\cos |x-a|$ remains differentiable at $x=a$

$\Rightarrow$ given $I(x)$ is non-differentiable at

$\begin{aligned} & x_0=0,1,2,3,4(5 \text { points }) \\\\ & (S) \rightarrow(3) \text { matching } \\\\ & (P) \rightarrow(2)(Q) \rightarrow(1)(R) \rightarrow(4)(S) \rightarrow(3)\end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1)=6 \\ & \lim _{x \rightarrow 0} \frac{\sin (\sin k x) \cdot \sin k x}{(\sin k x) k x} \cdot k+1-\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} \cdot x=3 \\ & k+1=3 \Rightarrow k=2 \end{aligned}$

Option-A : $\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}$

$\mathrm{f}(\mathrm{x})=0 \Rightarrow \sin \frac{\pi}{\mathrm{x}^2}=0 \Rightarrow \frac{\pi}{\mathrm{x}^2}=\mathrm{n} \pi, \mathrm{n} \in \mathrm{N}$

$\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}$

$\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}$, finite number of solutions

Option-B : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n}<\pi^2$, Number of solutions is 9

Option-C : $\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}$, Infinite number of solutions

Option-D : $\frac{1}{\pi^2}<\frac{1}{\sqrt{\mathrm{n}}}<\frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)$, Definitely more than 25 solutions

$f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; \quad x=0 \end{array} \quad g(x)=\left\{\begin{array}{cc} 1-2 x & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.\right.$

$g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{cc} 2 x & ; \quad 0 \leq \frac{1}{2}-x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}=\left\{\begin{array}{cc} 2 x ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right\}\right.$

$g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{ll} 1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 & ; \text { otherwise } \end{array}\right\}$

(P) $\quad$ Now $a=0, b=1, c=0, d=0$

$\because h(x)=g(x)+g\left(\frac{1}{2}-x\right)= \begin{cases}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{cases}$

Hence Range of h(x) is {0, 1}

(Q) $\mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=0$

$h(x)=f(x)=\left\{\begin{array}{cc} x|x| \sin \frac{1}{x} & ; x \neq 0 \\ 0 & ; x=0 \end{array}\right.$

$\begin{aligned} & R H D=\lim _{x \rightarrow 0} \frac{x^2 \sin \frac{1}{x}-0}{x}=0 \\ & \text { LHD }=\lim _{x \rightarrow 0} \frac{-x^2 \sin \frac{1}{x}-0}{x}=0 \end{aligned}$

Hence $h(x)$ is differentiable on $R$

(R) $\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=1, \mathrm{~d}=0$

$h(x)=x-g(x)=\left\{\begin{array}{cl} 3 x-1 & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

$\therefore \mathrm{h}(\mathrm{x}) \text { is ONTO }$

(S) $a=0, b=0, c=0, d=1$

$h(x)=g(x)=\left\{\begin{array}{cl} 1-2 x & ; 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise } \end{array}\right.$

Range of h(x) is [0, 1]

$ \begin{aligned} & 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}}+\ldots .+\frac{25 n-7 n^{2}}{7 n^{2}} \\\\ & =\left(\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+1\right)+\left(\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+1\right)+\ldots \ldots+\left(\frac{25 n-7 n^{2}}{7 n^{2}}+1\right) \\\\ & f(n)=\frac{9 n+16}{4 n+3 n^{2}}+\frac{9 n+32}{8 n+3 n^{2}}+\ldots . .+\frac{25 n}{7 n^{2}} \\\\ & =\sum_{r=1}^{n} \frac{9 n+16 r}{4 r n+3 n^{2}}=\frac{1}{n} \sum_{r=1}^{n} \frac{9+16\left(\frac{r}{n}\right)}{4\left(\frac{r}{n}\right)+3} \\\\ & \lim _{n \rightarrow \infty} f(n)=\int_{0}^{1} \frac{9+16 x}{4 x+3} d x \\\\ & =\int_{0}^{1} \frac{(16 x+12)-3}{4 x+3} d x \\\\ & =\left.\left(4 x-\frac{3}{4} \ln |4 x+3|\right)\right|_{0} ^{1} \\\\ & =4-\frac{3}{4} \ln \frac{7}{3} \end{aligned} $

(i) Given,

f₁ : R $ \to $ R and f₁(x) = sin$(\sqrt {1 - {e^{ - {x^2}}}} )$

$ \therefore $ f₁(x) is continuous at x = 0

Now,

${f_1}'(x) = \cos \sqrt {1 - {e^{ - {x^2}}}} .{1 \over {2\sqrt {1 - {e^{ - {x^2}}}} }}(2x{e^{ - {x^2}}})$

At x = 0

f₁'(x) does not exists.

$ \therefore $ f₁(x) is not differential at x = 0

Hence, option (2) for P.

(ii) Given, ${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {{{\tan }^{ - 1}}x'}}\,if\,x \ne 0 \hfill \cr 1,\,if\,x = 0 \hfill \cr} \right\}$

$ \Rightarrow {f_2}(x) = \left\{ \matrix{ {{ - \sin x} \over {{{\tan }^{ - 1}}x}}x < 0 \hfill \cr {{\sin x} \over {{{\tan }^{ - 1}}x}}x > 0 \hfill \cr 1\,x = 0 \hfill \cr} \right.$

Clearly, f₂(x) is not continuous at x = 0.

