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Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 15 of 20 · Click an option or "Show Solution" to reveal answer

Q141
If a=lim⁡x→01+1+x4−2x4\mathrm{a}=\lim\limits_{x \rightarrow 0} \dfrac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} and b=lim⁡x→0sin⁡2x2−1+cos⁡x\mathrm{b}=\lim\limits _{x \rightarrow 0} \dfrac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}, then the value of ab3a b^3 is :
A 36
B 25
C 32
D 30
Correct Answer
Option C
Solution
a=lim⁡x→01+1+x4−2x4=lim⁡x→01+x4−1x4(1+1+x4+2)=lim⁡x→0x4x4(1+1+x4+2)(1+x4+1)\begin{aligned} a= & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)} \end{aligned}

Applying limit

a=142\mathrm{a}=\frac{1}{4 \sqrt{2}}
b=lim⁡x→0sin⁡2x2−1+cos⁡x=lim⁡x→0(1−cos⁡2x)(2+1+cos⁡x)2−(1+cos⁡x)b=lim⁡x→0(1+cos⁡x)(2+1+cos⁡x)\begin{aligned} & b=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \\ & b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) \end{aligned}

Applying limits

b=2(2+2)=42b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}

Now,

ab3=142×(42)3=32a b^3=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^3=32
Q142
Consider the function f:(0,2)→Rf:(0,2) \rightarrow \mathbf{R} defined by f(x)=x2+2xf(x)=\dfrac{x}{2}+\dfrac{2}{x} and the function g(x)g(x) defined by $$g(x)=\left\{\begin{array}{ll} \min \lfloor f(t)\}, & 0
A gg is continuous but not differentiable at x=1x=1
B gg is continuous and differentiable for all x∈(0,2)x \in(0,2)
C gg is not continuous for all x∈(0,2)x \in(0,2)
D gg is neither continuous nor differentiable at x=1x=1
Correct Answer
Option A
Solution
f:(0,2)→R;f(x)=x2+2xf′(x)=12−2x2\begin{aligned} & f:(0,2) \rightarrow R ; f(x)=\frac{x}{2}+\frac{2}{x} \\ & f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2} \end{aligned}
∴f(x)\therefore \mathrm{f}(\mathrm{x})

is decreasing in domain.

g(x)={x2+2x0<x≤132+x1<x<2g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0 < x \leq 1 \\ \frac{3}{2}+x & 1 < x < 2\end{cases}
Q143
 If lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13, then 2α−β is equal to : \text{ If } \lim \limits_{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3} \text{, then } 2 \alpha-\beta \text{ is equal to : }
A 2
B 1
C 5
D 7
Correct Answer
Option C
Solution
lim⁡x→03+αsin⁡x+βcos⁡x+log⁡e(1−x)3tan⁡2x=13\lim \limits_{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}
⇒lim⁡x→03+α[x−x33!+….]+β[1−x22!+x44!….]+(−x−x22−x33…)3tan⁡2x=13\Rightarrow \lim \limits_{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots .\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots .\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3}
⇒lim⁡x→0(3+β)+(α−1)x+(−12−β2)x2+….3x2×x2tan⁡2x=13\Rightarrow \lim \limits_{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots .}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3}
⇒β+3=0,α−1=0 and −12−β23=13⇒β=−3,α=1⇒2α−β=2+3=5\begin{aligned} & \Rightarrow \beta+3=0, \alpha-1=0 \text{ and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} \\ & \Rightarrow \beta=-3, \alpha=1 \\ & \Rightarrow 2 \alpha-\beta=2+3=5 \end{aligned}
Q144
Consider the function f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbb{R} defined by f(x)=e−∣log⁡ex∣f(x)=e^{-\left|\log _e x\right|}. If mm and nn be respectively the number of points at which ff is not continuous and ff is not differentiable, then m+nm+n is
A 0
B 1
C 2
D 3
Correct Answer
Option B
Solution
f:(0,∞)→Rf(x)=e−∣log⁡ex∣\begin{aligned} & f:(0, \infty) \rightarrow R \\ & f(x)=e^{-\left|\log _e x\right|} \end{aligned}
f(x)=1e∣ln⁡x∣={1e−ln⁡x;0<x<11eln⁡x;x≥1\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\begin{array}{l} \frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \\ \frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathrm{x} \geq 1 \end{array}\right.
{11x=x;0<x<11x,x≥1\left\{\begin{array}{l} \frac{1}{\frac{1}{x}}=x ; 0< x<1 \\ \frac{1}{x}, x \geq 1 \end{array}\right.
m=0\mathrm{m}=0

