Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 16 of 20 · Click an option or "Show Solution" to reveal answer

Q151
If the function f(x)=sin⁡3x+αsin⁡x−βcos⁡3xx3,x∈Rf(x)=\dfrac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}, is continuous at x=0x=0, then f(0)f(0) is equal to :
A 4
B −-2
C −-4
D 2
Correct Answer
Option C
Solution
lim⁡x→0f(x)=f(0) (continuous at x=0)lim⁡x→0sin⁡3x+αsin⁡x−βcos⁡3xx3\begin{aligned} & \lim \limits_{x \rightarrow 0} f(x)=f(0) \quad \text{ (continuous at } x=0) \\ & \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3} \end{aligned}

For limit to exist

β=0\beta=0
⇒lim⁡x→0sin⁡3x+αsin⁡xx3⇒lim⁡x→0(3+α)sin⁡x−4sin⁡3xx3\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3} \end{aligned}

For limit to exist

α+3=0⇒α=−3\alpha+3=0 \Rightarrow \alpha=-3
⇒lim⁡x→0−4sin⁡3xx3=−4=f(0)\Rightarrow \lim \limits_{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)
Q152
Let ,f:[−1,2]→Rf:[-1,2] \rightarrow \mathbf{R} be given by f(x)=2x2+x+[x2]−[x]f(x)=2 x^2+x+\left[x^2\right]-[x], where [t][t] denotes the greatest integer less than or equal to tt. The number of points, where ff is not continuous, is :
A 5
B 6
C 4
D 3
Correct Answer
Option C
Solution
f(x)=2x2+x+[x2]−[x]=2x2+[x2]+{x}f(−1)=2+1+0=3f(−1+)=2+0+0=2f(0−)=0+1=1f(0+)=0+0+0=0f(1+)=2+1+0=3\begin{aligned} & f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\ & f(-1)=2+1+0=3 \\ & f\left(-1^{+}\right)=2+0+0=2 \\ & f\left(0^{-}\right)=0+1=1 \\ & f\left(0^{+}\right)=0+0+0=0 \\ & f\left(1^{+}\right)=2+1+0=3 \end{aligned}
f(1−)=2+0+1=3f(2−)=8+3+1=12f(2+)=8+4+0=12\begin{aligned} & f\left(1^{-}\right)=2+0+1=3 \\ & f\left(2^{-}\right)=8+3+1=12 \\ & f\left(2^{+}\right)=8+4+0=12 \end{aligned}

∴\therefore discontinuous at

x=0,2,3,−1x=0, \sqrt{2}, \sqrt{3},-1
Q153
If ∑r=1nTr=(2n−1)(2n+1)(2n+3)(2n+5)64\sum\limits_{r=1}^n T_r=\dfrac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)}{64}, then lim⁡n→∞∑r=1n(1Tr)\lim \limits_{n \rightarrow \infty} \sum\limits_{r=1}^n\left(\dfrac{1}{T_r}\right) is equal to :
A 23\dfrac{2}{3}
B 13\dfrac{1}{3}
C 1
D 0
Correct Answer
Option A
Solution

Tn=Sn−Sn−1⇒Tn=18(2n−1)(2n+1)(2n+3)⇒1Tn=8(2n−1)(2n+1)(2n+3)lim⁡n→∞∑r=1n1Tr=lim⁡n→∞8∑r=1n1(2n−1)(2n+1)(2n+3)=lim⁡n→∞84∑(1(2n−1)(2n+1)−1(2n+1)(2n+3))=lim⁡n→∞2[(11.3−13.5)+(13.5−15.7)+…]=23\begin{aligned} & T_n=S_n-S_{n-1} \\\\ & \Rightarrow T_n=\dfrac{1}{8}(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3) \\\\ & \Rightarrow \dfrac{1}{T_{\mathrm{n}}}=\dfrac{8}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\\\ & \lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}} \dfrac{1}{T_{\mathrm{r}}}=\lim _{\mathrm{n} \rightarrow \infty} 8 \sum_{\mathrm{r}=1}^{\mathrm{n}} \dfrac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)} \\\\ & =\lim _{\mathrm{n} \rightarrow \infty} \dfrac{8}{4} \sum\left(\dfrac{1}{(2 \mathrm{n}-1)(2 \mathrm{n}+1)}-\dfrac{1}{(2 \mathrm{n}+1)(2 \mathrm{n}+3)}\right) \\\\ & =\lim _{\mathrm{n} \rightarrow \infty} 2\left[\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}\right)+\left(\dfrac{1}{3.5}-\dfrac{1}{5.7}\right)+\ldots\right] \\\\ & =\dfrac{2}{3}\end{aligned}