$ \therefore $ Option (1) for Q.

(iii) Given, f₃(x) = [sin(log_e(x + 2))], where [ ] is G.I.F.

and f₃ : ($-$1, e^{$\pi $/2} $-$ 2) $ \to $ R

It is given,

$ - 1 < x < {e^{\pi /2}} - 2$

$ \Rightarrow - 1 + 2 < x + 2 < {e^{\pi /2}} - 2 + 2$

$ \Rightarrow 1 < x + 2 < {e^{\pi /2}}$

$ \Rightarrow {\log _e}1 < {\log _e}(x + 2) < {\log _e}{e^{\pi /2}}$

$ \Rightarrow 0 < {\log _e}(x + 2) < {\pi \over 2}$

$ \Rightarrow \sin 0 < \sin {\log _e}(x + 2) < \sin {\pi \over 2}$

$ \Rightarrow 0 < \sin {\log _e}(x + 2) < 1$

$ \therefore $ $[\sin {\log _e}(x + 2)] = 0$

$ \therefore $ ${f_3}(x) = 0,\,f{'_3}(x) = {f_3}''(x) = 0$

It is differentiable and continuous at x = 0.

$ \therefore $ Option (4) for R

(iv) Given, ${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right),\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$

Now, $\mathop {\lim }\limits_{x \to 0} {f_4}(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \left( {{1 \over x}} \right) = 0$

${f_4}'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$

For $x = 0,\,{f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} {{{h^2}\sin \left( {{1 \over h}} \right) - 0} \over h}$

$ \Rightarrow {f_4}'(x) = \mathop {\lim }\limits_{h \to 0} h\sin \left( {{1 \over h}} \right) = 0$

Thus,

${f_4}'(x) = \left\{ \matrix{ 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right),\,x \ne 0 \hfill \cr 0,\,x = 0 \hfill \cr} \right.$

Again, $\mathop {\lim }\limits_{x \to 0} $

$f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right)$

does not exists.

Since, $\mathop {\lim }\limits_{x \to 0} $ cos${\left( {{1 \over x}} \right)}$ does not exists.

Hence, f'(x) is not continuous at x = 0.

$ \therefore $ Option (3) for S.

f'(x) is increasing

For some x in $\left( {{1 \over 2},\,1} \right)$

f'(x) = 1 [LMVT]

$ \therefore $ f'(1) > 1

$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + x + 1} \over {x + 1}} - ax - b} \right) = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2} + x + 1 - a{x^2} - ax - bx - b} \over {x + 1}} = 4$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{{x^2}(1 - a) + x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

Here, we make degree of N^r = degree of D^r

$\therefore$ $1 - a = 0$

and $\mathop {\lim }\limits_{x \to \infty } {{x(1 - a - b) + (1 - b)} \over {x + 1}} = 4$

$ \Rightarrow 1 - a - b = 4$

$ \Rightarrow b = - 4$

To check differentiable at x = 0

$R\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos {\pi \over h}} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} h.\,\left| {\cos {\pi \over h}} \right| = 0$

$L\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos \left( { - {\pi \over h}} \right)} \right| - 0} \over { - h}}$

$ = 0$

So, f(x) is differentiable at x = 0

To check differentiability at x = 2

$R\{ f'(2)\} = \mathop {\lim }\limits_{h \to 0} {{f(2 + h) - f(2)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}\left| {\cos \left( {{\pi \over {2 + h}}} \right)} \right| - 0} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\cos \left( {{\pi \over {2 + h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 + h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{{\pi h} \over {2(2 + h)}}} \right)} \over {h.{\pi \over {2(2 + h)}}.{\pi \over {2(2 + h)}}}}$

$ \Rightarrow R\left( {f'(2)} \right) = \pi $

$L\left( {f'(2)} \right) = \mathop {\lim }\limits_{h \to 0} {{f(2 - h) - f(2)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\left| {\cos {\pi \over {2 - h}}} \right| - {2^2}.\left| {\cos {\pi \over 2}} \right|} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2} - \left( { - \cos {\pi \over {2 - h}}} \right) - 0} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{ - {{(2 - h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 - h}}} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\sin \left( { - {{\pi h} \over {2(2 - h)}}} \right)} \over {h \times {{ - \pi } \over {2(2 - h)}} \times {{ - \pi } \over {2(2 - h)}}}}$

$ \Rightarrow L(f'(2)) = - \pi $

Thus, f(x) is differentiable at x = 0 but not at x = 2.

Here, $\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}}$ [1^$\infty$ from]

$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \{ x\log (1 + {b^2})\} \,.\,{1 \over x}}}$

$ \Rightarrow {e^{\log (1 + {b^2})}} = {(1 + b)^2}$ ..... (i)

Given,

$\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}} = 2b{\sin ^2}\theta $

$ \Rightarrow (1 + {b^2}) = 2b{\sin ^2}\theta $

$\therefore$ ${\sin ^2}\theta = {{1 + {b^2}} \over {2{b^2}}}$ ..... (ii)

By $AM \ge GM$

${{b + {1 \over b}} \over 2} \ge {\left( {b\,.\,{1 \over b}} \right)^{1/2}} \Rightarrow {{{b^2} + 1} \over {2b}} \ge 1$ ...... (iii)