(No point at which function is not continuous)

n=1\mathrm{n}=1

(Not differentiable)

∴m+n=1\therefore \mathrm{m}+\mathrm{n}=1
Q145
lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1x2\lim \limits_{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}
A is equal to 1
B does not exist
C is equal to −1-1
D is equal to 2
Correct Answer
Option D
Solution
lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1x2lim⁡x→0e2∣sin⁡x∣−2∣sin⁡x∣−1∣sin⁡x∣2×sin⁡2xx2\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\ & \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2} \end{aligned}

Let

∣sin⁡x∣=t|\sin \mathrm{x}|=\mathrm{t}
lim⁡t→0e2t−2t−1t2×lim⁡x→0sin⁡2xx2=lim⁡t→02e2t−22t×1=2×1=2\begin{aligned} & \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \\ & =\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2 \end{aligned}
Q146
Let g(x)g(x) be a linear function and f(x)={g(x),x≤0(1+x2+x)1x,x>0f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right., is continuous at x=0x=0. If f′(1)=f(−1)f^{\prime}(1)=f(-1), then the value g(3)g(3) is
A log⁡e(49)−1\log _e\left(\dfrac{4}{9}\right)-1
B 13log⁡e(49e1/3)\dfrac{1}{3} \log _e\left(\dfrac{4}{9 e^{1 / 3}}\right)
C log⁡e(49e1/3)\log _e\left(\dfrac{4}{9 e^{1 / 3}}\right)
D 13log⁡e(49)+1\dfrac{1}{3} \log _e\left(\dfrac{4}{9}\right)+1
Correct Answer
Option C
Solution

Let

g(x)=ax+bg(x)=a x+b

Now function

f(x)\mathrm{f}(\mathrm{x})

in continuous at

x=0\mathrm{x}=0
∴lim⁡x→0+f(x)=f(0)⇒lim⁡x→0(1+x2+x)1x=b⇒0=b∴g(x)=ax\begin{aligned} & \therefore \lim \limits_{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \\ & \Rightarrow \lim \limits_{\mathrm{x} \rightarrow 0}\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}}=\mathrm{b} \\ & \Rightarrow 0=\mathrm{b} \\ & \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax} \end{aligned}

Now, for

x>0\mathrm{x}>0
f′(x)=1x⋅(1+x2+x)1x−1⋅1(2+x)2+(1+x2+x)1x⋅ln⁡(1+x2+x)⋅(−1x2)∴f′(1)=19−23⋅ln⁡(23) And f(−1)=g(−1)=−a∴a=23ln⁡(23)−19∴g(3)=2ln⁡(23)−13=ln⁡(49⋅e1/3)\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \\ & +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \\ & \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \\ & \text{ And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \\ & \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \\ & \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \\ & =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right) \end{aligned}
Q147
lim⁡x→0e−(1+2x)12xx\lim \limits_{x \rightarrow 0} \dfrac{e-(1+2 x)^{\dfrac{1}{2 x}}}{x} is equal to
A −2e\dfrac{-2}{e}
B e−e2e-e^2
C 0
D ee
Correct Answer
Option D
Solution
lim⁡x→0e−(1+2x)12xx\lim \limits_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}

Using expansion

=lim⁡x→0e−e[1−2x2+11×4x224+…]x=lim⁡x→0(e−11x6e+…)=e\begin{aligned} & =\lim \limits_{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\ & =\lim \limits_{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e \end{aligned}
Q148
Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a function given by f(x)={1−cos⁡2xx2,x0f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x 0\end{cases} where α,β∈R\alpha, \beta \in \mathbf{R}. If ff is continuous at x=0x=0, then α2+β2\alpha^2+\beta^2 is equal to :
A 48
B 6
C 3
D 12
Correct Answer
Option D
Solution
f(x)={1−cos⁡2xx2,x0f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^2}, & x0 \end{array}\right.
f(x)f(x)