Q154
If lim⁡x→∞((e1−e)(1e−x1+x))x=α\lim \limits_{x \rightarrow \infty}\left(\left(\dfrac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\dfrac{1}{\mathrm{e}}-\dfrac{x}{1+x}\right)\right)^x=\alpha, then the value of log⁡eα1+log⁡eα\dfrac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha} equals :
A e−2e^{-2}
B e2\mathrm{e}^2
C ee
D e−1e^{-1}
Correct Answer
Option C
Solution
α=lim⁡x→∞[(e1−e)(1e−x1+x)]x.\begin{aligned} \alpha &= \lim_{x \to \infty} \left[\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right]^x. \end{aligned}

Begin by rewriting the expression inside the limit. Notice that

x1+x=1−11+x,\frac{x}{1+x} = 1 - \frac{1}{1+x},

so

1e−x1+x=1e−(1−11+x)=1e−1+11+x.\frac{1}{e} - \frac{x}{1+x} = \frac{1}{e} - \left(1 - \frac{1}{1+x}\right) = \frac{1}{e} - 1 + \frac{1}{1+x}.

Express the constant part as

1−1e=e−1e,1 - \frac{1}{e} = \frac{e-1}{e},

to obtain

1e−x1+x=−e−1e+11+x.\frac{1}{e} - \frac{x}{1+x} = -\frac{e-1}{e} + \frac{1}{1+x}.

Multiplying by the prefactor, we have

(e1−e)(1e−x1+x)=e1−e[−e−1e+11+x].\left(\frac{e}{1-e}\right)\left(\frac{1}{e} - \frac{x}{1+x}\right) = \frac{e}{1-e}\left[-\frac{e-1}{e} + \frac{1}{1+x}\right].

Split the expression:

e1−e⋅(−e−1e)+e1−e⋅11+x=−e−11−e+e(1−e)(1+x).\frac{e}{1-e}\cdot\left(-\frac{e-1}{e}\right) + \frac{e}{1-e}\cdot\frac{1}{1+x} = -\frac{e-1}{1-e} + \frac{e}{(1-e)(1+x)}.

Since

1−e=−(e−1),1-e = -(e-1),

it follows that

−e−11−e=−e−1−(e−1)=1.-\frac{e-1}{1-e} = -\frac{e-1}{-(e-1)} = 1.

Hence, the expression simplifies to

1+e(1−e)(1+x).1 + \frac{e}{(1-e)(1+x)}.

Again, replacing 1−e1-e by −(e−1)-(e-1),

1+e−(e−1)(1+x)=1−e(e−1)(1+x).1 + \frac{e}{-(e-1)(1+x)} = 1 - \frac{e}{(e-1)(1+x)}.

Thus, the limit becomes

α=lim⁡x→∞(1−e(e−1)(1+x))x.\alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)(1+x)}\right)^x.

For large xx, note that

1+x∼x,1+x \sim x,

so we approximate

α=lim⁡x→∞(1−e(e−1)x)x.\alpha = \lim_{x \to \infty} \left(1 - \frac{e}{(e-1)x}\right)^x.

Recall the limit

lim⁡x→∞(1−ax)x=e−a,\lim_{x \to \infty} \left(1 - \frac{a}{x}\right)^x = e^{-a},

with

a=ee−1.a = \frac{e}{e-1}.

Therefore,

α=e−ee−1.\alpha = e^{-\frac{e}{e-1}}.

Taking the natural logarithm,

ln⁡α=−ee−1.\ln \alpha = -\frac{e}{e-1}.

Now, compute

ln⁡α1+ln⁡α=−ee−11−ee−1.\frac{\ln \alpha}{1 + \ln \alpha} = \frac{-\frac{e}{e-1}}{1 - \frac{e}{e-1}}.

Express the denominator with a common denominator:

1−ee−1=e−1e−1−ee−1=−1e−1.1 - \frac{e}{e-1} = \frac{e-1}{e-1} - \frac{e}{e-1} = -\frac{1}{e-1}.

