Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 13 of 20 · Click an option or "Show Solution" to reveal answer

Q121
Let f,gf, g and hh be the real valued functions defined on R\mathbb{R} as f(x)={x∣x∣,x≠01,x=0f(x)=\left\{\begin{array}{cc}\dfrac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right. g(x)={sin⁡(x+1)(x+1),x≠−11,x=−1g(x)=\left\{\begin{array}{cc}\dfrac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right. and h(x)=2[x]−f(x)h(x)=2[x]-f(x), where [x][x] is the greatest integer ≤x\leq x. Then the value of lim⁡x→1g(h(x−1))\lim\limits_{x \rightarrow 1} g(h(x-1)) is :
A 1
B −1-1
C sin⁡(1)\sin (1)
D 0
Correct Answer
Option A
Solution
f(x)=sgn(x)f(x) = {\mathop{\rm sgn}} (x)
h(x)=2[x]−sgn(x)h(x) = 2[x] - {\mathop{\rm sgn}} (x)

If

x→1+x \to {1^ + }

then

h(x−1)=2[x]−2−sgn(x−1)h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)
=0−1=−1= 0 - 1 = - 1

& if

x→1−x \to {1^ - }

then

h(x−1)=2[x]−2−sgn(x−1)h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)
=−2+1=−1= - 2 + 1 = - 1

∴\therefore

lim⁡x→1+g(h(x−1))=lim⁡x→1+sin⁡(h(x+1)+1)h(x−1)+1=1\mathop {\lim }\limits_{x \to {1^ + }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ + }} {{\sin (h(x + 1) + 1)} \over {h(x - 1) + 1}} = 1
lim⁡x→1−g(h(x−1))=lim⁡x→1−sin⁡(h(x−1)+1)h(x−1)+1=1\mathop {\lim }\limits_{x \to {1^ - }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ - }} {{\sin (h(x - 1) + 1)} \over {h(x - 1) + 1}} = 1
Q122
Suppose f:R→(0,∞)f: \mathbb{R} \rightarrow(0, \infty) be a differentiable function such that 5f(x+y)=f(x)⋅f(y),∀x,y∈R5 f(x+y)=f(x) \cdot f(y), \forall x, y \in \mathbb{R}. If f(3)=320f(3)=320, then ∑n=05f(n)\sum\limits_{n=0}^{5} f(n) is equal to :
A 6875
B 6525
C 6575
D 6825
Correct Answer
Option D
Solution
5f(x+y)=f(x).f(y)5f(x + y) = f(x).f(y)
5f(3)=f(1).f(2)5f(3) = f(1).f(2)
5f(2)=(f(1))25f(2) = {(f(1))^2}
f(10)=5f(10) = 5
f(1)=20f(1) = 20
⇒f(1).(f(1))25=1600\Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600
∑n=05f(n)=f(0)+20+80+320+1280+5120\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120}
=1750+5120=6825= 1750 + 5120 = 6825
Q123
Let x=2x=2 be a root of the equation x2+px+q=0x^2+px+q=0 and f(x)={1−cos⁡(x2−4px+q2+8q+16)(x−2p)4,x≠2p0,x=2pf(x) = \left\{ \begin{array}{ll}{{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x \ne 2p} \\ {0,} & {x = 2p} \end{array} \right. Then lim⁡x→2p+[f(x)]\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)], where [.]\left[ . \right] denotes greatest integer function, is
A 2
B 1
C 0
D −1-1
Correct Answer
Option C
Solution

lim⁡x→2+(1−cos⁡(x2−4px+q2+8q+16)(x2−4px+q2+8q+16)2)((x2−4px+q2+8q+16)2(x−2p)2)\lim \limits_{x \rightarrow 2^{+}}\left(\dfrac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\dfrac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right) lim⁡h→012((2p+h)2−4p(2p+h)+q2+82+16h2)2=12\lim \limits_{h \rightarrow 0} \dfrac{1}{2}\left(\dfrac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\dfrac{1}{2} Using L'Hospital's lim⁡x→2p+[f(x)]=0\lim _{x \rightarrow 2 p^{+}}[f(x)]=0