From Eqs. (ii) and (iii), we get ${\sin ^2}\theta = 1$

$ \Rightarrow \theta = \pm {\pi \over 2}$ as $\theta \in ( - \pi ,\pi ]$

As when $x \in ( - 1,1),f'(x) < 0$

So, f(x) is decreasing on ($-1,1$) at x = 1

$f''(1) = {{4a} \over {{{(a + 2)}^2}}} > 0$

So, local minima at x = 1

Given that,

$g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$ for $u \in ( - \infty ,\infty )$

$g( - u) = 2{\tan ^{ - 1}}{e^{ - u}} - {\pi \over 2} = 2{\cot ^{ - 1}}({e^u}) - {\pi \over 2}$

$ = 2\left( {{\pi \over 2} - {{\tan }^{ - 1}}({e^u}) - {\pi \over 2}} \right)$

$ = - 2{\tan ^{ - 1}}({e^u}) + {\pi \over 2}$

$ = - g(u)$

$\therefore$ $g(u) = - g(u)$

$ \Rightarrow g(u)$ is an odd function

We have, $g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$

$g'(u) = {{2{e^u}} \over {1 + 2{e^{2u}}}}$

$g'(u) > 0,\forall u \in R$ [$\because$ ${e^u} > 0$]

So, $g'(u)$ is increasing function.

Given, $g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}},0 < x < 2,m \ne 0,n > 0,m,n$ are integer.

Left hand derivative (L.H.D.) of $|x - 1|$ at $x = 1$ is P.

As $|x - 1| = \left\{ {\matrix{ {x - 1,} & {x \ge 1} \cr { - (1 - x),} & {x < 1} \cr } } \right.$

$\therefore$ L.H.D. at $x = 1$ will be $-1$.

[$\therefore$ L.H.D. at $x = 1$ is $\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$]

$p = - 1$

Also, $\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(1 + h - 1)}^n}} \over {\log {{\cos }^m}(1 + h - 1)}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {\log {{\cos }^m}h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {m\log \cos \,h}} = - 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{n\,.\,{h^{n - 1}}} \over {m\log \cos \,h}} = - 1$ [using L. Hospital rule]

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {{{ - n} \over m}} \right)\,.\,{{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = - 1$

$ \Rightarrow \left( {{n \over m}} \right)\mathop {\lim }\limits_{h \to 0} {{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = 1$

$n - 2 = 0$ and ${n \over m} = 1 \Rightarrow m = n = 2$

or

$[f(x) = |x - 1| = - x + 1,x < 1f'(x) = - 1$

So, L.H.D. at $x = - 1$ is $-1$

$\therefore$ $P = - 1$

Now, $\mathop {\lim }\limits_{x \to {1^ + }} {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}} = p$ .... (i)

Let $x - 1 = t$ in L.H.S. of eq. (i)

$\mathop {\lim }\limits_{t \to } {{{t^n}} \over {m\log \cos \,t}},\left( {{0 \over 0}\,\mathrm{form}} \right)$ [using L' Hospital rule]

We get, $\mathop {\lim }\limits_{t \to 0} {{n{t^{n - 1}}} \over {m\tan \,t}} = {{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{{t^{n - 1}}} \over {\tan \,t}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

Again using L' Hospital rule,

${{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = P$

$ \Rightarrow {{ + n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = - 1$

For non-zero answer

$n - 2 = 0 \Rightarrow n = 2$ Also $m = n = 2$.

$f(x)=2+\cos x$ is differentiable everywhere and $f'(x)=-\sin x$

$\therefore f'(c)=0 \Rightarrow c=0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots \ldots$.

$\therefore$ for each real $t,[t, t+\pi]$ contains a value from $(0, \pm \pi, \pm 2 \pi, \pm 3 \pi \ldots)$

$\therefore$ Statement 1 is true.

$f(t+2 \pi)=2+\cos (t+2 \pi)=2+\cos t=f(t)$

So, statement 2 is true.

Therefore, $f(x)$ is a periodic function of period $2 \pi$.

$f'(x)=k e^{x}-1 < 0$ if $k \leq 0$

$\therefore f(x)$ is decreasing for all $x$.

Now, $\lim_\limits{x \rightarrow+\infty} f(x)=\lim_\limits{x \rightarrow+\infty} k e^{x}-x=-\infty$

Also, $f(k)=k\left(e^{k}-1\right) \geq 0$

$\left[\therefore k \leq 0 \Rightarrow e^{k}-1 \leq 0\right]$

$\therefore f(x)=0$ has exactly one root in $[k, \infty)$.

$\therefore \quad y=x$ meets $y=k e^{x}$ for $k \leq 0$ at one point.

Let’s call the function $ f(x) = k e^{x} - x $.

To find when it has only one root, check where its graph just touches the x-axis. This happens when $ f(x) = 0 $ and the slope is zero at the same point.

Find the derivative: $ f'(x) = k e^{x} - 1 $.

Set $ f'(x) = 0 $ to find the minimum or maximum point: $ k e^{x} - 1 = 0 $. This gives $ x = -\log k $.

Now check the second derivative: $ f''(x) = k e^{x} $.

Plugging in $ x = -\log k $, we get $ f''(-\log k) = k e^{-\log k} $.

The value of $ f''(-\ln k) = 1 $, which is greater than 0, so this is a minimum point.

The function will only touch the x-axis once if the minimum value is zero. So set $ f(-\ln k) = 0 $:

$ f(-\ln k) = k e^{-\ln k} - (-\ln k) = 1 + \ln k $.