Thus,

−ee−1−1e−1=e1=e.\frac{-\frac{e}{e-1}}{-\frac{1}{e-1}} = \frac{e}{1} = e.

The final result is

ln⁡α1+ln⁡α=e.\frac{\ln\alpha}{1+\ln\alpha} = e.
Q155
If the function f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},x0 f(x)=\left\{\begin{array}{l} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x0 \end{array}\right. is continuous at x=0x=0, then k12+k22k_1^2+k_2^2 is equal to :
A 5
B 10
C 20
D 8
Correct Answer
Option B
Solution
lim⁡x→0−2x{sin⁡(k1+1)x+sin⁡(k2−1)x}=4⇒2(k1+1)+2(k2−1)=4⇒k1+k2=2⇒lim⁡x→0+2xln⁡(2+k1x2+k2x)=4⇒lim⁡x→0+1xln⁡(1+(k1−k2)x2+k2x)=2⇒k1−k22=2⇒k1−k2=4∴k1=3,k2=−1k12+k22=9+1=10\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\ & \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\ & \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\ & \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\ & \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\ & \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\ & \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10 \end{aligned}
Q156
lim⁡x→∞(2x2−3x+5)(3x−1)x2(3x2+5x+4)(3x+2)x\lim \limits_{x \rightarrow \infty} \dfrac{\left(2 x^2-3 x+5\right)(3 x-1)^{\dfrac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}} is equal to :
A 2e3\dfrac{2 e}{3}
B 23e\dfrac{2}{3 \sqrt{\mathrm{e}}}
C 2e3\dfrac{2 \mathrm{e}}{\sqrt{3}}
D 23e\dfrac{2}{\sqrt{3 \mathrm{e}}}
Correct Answer
Option B
Solution
lim⁡x→∞(2−3x+5x2)(1−13x)x/2(3+5x+4x2)(1+23x)x/2=lim⁡x→∞23⋅ex2(1−13x−1)ex2(1+23x−1)=23⋅e−16e1/3=23e−12\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right)\left(1-\frac{1}{3 x}\right)^{x / 2}}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right)\left(1+\frac{2}{3 x}\right)^{x / 2}} \\ & =\lim _{x \rightarrow \infty} \frac{2}{3} \cdot \frac{e^{\frac{x}{2}\left(1-\frac{1}{3 x}-1\right)}}{e^{\frac{x}{2}\left(1+\frac{2}{3 x}-1\right)}} \\ & =\frac{2}{3} \cdot \frac{e^{-\frac{1}{6}}}{e^{1 / 3}}=\frac{2}{3} e^{-\frac{1}{2}} \end{aligned}
Q157
lim⁡x→0cosec⁡x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)\lim \limits_{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right) is:
A 115\dfrac{1}{\sqrt{15}}
B 125\dfrac{1}{2 \sqrt{5}}
C 00
D −125-\dfrac{1}{2 \sqrt{5}}
Correct Answer
Option D
Solution
lim⁡x→0cosec⁡x(2cos⁡2x+3cos⁡x−cos⁡2x+sin⁡x+4)lim⁡x→0cosec⁡(cos⁡2x+3cos⁡x−sin⁡x−4)(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→01sin⁡x(cos⁡2x+3cos⁡x−4)−sin⁡x(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0(cos⁡x+4)(cos⁡x−1)−sin⁡xsin⁡x(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0−2sin⁡2x2(cos⁡x+4)−2sin⁡x2cos⁡x22sin⁡x2cos⁡x2(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)lim⁡x→0−(sin⁡xx2(cos⁡x+4)+cos⁡x2)cos⁡x2(2cos⁡2x+3cos⁡x+cos⁡2x+sin⁡x+4)−125\begin{aligned} & \lim _{x \rightarrow 0} \operatorname{cosec} x\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right) \\ & \lim _{x \rightarrow 0} \frac{\operatorname{cosec}\left(\cos ^2 x+3 \cos x-\sin x-4\right)}{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{1}{\sin x} \frac{\left(\cos ^2 x+3 \cos x-4\right)-\sin x}{\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{\left(\cos ^x+4\right)(\cos x-1)-\sin x}{\sin x\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{-2 \sin ^2 \frac{x}{2}(\cos x+4)-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & \lim _{x \rightarrow 0} \frac{-\left(\sin ^x \frac{x}{2}(\cos x+4)+\cos \frac{x}{2}\right)}{\cos \frac{x}{2}\left(\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4}\right)} \\ & -\frac{1}{2 \sqrt{5}} \end{aligned}
Q158
Let f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R} be a function such that f(x)−6f(1x)=353x−52f(x)-6 f\left(\dfrac{1}{x}\right)=\dfrac{35}{3 x}-\dfrac{5}{2}. If the lim⁡x→0(1αx+f(x))=β;α,β∈R\lim\limits _{x \rightarrow 0}\left(\dfrac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in \mathbb{R}, then α+2β\alpha+2 \beta is equal to
A 6
B 5
C 3
D 4
Correct Answer
Option D
Solution