Q124
If the function f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 is continuous at x = {\pi \over 2},then, then 9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$$ is equal to
A 11
B 10
C 8
D 2e4^4 + 8
Correct Answer
Option B
Solution
lim⁡x→π−2(1+∣cos⁡x∣)λ∣cos⁡x∣=eλ\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^ \frac{\lambda}{|\cos x|}=e^\lambda

And,

lim⁡x→π+2ecot⁡6xcot⁡4x=lim⁡x→π+2esin⁡4x.cos⁡6xsin⁡6x.cos⁡4x=e23\begin{aligned} &\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}\\\\ &= \mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} {e^{{{\sin 4x.\cos 6x} \over {\sin 6x.\cos 4x}}}} \\\\ & =e^{\frac{2}{3}} \end{aligned}

Also,

f(π/2)=μf(\pi / 2)=\mu

For continuous function, e2/3=eλ=μ \mathrm{e}^{2 / 3}=\mathrm{e}^\lambda=\mu

λ=23,μ=e2/3\lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3}

∴\therefore 9λ+6ln⁡μ+μ6−e6λ9 \lambda+6 \ln \mu+\mu^{6}-e^{6 \lambda} =6+4+e4−e4=10=6+4+e^{4}-e^{4}=10

Q125
The value of lim⁡n→∞1+2−3+4+5−6 + ..... + (3n−2)+(3n−1)−3n2n4+4n+3−n4+5n+4\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }} is :
A 322{3 \over {2\sqrt 2 }}
B 32(2+1){3 \over 2}(\sqrt 2 + 1)
C 3(2+1)3(\sqrt 2 + 1)
D 2+12{{\sqrt 2 + 1} \over 2}
Correct Answer
Option B
Solution
I=lim⁡n→∞(1+2+3+…+3n)−2(3+6+9+..+3n)2n4+4n+3−n4+5n+4=lim⁡n→∞3n(3n+1)2−6n(n+1)2(2n4+4n+3−n4+5n+4)=lim⁡n→∞3n(n−1)[2n4+4n+3+n4+5n+4]2⋅[(2n4+4n−3)−(n4+5n+4)]=lim⁡n→∞3⋅1⋅(1−1n)[2+4n3+3n4+1+5n3+4n4]2[1−1n3−7n4]=3(2+1)2\begin{aligned} & I=\lim _{n \rightarrow \infty} \frac{(1+2+3+\ldots+3 n)-2(3+6+9+. .+3 n)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\\\ & =\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)} \\\\ & =\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]} \\\\ & =\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]} \\\\ & =\frac{3(\sqrt{2}+1)}{2} \\ & \end{aligned}
Q126
The set of all values of aa for which lim⁡x→a([x−5]−[2x+2])=0\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0, where [α\alpha] denotes the greatest integer less than or equal to α\alpha is equal to
A [−7.5,−6.5][-7.5,-6.5]
B (−7.5,−6.5](-7.5,-6.5]
C [−7.5,−6.5)[-7.5,-6.5)
D (−7.5,−6.5)(-7.5,-6.5)
Correct Answer
Option D
Solution
lim⁡x→a([x−5]−[2x+2])=0\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0
⇒lim⁡x→a([x]−5−[2x]−2)=0\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0
⇒lim⁡x→a[x]−[2x]−7=0\Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0

Case 1 : If

a∈[n,n+12)a \in \left[ {n,n + {1 \over 2}} \right)

then

lim⁡x→a([x]−[2x]−7)=0\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0
⇒n−2n−7=0\Rightarrow n - 2n - 7 = 0
⇒n=−7\Rightarrow n = - 7

∴\therefore

a∈[−7,−6.5)a \in \left[ { - 7, - 6.5} \right)