Set $ 1 + \log k = 0 $ to get the value of $ k $:

$ k = \frac{1}{e}$.

We want the equation $ k e^{x} - x = 0 $ to have two different (distinct) solutions for $ x $.

Let’s call the function $ f(x) = k e^{x} - x $. For this equation to have two different solutions, the graph of $ f(x) $ must cross the x-axis at two points.

This will happen if the lowest point (minimum) of $ f(x) $ is below zero, or $ f(x)_{\min} < 0 $.

To find the minimum, take the derivative: $ f'(x) = k e^{x} - 1 $. Set this to zero: $ k e^{x} - 1 = 0 \implies e^{x} = \frac{1}{k} \implies x = -\log k $.

Now, plug $ x = -\log k $ back into $ f(x) $:
$ f(-\log k) = k e^{-\log k} - (-\log k) = k \cdot \frac{1}{k} + \log k = 1 + \log k $.

For two distinct roots, this minimum must be less than zero:
$ 1 + \log k < 0 $

This means $ \log k < -1 $, or $ k < \frac{1}{e} $. We also need $ k > 0 $.

So the possible values for $ k $ are $ 0 < k < \frac{1}{e} $.

Therefore, $ k \in \left(0, \frac{1}{e}\right) $.

First, factor the numerator and denominator of $f(x)$:

$f(x)=\frac{x^{2}-6x+5}{x^{2}-5x+6} = \frac{(x-5)(x-1)}{(x-2)(x-3)}$

(A) When $-1 < x < 1$:

Let's check the sign of each factor in this interval:

For $x$ between $-1$ and $1$:

• $x-5 < 0$ (negative)
• $x-1 < 0$ (negative)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, multiplying two negatives in the numerator and two negatives in the denominator: $f(x) = \frac{(negative) \times (negative)}{(negative) \times (negative)} = \frac{positive}{positive} > 0$

Now check if $f(x) < 1$. Let's look at $f(x) - 1$: $f(x)-1 = \frac{-(x+1)}{(x-2)(x-3)}$

For $-1 < x < 1$, $(x+1) > 0$ and $(x-2), (x-3) < 0$. So numerator is negative, denominator is positive (since two negatives). Therefore, $f(x)-1 < 0 \implies f(x) < 1$.

So, for $-1 < x < 1$: $0 < f(x) < 1$. This matches option (P).

(B) When $1 < x < 2$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 < 0$ (negative)
• $x-3 < 0$ (negative)

So, numerator: $negative \times positive = negative$. Denominator: negative $\times$ negative = positive.

So $f(x) = \frac{negative}{positive} = negative$, i.e., $f(x) < 0$. This matches (Q).

(C) When $3 < x < 5$:

• $x-5 < 0$ (negative)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $negative \times positive = negative$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{negative}{positive} = negative$, so $f(x) < 0$. This matches (Q).

(D) When $x > 5$:

• $x-5 > 0$ (positive)
• $x-1 > 0$ (positive)
• $x-2 > 0$ (positive)
• $x-3 > 0$ (positive)

Numerator: $positive \times positive = positive$.
Denominator: $positive \times positive = positive$.

So $f(x) = \frac{positive}{positive} = positive$, i.e., $f(x) > 0$.

Also, let's check $f(x) < 1$ again. Use previous result: $f(x) - 1 = \frac{-(x+1)}{(x-2)(x-3)}$ For $x > 5$, numerator is negative, denominator is positive, so $f(x)-1 < 0$, so $f(x) < 1$.

Therefore, for $x > 5$, $0 < f(x) < 1$, which matches (P).

(A) $y = x|x| = \left\{ {\matrix{ { - {x^2}} & {if\,x < 0} \cr {{x^2}} & {if\,x \ge 0} \cr } } \right.$

From graph $y=x|x|$ is (continuous in ($-1,1$) (p) differentiable in ($-1,1$) (q) Strictly increasing in ($-1,1$) (r)

(B) $y = \sqrt {|x|} $

$ = \left\{ {\matrix{ {\sqrt { - x} } & {if\,x < 0} \cr {\sqrt x } & {if\,x \ge 0} \cr } } \right.$

${y^2} = - x,x < 0$ {where y can take only +ve values}

and ${y^2} = x,x \ge 0$

$\therefore$ Graph is as follows:

From graph $y = \sqrt {|x|} $ is continuous in ($-1,1$) (P) not differentiable at x = 0 (s)

(C) Note this step

$y = x + [x] = \left\{ {\matrix{ {x - 1,} & { - 1 \le x < 0} \cr {x,} & {0 \le x < 1} \cr {x + 1,} & {x + 1} \cr } } \right.$

$\therefore$ Graph of $y = x + [x]$ is

From graph, $y = x + [x]$ is neither continuous, nor differentiable at x = 0 and hence in ($-1,1$) (s).

(D) $ = |x - 1| + |x + 1|$

$ = \left\{ {\matrix{ { - 2x} & {x < - 1} \cr 2 & { - 1 \le x < 1} \cr {2x} & {x \ge 1} \cr } } \right.$

Graph is as follows:

From graph, $y=f(x)$ is continuous (p) differentiable (q) in ($-1,1$) but not strictly increasing in ($-1,1$).