is continuous at

x=0x=0
⇒f(0)=lim⁡x→0−f(x)=lim⁡x→0+f(x)\Rightarrow f(0)=\lim \limits_{x \rightarrow 0^{-}} f(x)=\lim \limits_{x \rightarrow 0^{+}} f(x)
lim⁡x→0−f(x)=αlim⁡x→0−(1−cos⁡2xx2)=α⇒lim⁡x→0−2sin⁡2hx2=α⇒lim⁡h→02sin⁡2h2h=α⇒α=2 Also, lim⁡x→0+f(x)=f(0)⇒lim⁡x→0+β1−cos⁡xx=2\begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\alpha \\ & \lim _{x \rightarrow 0^{-}}\left(\frac{1-\cos 2 x}{x^2}\right)=\alpha \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 h}{x^2}=\alpha \\ & \Rightarrow \lim _{h \rightarrow 0} \frac{2 \sin ^2}{h^2} h=\alpha \\ & \Rightarrow \alpha=2 \\ & \end{aligned}\\ &\begin{aligned} & \text{ Also, } \lim _{x \rightarrow 0^{+}} f(x)=f(0) \\ & \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\beta \sqrt{1-\cos x}}{x}=2 \end{aligned} \end{aligned}
⇒lim⁡h→0β1−cos⁡hh2h2h=2\Rightarrow \lim \limits_{h \rightarrow 0} \frac{\beta \sqrt{\frac{1-\cos h}{h^2}} h^2}{h}=2
⇒β2=2⇒β=22⇒α2+β2=4+8=12\begin{aligned} & \Rightarrow \quad \frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \quad \beta=2 \sqrt{2} \\ & \Rightarrow \quad \alpha^2+\beta^2=4+8 \\ & \quad=12 \end{aligned}

∴\therefore Option (1) is correct

Q149
If the function f(x)={72x−9x−8x+12−1+cos⁡x,x≠0alog⁡e2log⁡e3,x=0f(x)= \begin{cases}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{cases} is continuous at x=0x=0, then the value of a2a^2 is equal to
A 968
B 1250
C 1152
D 746
Correct Answer
Option C
Solution
f(x)={72x−9x−8x+12−1+cos⁡x, x≠0alog⁡e2log⁡e3      ,  x=0f(x) = \left\{ \begin{array}{ll}{{{{72}^x} - {9^x} - {8^x} + 1} \over {\sqrt 2 - \sqrt {1 + \cos x} }},\,x \ne 0 \\ a{\log _e}2{\log _e}3\,\,\,\,\,\,,\,\,x = 0 \end{array} \right.
∵f(x)\because f(x)

is continuous at

x=0x=0
⇒lim⁡x→072x−9x−8x+12−1+cos⁡xlim⁡x→0(9x−1)(8x−1)(2+1+cos⁡x)(1−cos⁡x)x2×x2=(ln⁡9⋅ln⁡8)(22)×2=42×2×3ln⁡2⋅ln⁡3242⋅ln⁡2⋅ln⁡3⇒a=242a2=1152\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} \\ & \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2} \times x^2} \\ & =(\ln 9 \cdot \ln 8)(2 \sqrt{2}) \times 2 \\ & =4 \sqrt{2} \times 2 \times 3 \ln 2 \cdot \ln 3 \\ & 24 \sqrt{2} \cdot \ln 2 \cdot \ln 3 \\ & \Rightarrow \quad a=24 \sqrt{2} \\ & \quad a^2=1152 \end{aligned}
Q150
For a,b>0\mathrm{a}, \mathrm{b}>0, let f(x)={tan⁡((a+1)x)+btan⁡xx,x0f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x 0\end{cases} be a continuous function at x=0x=0. Then ba\dfrac{\mathrm{b}}{\mathrm{a}} is equal to :
A 4
B 5
C 8
D 6
Correct Answer
Option D
Solution
f(x)={tan⁡((a+1)x)+btan⁡xx,x0f(x)=\left\{\begin{array}{cc} \frac{\tan ((a+1) x)+b \tan x}{x}, & x0 \end{array}\right.
f(x)f(x)

is continuous at

x=0x=0
⇒lim⁡x→0−f(x)=f(0)=lim⁡x→0+f(x)lim⁡x→0−f(x)=3⇒lim⁡x→0−tan⁡((a+1)x)+b+axx=3⇒a+1+b=3⇒a+b=2..... (1)\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\ & \lim _{x \rightarrow 0^{-}} f(x)=3 \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\ & \Rightarrow a+1+b=3 \\ & \Rightarrow a+b=2 \quad \text{..... (1)} \end{aligned}

also,

lim⁡x→0+ax+b2x2−axbaxx=3\lim \limits_{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3
=lim⁡x→0+ah+b2h2−ahba×hh=3=lim⁡h→0a+b2h2−abah×a+b2h+aa+b2h+a=3=lim⁡h→0a+b2h−abah(a+b2h+a)=3⇒b2ba(2a)=3⇒b2a=3⇒ba=6..... (2)\begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\ & =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\ & \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\ & \Rightarrow \frac{b}{2 a}=3 \\ & \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)} \end{aligned}
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