Case 2 : If

a∈[n+12,n+1)a \in \left[ {n + {1 \over 2},n + 1} \right)

then

lim⁡x→a([x]−[2x]−7)=0\mathop {\lim }\limits_{x \to a} \left( {[x] - [2x] - 7} \right) = 0
⇒n−(2n+1)−7=0\Rightarrow n - (2n + 1) - 7 = 0
⇒−n−8=0\Rightarrow - n - 8 = 0
⇒n=−8\Rightarrow n = - 8

∴\therefore

a∈[−7.5,−7)a \in \left[ { - 7.5, - 7} \right)
R.H.L=lim⁡x→−7.5+([x]−[2x]−7)R.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ + }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right)
=−8−(−15)−7= - 8 - \left( { - 15} \right) - 7
=−8+15−7=0= - 8 + 15 - 7 = 0
L.H.L=lim⁡x→−7.5−([x]−[2x]−7)=−8−(−16)−7=−8+16−7=1\begin{aligned}& L.H.L = \mathop {\lim }\limits_{x \to - {{7.5}^ - }} \left( {\left[ x \right] - \left[ {2x} \right] - 7} \right) \\ \\ & = - 8 - \left( { - 16} \right) - 7 \\ \\ & = - 8 + 16 - 7 = 1\end{aligned}

∴\therefore R.H.L ≠\ne L.H.L ∴\therefore At x = -7.5, limit does not exists. ∴\therefore

a∈(−7.5,−6.5)a \in \left( { - 7.5, - 6.5} \right)
Q127
Let f(x)={x2sin⁡(1x), x≠00, x=0f(x) = \left\{ \begin{array}{ll}{{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \\ 0 & {,\,x = 0} \end{array} \right. Then at x=0x=0
A ff is continuous but f′f' is not continuous
B ff and f′f' both are continuous
C ff is continuous but not differentiable
D f′f' is continuous but not differentiable
Correct Answer
Option A
Solution

Given,

f(x)={x2sin⁡(1x),x≠00,x=0f(x) = \left\{ \begin{array}{ll}{{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \\ {0,} & {x = 0} \end{array} \right.

∴\therefore

f′(x)=2xsin⁡(1x)−cos⁡(1x)f'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)

Now,

lim⁡x→0f′(x)=lim⁡x→0[2xsin⁡(1x)−cos⁡(1x)]\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]
=0−lim⁡x→0cos⁡(1x)= 0 - \mathop {\lim }\limits_{x \to 0} \cos \left( {{1 \over x}} \right)

= Does not exist ∴\therefore

f′(x)f'(x)

is discontinuous function at

x=0x = 0

. L.H.D.

=lim⁡h→0+f(0−h)−f(0)−h= \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}
=lim⁡h→0+f(−h)−h= \mathop {\lim }\limits_{h \to {0^ + }} {{f( - h)} \over { - h}}
=lim⁡h→0+−h2sin⁡(1h)−h=0= \mathop {\lim }\limits_{h \to {0^ + }} {{ - {h^2}\sin \left( {{1 \over h}} \right)} \over { - h}} = 0

R.H.D.

=lim⁡h→0+f(0+h)−f(0)h= \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}
=lim⁡h→0+f(h)h= \mathop {\lim }\limits_{h \to {0^ + }} {{f(h)} \over h}
=lim⁡h→0+h2sin⁡(1h)h=0= \mathop {\lim }\limits_{h \to {0^ + }} {{{h^2}\sin \left( {{1 \over h}} \right)} \over h} = 0

∴\therefore L.H.D. = R.H.D.

⇒f(x)\Rightarrow f(x)

is differentiable at

x=0x = 0

. So, f(x) is continuous.