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

Evaluate the given indeterminate form i.e,

$\quad$ let $y=\lim\limits_{x \rightarrow 0}\left(\frac{1}{x}\right) \sin x$

Taking $\ln$ on both sides

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \sin x \log \left(\frac{1}{x}\right) \\ & =\lim _{x \rightarrow 0} \sin x(-\log x) \\ & =\lim _{x \rightarrow 0}\left(\frac{-\log x}{\operatorname{cosec} x}\right) \end{aligned} $

Applying L-Hopital's rule.

$ \begin{aligned} \ln y & =\lim _{x \rightarrow 0} \frac{\frac{-1}{x}}{-\operatorname{cosec} x \cot x} \\ \Rightarrow \quad \ln y & =\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \tan x \end{aligned} $

$ \begin{aligned} & =1 \times 0 & \therefore & \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ & =0 . & & \lim _{x \rightarrow 0} \tan x=0 \\ \Rightarrow \quad y & =e^0=1 & & \end{aligned} $

Here, $\quad f(x-y)=f(x) \cdot g(y)-f(y) \cdot g(x)$

Put, $\quad x=y$ then

$f(0)=f(x) \cdot g(x)-f(x) \cdot g(x)=0$ ..... (i)

Also, $\quad g(x-y)=g(x) \cdot g(y)+f(x) \cdot f(y)$

Now put $\quad y=0$ in above, we have

$g(x)=g(x) \cdot g(0)+f(x) \cdot f(0)$

$=g(x) \cdot g(0)+0 \text { using }(\mathrm{i})$

$g(x)=g(x) g(0) \quad \therefore g(0)=1$

Again put $\quad x=y$ in $g(x-y)$, we have

$g(0)=g^{2}(y)+f^{2}(x)$

$\Rightarrow \quad g^{2}(y)+f^{2}(x)=1$ ...... (ii)

Since, RHD is exists then

$\begin{aligned} \text { RHD }=f^{\prime}(0 \text { th }) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(0) \cdot g(h)-f(-h) \cdot g(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-f(-h)}{h} \\ & =\frac{f(0)-f(-h)}{-h}=\text { L.H.D } \end{aligned}$

Hence, $f$ is differentiable at $x=0$

$f^{2}(x)+g^{2}(x)=1$

$\Rightarrow 2 f^{\prime}(x) \cdot f(x)+2 g^{\prime}(x) \cdot g(x)=0$

$\Rightarrow \quad 2 f(0) \cdot f^{\prime}(0)+2 g^{\prime}(0) \cdot g(0)=0$

$\Rightarrow \quad 0+2 g^{\prime}(0) \cdot 1=0 \quad \because f(0)=0$ and $g(0)=1$

$\Rightarrow \quad g^{\prime}(0)=0$

$\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sin x^2 \cdot\left(\log _e x\right)^\alpha \cdot \sin \frac{1}{x^2}}{x^{\alpha \beta} \cdot\left(\log _c(1+x)\right)^\beta}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^\alpha}{\left(\log _e(x+1)\right)^\beta \cdot x^{\alpha \beta+2}}=0 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{\log _c x}{\log _e(x+1)}\right)^\beta \cdot \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t \\ & \lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0 \end{aligned}$

As we know $\lim _\limits{x \rightarrow \infty} \frac{x}{e^x}=0$

$\alpha \beta+2>0 \Rightarrow \alpha \beta>-2$

$ f(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ \left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{4} \leq x<\frac{1}{2} \\\\ 2\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{2} \leq x<\frac{3}{4} \\\\ 3\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{3}{4} \leq x<1 \end{array}\right. $

$ f(x) \text { is discontinuous at } x=\frac{3}{4} \text { only } $

$ f^{\prime}(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ 2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4}< x <\frac{1}{2} \\\\ 4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \frac{1}{2} < x <\frac{3}{4} \\\\ 6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \frac{3}{4} < x < 1 \end{array}\right. $

$f(x)$ is non-differentiable at $x=\frac{1}{2}$ and $\frac{3}{4}$

Minimum values of $f(x)$ occur at $x=\frac{5}{12}$ whose value is $-\frac{1}{432}$

Given,

$f(x) = {{{x^2} - 3x - 6} \over {{x^2} + 2x + 4}}$ .... (i)

$ \Rightarrow f'(x) = {{({x^2} + 2x + 4)(2x - 3) - ({x^2} - 3x - 6)(2x + 2)} \over {{{({x^2} + 2x + 4)}^2}}}$

$ \Rightarrow f'(x) = {{5x(x + 4)} \over {{{({x^2} + 2x + 4)}^2}}}$


Sign scheme for f'(x)

Here, f is decreasing in the interval ($-$2, $-$1) and f is increasing in the interval (1, 2).

Now, $f( - 4) = {{11} \over 6},f(0) = {{ - 3} \over 2}$ [from Eq. (i)]

and $\mathop {\lim }\limits_{x \to \pm \,\infty } f(x) = 1$

$\therefore$ Range $ = \left[ {{{ - 3} \over 2},{{11} \over 6}} \right]$

Hence, f(x) is into.


f(x) has local maxima at x = $-$4

and local minima at x = 0.