F(x)−6f(1/x)=353x−52..... (1)F(x)-6 f(1 / x)=\dfrac{35}{3 x}-\dfrac{5}{2}\quad\text{..... (1)}

\begin{aligned} &\begin{aligned} & \text{ Replace } x \rightarrow \frac{1}{x} \\ & F(1 / x)-6(x)=\frac{35 x}{3}-\frac{5}{2}\quad\text{..... (2)} \end{aligned}\\ &\text{ Using (1) & (2) }\\ &\begin{aligned} & f(x)=-2 x-\frac{1}{3 x}+\frac{1}{2} \\ & B=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right) \end{aligned} \end{aligned}
=lim⁡x→0(1αx−2x−13x+12)α=3,B=1/2 So, α+2B=3+1=4\begin{aligned} &= \lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}-2 x-\frac{1}{3 x}+\frac{1}{2}\right) \\ & \alpha=3, \quad B=1 / 2 \\ & \text{ So, } \alpha+2 B=3+1=4 \end{aligned}
Q159
ff is defined in [−5,5]\left[ { - 5,5} \right] as f(x)=xf\left( x \right) = x if xx is rational              \,\,\,\,\,\,\,\,\,\,\,\,\, =−x = - x if xx is irrational. Then
A f(x)f(x) is continuous at every x, except x=0x = 0
B f(x)f(x) is discontinuous at every x,x, except x=0x = 0
C f(x)f(x) is continuous everywhere
D f(x)f(x) is discontinuous everywhere
Correct Answer
Option B
Solution

Let a is a rational number other than

0,0,

in

[−5,5],\left[ { - 5,5} \right],

then

f(a)=af\left( a \right) = a

and

lim⁡x→a  f(x)=−a\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a

[ As in the immediate neighbourhood of a rational - number, we find irrational numbers ] ∴\therefore

f(x)f(x)

is not continuous at any rational number If a is irrational number, then

 f(a)=−a\,f\left( a \right) = - a

and

lim⁡x→a f(x)=a\mathop {\lim }\limits_{x \to a} \,f\left( x \right) = a

∴\therefore

f(x)f(x)

is not continuous at any irrational number clearly

lim⁡x→0f(x)=f(0)=0\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0

∴\therefore

f(x)f(x)

is continuous at

x=0x=0
Q160
Given below are two statements: Statement I: lim⁡x→0(tan⁡−1x+log⁡e1+x1−x−2xx5)=25 \lim\limits_{x \to 0} \left( \dfrac{\tan^{-1} x + \log_e \sqrt{\dfrac{1+x}{1-x}} - 2x}{x^5} \right) = \dfrac{2}{5} Statement II: lim⁡x→1(x21−x)=1e2 \lim\limits_{x \to 1} \left( x^{\dfrac{2}{1-x}} \right) = \dfrac{1}{e^2} In the light of the above statements, choose the correct answer from the options given below:
A Statement I is false but Statement II is true
B Both Statement I and Statement II are false
C Both Statement I and Statement II are true
D Statement I is true but Statement II is false
Correct Answer
Option C
Solution
lim⁡x→0tan⁡−1x+12[ln⁡(1+x)−ln⁡(1−x)]−2xx5=lim⁡x→0(x−x33+x55…)+12[x−x22+x33…−(−x−x22−x33…)]−2xx5=lim⁡x→02x+2x55….−2xx5=25lim⁡x→1x2(1−x)=elim⁡x→1(21−x)(x−1)=e−2⇒ Both statements correct \begin{aligned} &\begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots .-2 x}{x^5}=\frac{2}{5} \\ & \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=e^{-2} \end{aligned}\\ &\Rightarrow \text{ Both statements correct } \end{aligned}
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