The given function f : R $ \to $ R is satisfying as

$f(x + y) = f(x) + f(y) + f(x)f(y)$

So, $f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))$

$ \because $ $f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}$

$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1$ (given)

$ \therefore $ $f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1$

$ \Rightarrow {\log _e}(1 + f(x)) = x + c$

$ \Rightarrow 1 + f(x) = {e^{x + c}}$

$ \Rightarrow f(x) = {e^{x + c}} - 1$

$ \because $ $f(0) = 0 \Rightarrow c = 0$

Therefore, $f(x) = {e^x} - 1$ is differentiable at every $x \in R$.

And $f'(x) = {e^x} \Rightarrow f'(0) = 1$

Now, $g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}$ and if g(0) = 1

LHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}$

and, RHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}$

So, if g(0) = 1, then g is differentiable at every x$ \in $R.

Given functions f : R $ \to $ R be defined

by f(x) = x³ $-$ x² + (x + 1) sin x and g : R $ \to $ R be an arbitrary function.

Now, let g is continuous at x = 1, then

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}$

$ \because $ $(fg)(x) = f(x)\,.\,g(x)$ (given)

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}$

{$ \because $ f(1) = 0 and g is continuous at x = 1, so g(1 $-$ h) = g(1)}

$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^2}( - h) + ( - h)\sin (1 - h)} \over { - h}}$

$ = (1 + \sin 1)g(1)$

Similarly, $\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}$

$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2}(h) + h\sin (1 + h)} \over h}$

$ = (1 + \sin 1)g(1)$

$ \because $ RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1

Now, let (fg)(x) is differentiable at x = 1, so

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}$

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}}$ {$ \because $ f(1) = 0}

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}$

$\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h) $

$= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)$

It does not mean that g(x) is continuous or differentiable at x = 1.

But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.

$\mathop {\lim }\limits_{n \to \infty } \left[ {{{1 + \root 3 \of 2 + \root 3 \of 3 + ...\root 3 \of n } \over {{n^{7/3}}\left( {{1 \over {{{(an + 1)}^2}}} + {1 \over {{{(an + 2)}^2}}} + ... + {1 \over {{{(an + n)}^2}}}} \right)}}} \right]$, $a \in R,\,|a|\, > 1$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {({r^{1/3}})} } \over {{n^{7/3}}\sum\limits_{r = 1}^n {{1 \over {{{(an + r)}^2}}}} }}$

$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^{1/3}}{1 \over n}} } \over {\sum\limits_{r = 1}^n {{1 \over {{{(a + {r \over n})}^2}}}{1 \over n}} }}$

$ = {{\int\limits_0^1 {{x^{1/3}}dx} } \over {\int\limits_0^1 {{{dx} \over {{{(a + x)}^2}}}} }} = 54$, (given)

$ \Rightarrow {{{3 \over 4}[{x^{4/3}}]_0^1} \over {\left[ { - {1 \over {x + a}}} \right]_0^1}} = 54$

$ \Rightarrow {{{3 \over 4}} \over { - {1 \over {a + 1}} + {1 \over a}}} = 54$

$ \Rightarrow {3 \over {4 \times 54}} = {1 \over {a(a + 1)}} $

$\Rightarrow {a^2} + a = 72$

$ \Rightarrow {a^2} + 9a - 8a - 72 = 0$

$ \Rightarrow a(a + 9) - 8(a + 9) = 0$

$ \Rightarrow (a - 8)(a + 9) = 0$

$ \Rightarrow a = 8$ or $ - 9$

Hence, options (c) and (d) are correct.

It is given, that f : R $ \to $ R and

Property 1 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {\sqrt {|h|} }}$ exists and finite, and


Property 2 : $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ exists and finite.

Option A,

P2 : $\mathop {\lim }\limits_{h \to 0} {{\sin h - \sin 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {1 \over h}\left( {{{\sin \,h} \over h}} \right)$ = doesn't exist.

Option B,

P1 : $\mathop {\lim }\limits_{h \to 0} {{{h^{2/3}} - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} {h^{2/3 - 1/2}} = \mathop {\lim }\limits_{h \to 0} {h^{1/6}} = 0$ exists and finite.

Option C,

P1 : $\mathop {\lim }\limits_{h \to 0} {{|h| - 0} \over {\sqrt {|h|} }} = \mathop {\lim }\limits_{h \to 0} \sqrt {|h|} = 0$, exists and finite.

Option D,

P2 : $\mathop {\lim }\limits_{h \to 0} {{h|h| - 0} \over {{h^2}}} = \mathop {\lim }\limits_{h \to 0} {{|h|} \over h} = \left\{ {\matrix{ 1 & {if\,h \to {0^ + }} \cr { - 1} & {if\,h \to {0^ - }} \cr } } \right.$

So, $\mathop {\lim }\limits_{h \to 0} {{f(h) - f(0)} \over {{h^2}}}$ does not exist.

Hence, options (b) and (c) are correct.

Given function f : R $ \to $ R is

$f(x) = \left\{ {\matrix{ {{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 3x + 1,} & {x < 0;} \cr {{x^2} - x + 1,} & {0 \le x < 1;} \cr {{2 \over 3}{x^3} - 4{x^2} + 7x - {8 \over 3},} & {1 \le x < 3;} \cr {(x - 2){{\log }_e}(x - 2) - x + {{10} \over 3},} & {x \ge 3;} \cr } } \right\}$

So,

$f'(x) = \left\{ {\matrix{ {{5x^4} + 20{x^3} + 30{x^2} + 20x + 3,} & {x < 0;} \cr {2x - 1,} & {0 \le x < 1;} \cr {2{x^2} + 8x + 7,} & {1 \le x < 3;} \cr {{{\log }_e}(x - 2),} & {x \ge 3;} \cr } } \right\}$

At x = 1, f"(1^-) = 2 > 0 and f"(1⁺) = 4$ - $8 = $ - $4 < 0

$ \therefore $ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.

For x $ \in $ ($ - $$\infty $, 0)

f'(x) = 5x⁴ + 20x³ + 30x² + 20x + 3

and since

f'($ - $1) = 5$ - $20 + 20 + 30 $ - $ 20 + 3 = $ - $2 < 0

So, f(x) is not increasing on x $ \in $($ - $$\infty $, 0).

Now, as the range of function f(x) is R, so f is onto function.

Hence, options (b), (c) and (d) are correct.

Given,

$\mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}} = {\sin ^2}x$

Using L' Hospital rules

$\mathop {\lim }\limits_{t \to x} {{f(x)\cos t - f'(t)\sin x} \over 1} = {\sin ^2}x$

$ \Rightarrow f(x)\cos x - f'(x)\sin x = {\sin ^2}x$

$ \Rightarrow f'(x)\sin x - f(x)\cos x = - {\sin ^2}x$

$ \Rightarrow {{f'(x)\sin x - f(x)\cos x} \over {{{\sin }^2}x}} = - 1$

$ \Rightarrow d\left( {{{f(x)} \over {\sin x}}} \right) = - 1$

On integrating, we get

${{f(x)} \over {\sin x}} = - x + C$

$ \Rightarrow f(x) = - x\sin x + C\sin x$

It is given that $x = {\pi \over 6},\,f\left( {{\pi \over 6}} \right) = - {\pi \over {12}}$

$ \therefore $ $f\left( {{\pi \over 6}} \right) = - {\pi \over 6}\sin {\pi \over 6} + C\sin {\pi \over 6}$

$ = - {\pi \over {12}} = - {\pi \over {12}} + {1 \over 2}C$

$ \Rightarrow C = 0$

$ \therefore $ $f(x) = - x\sin x$

(a) $f(x) = - x\sin x$

$f\left( {{\pi \over 4}} \right) = - {\pi \over 4}\sin {\pi \over 4} = - {\pi \over {4\sqrt 2 }}$ false

(b) f(x) = $-$ x sin x

$\sin x > x - {{{x^3}} \over 6},\,\forall x \in (0,\pi )$

$ \Rightarrow - x\sin x < - {x^2} + {{{x^4}} \over 6}$

$ \Rightarrow f(x) < {{{x^4}} \over 6} - {x^2},\,\forall x \in (0,\,\pi )$

It is true.

(c) f(x) = $-$x sin x

f'(x) = $-$ sin x $-$ x cos x

f'(x) = 0

$ \Rightarrow $ $-$ sin x $-$ x cos x = 0

tan x = $-$ x



$ \Rightarrow $ Their exists $\alpha $$ \in $(0, $\pi $) for which f'($\alpha $) = 0

It is true.

(d) $f(x) = - x\sin x$

$f'(x) = - \sin x - x\cos x$

$f''(x) = - 2\cos x + x\sin x$

$f''\left( {{\pi \over 2}} \right) = {\pi \over 2},\,f\left( {{\pi \over 2}} \right) = - {\pi \over 2}$

$ \therefore $ $f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0$

It is true.

We have,

${(f(0))^2} + {(f'(0))^2} = 85$

and $f:R \to [ - 2,2]$

(a) Since, f is twice differentiable function, so f is continuous function.

$ \therefore $ This is true for every continuous function.

Hence, we can always find x $ \in $ (r, s), where f(x) is one-one.

$ \therefore $ This statement is true.

(b) By L.M.V.T.

$f'(c) = {{f(b) - f(a)} \over {b - a}}$

$ \Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$

$ \Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$

$ = \left| {{{f(0) - f( - 4)} \over 4}} \right|$

Range of f is [$-$2, 2]

$ \therefore $ $ - 4 \le f(0) - f( - 4) \le 4$

$ \Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$

Hence, |f'(x₀)| = 1.

Hence, statement is true.

(c) As no function is given, then we assume

$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$

$ \therefore $ $f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$

Now, ${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$

${(f(0))^2} + {(f'(0))^2} = 85$

and $\mathop {\lim }\limits_{x \to \infty } f(x)$ does not exists.

Hence, statement is false.

(d) From option b, $|f'({x_0})|\, \le 1$

${x_0} \in ( - 4,0)$

$ \therefore $ ${(f'({x_0}))^2}\, \le 1$

Hence, $g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$

[$ \because $ f(x₀) $ \in $ [$-$2, 2]

$ \Rightarrow g({x_0})\, \le 5$

Now, let p $ \in $ ($-$4, 0) for which g(p) = 5

Similarly, let q be smallest positive number q $ \in $ (0, 4)

such that g(q) = 5

Hence, by Rolle's theorem is (p, q)

g'(c) = 0 for $\alpha $ $ \in $ ($-$4, 4) and since g(x) is greater than 5 as we move from x = p to x = q

and f(x))² $ \le $ 4

$ \Rightarrow $ (f'(x))² $ \ge $ 1 in (p, q)

Thus, g'(c) = 0

$ \Rightarrow $ f'f + f'f" = 0

So, f($\alpha $) + f"($\alpha $) = 0 and f'($\alpha $) $ \ne $ 0

Hence, statement is true.

We have,

$f'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R$

$ \Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)$

$ \Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}$

$ \Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)$

On integrating both side, we get

${e^{ - f(x)}}$ = ${e^{ - g(x)}}$ + C

At x = 1

${e^{ - f(1)}}$ = ${e^{ - g(1)}}$ + C

${e^{ - 1}} = {e^{ - g(1)}} + C$ [$ \because $ f(1) = 1] ...(i)

At x = 2

${e^{ - f(2)}}$ = ${e^{ - g(2)}}$ + C

$ \Rightarrow $ ${e^{ - f(2)}}$ = ${e^{ - 1}}$ + C [$ \because $ g(2) = 1] .... (ii)

From Eqs. (i) and (ii)

${e^{ - f(2)}}$ = $2{e^{ - 1}}$ $-$ ${{e^{ - g(1)}}}$ ...(iii)

$ \Rightarrow $ ${e^{ - f(2)}}$ > $2{e^{ - 1}}$

We know that, e^$-$x is decreasing

$ \therefore $ $-$f(2) < log_e 2$-$1

f(2) > 1 $-$ log_e 2

$ \Rightarrow $ ${{e^{ - g(1)}}}$ + ${e^{ - f(2)}}$ = $2{e^{ - 1}}$ [from Eq. (iii)]

$ \Rightarrow $ ${{e^{ - g(1)}}}$ < $2{e^{ - 1}}$ $-$g(1) < log_e 2 $-$1

$ \Rightarrow $ g(1) >$-$log_e 2

$f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)$

Now, $\mathop {\lim }\limits_{x \to {1^ - }} f(x)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0$

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $

${{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)$

$ = \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right)$, which does not exist.

(a) $ \because $ ${e^x} \in (1,e)$ in (0, 1) and $\int_0^x {f(t)\sin t\,dt \in } $ (0, 1) in (0, 1)

$ \therefore $ ${e^x} - \int_0^x {f(t)\sin t\,dt} $ cannot be zero.

So, option (a) is incorrect.

(b) $f(x) + \int_0^{{\pi \over 2}} {f(t)\sin t\,dt} $ always positive

$ \therefore $ Option (b) is incorrect.

(c) Let $h(x) = x - \int_0^{{\pi \over 2} - x} {f(t)\cos t\,dt} $,

$h(0) = - \int_0^{{\pi \over 2}} {f(t)\cos t\,dt} < 0$

$h(1) = 1 - \int_0^{{\pi \over 2} - 1} {f(t)\cos t\,dt} > 0$

$ \therefore $ Option (c) is correct.

(d) Let $g(x) = {x^9} - f(x)$

$g(0) = - f(0) < 0$, $g(1) = 1 - f(1) > 0$

$ \therefore $ Option (d) is correct.

$f(x) = x\cos (\pi (x + [x]))$

At x = 0

$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} x\cos (\pi (x + [x]) = 0$

and f(x) = 0

$ \therefore $ It is continuous at x = 0 and clearly discontinuous at other integer points.

$f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

If ${x^3} - x \ge 0$

$ \Rightarrow \cos \left| {{x^3} - x} \right| = \cos ({x^3} - x)$

and if ${x^3} - x \le 0$

$ \Rightarrow \cos |{x^3} - x| = \cos ({x^3} - x)$

$\therefore$ $\cos (|{x^3} - x|) = \cos ({x^3} - x),\,\forall x \in R$ ..... (i)

Again, if ${x^3} + x \ge 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = x\sin ({x^3} + x)$

and if ${x^3} + x \le 0$

$ \Rightarrow |x|\sin (|{x^3} + x|) = - x\sin \{ - ({x^3} + x)\} $

$\therefore$ $|x|\sin (|{x^3} + x|) = x\sin ({x^3} + x),\,\forall x \in R$ ...... (ii)

$ \Rightarrow f(x) = a\cos (|{x^3} - x|) + b|x|\sin (|{x^3} + x|)$

$\therefore$ $f(x) = a\cos ({x^3} - x) + bx\sin ({x^3} + x)$ ..... (iii)

For a function to be differentiable at x = 0, the function must be continuous.

$f(0) = a\cos (0 - 0) + b(0)\sin (0) = a$

Therefore,

$f({0^ + }) = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

$f({0^ - }) = \mathop {\lim }\limits_{h \to \infty } [a(\cos ( - {h^3} + h) + b( - h)\sin ( - {h^3} - h)]$

$ = \mathop {\lim }\limits_{h \to \infty } [a\cos ({h^3} - h) + bh\sin ({h^3} + h)] = 0$

which is continuous at x = 0; hence, f(x) is differentiable for all values of a and b.

Therefore,

$f(1) = a\cos (1 - 1) + 1\sin (1 + 1) = a + b\sin 2$

$f({1^ + }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 + h)^3} - (1 + h) + b(1 + h)\sin {(1 + h)^3} + (1 + h)$

$ = f(1)$

$f({1^ - }) = \mathop {\lim }\limits_{h \to 0} a\cos {(1 - h)^3} - (1 - h) + b(1 - h)\sin {(1 - h)^3} + (1 - h)$

$ = f(1)$

Thus, f(x) is continuous and we can also see that f is differentiable at x = 0 and x = 1.