Q128
Let [x][x] denote the greatest integer function and f(x)=max⁡{1+x+[x],2+x,x+2[x]},0≤x≤2f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2. Let mm be the number of points in [0,2][0,2], where ff is not continuous and nn be the number of points in (0,2)(0,2), where ff is not differentiable. Then (m+n)2+2(m+n)^{2}+2 is equal to :
A 3
B 6
C 2
D 11
Correct Answer
Option A
Solution
 Let g(x)=1+x+[x]={1+x;x∈[0,1) 2+x;x∈[1,2)5;x=2λ(x)=x+2[x]={x;x∈[0,1)x+2;x∈[1,2)6;x=2r(x)=2+xf(x)={2+x;x∈[0,2)6;x=2\begin{aligned} & \text{ Let } g(x)=1+x+[x]=\left\{\begin{array}{cc} 1+x ; & x \in[0,1) \\\ 2+x ; & x \in[1,2) \\ 5 ; & x=2 \end{array}\right. \\\\ & \lambda(x)=x+2[x]=\left\{\begin{array}{cc} x ; & x \in[0,1) \\ x+2 ; & x \in[1,2) \\ 6 ; & x=2 \end{array}\right. \\\\ & r(x)=2+x \\\\ & f(x)=\left\{\begin{array}{cc} 2+x ; & x \in[0,2) \\ 6 ; & x=2 \end{array}\right. \end{aligned}

f(x)\mathrm{f}(\mathrm{x}) is discontinuous only at x=2⇒m=1x=2 \Rightarrow \mathrm{m}=1 f(x)\mathrm{f}(\mathrm{x}) is differentiable in (0,2)⇒n=0(0,2) \Rightarrow \mathrm{n}=0

(m+n)2+2=3(m+n)^2+2=3
Q129
If lim⁡x→0eax−cos⁡(bx)−cxe−cx21−cos⁡(2x)=17\lim\limits_{x \rightarrow 0} \dfrac{e^{a x}-\cos (b x)-\dfrac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17, then 5a2+b25 a^{2}+b^{2} is equal to
A 64
B 68
C 72
D 76
Correct Answer
Option B
Solution

The Taylor series for exe^x, cos⁡x\cos x, and e−xe^{-x} around x=0x=0 are:

eax=1+ax+(ax)22!+…,e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,
cos⁡(bx)=1−(bx)22!+…,\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,
e−cx=1−cx+(cx)22!−….e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.

Substituting these into the limit and simplifying:

lim⁡x→0(1+ax+(ax)22!)−(1−(bx)22!)−cx2(1−(cx)+(cx)22!)1−cos⁡2x.\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.

This simplifies to:

lim⁡x→0(a−c2)x+(a2+b2+c22)x2+…(1−cos⁡2x(2x)2)×4x2.\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.

Let's evaluate this expression:

(1−cos⁡2x(2x)2)×4x2.(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.

We know that 1−cos⁡2x1 - \cos 2x can be rewritten using the double-angle formula for cosine, which is cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2x.

So, 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^2x.

Substituting this into our expression we get:

(2sin⁡2x(2x)2)×4x2.(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.

This simplifies to:

12{1 \over 2}
(sin⁡xx)2×4x2=12×4x2=2x2.(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.

So,

(1−cos⁡2x(2x)2)×4x2=2x2.(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.

∴\therefore

lim⁡x→0(a−c2)x+(a2+b2+c22)x2+…2x2=17.\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.

Now, for the limit to exist, the coefficient of xx in the numerator must be zero, as the limit would be undefined otherwise.

This gives us the equation:

a−c2=0⇒c=2a.a-\frac{c}{2} = 0 \Rightarrow c=2a.

Then, for the limit to equal 1717, the coefficient of x2x^2 in the numerator must equal 17×2=3417 \times 2 = 34.

This gives us the equation:

a2+b2+c22=34⇒a2+b2+c2=68.\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.

Since c=2ac=2a, we can substitute cc in the equation to get:

a2+b2+(2a)2=68⇒5a2+b2=68.a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.
Q130
Let ff and gg be two functions defined by f(x)=\left\{\begin{array}{cc}x+1, & x Then (g \circ f)(x)$$ is :
A continuous everywhere but not differentiable at x=1x=1
B differentiable everywhere
C not continuous at x=−1x=-1
D continuous everywhere but not differentiable exactly at one point
Correct Answer
Option D
Solution
 Sol. f(x)={x+1,x<0 1−x,0≤x<1x−1,1≤xg(x)={x+1,x<01,x≥0g(f(x))={x+2,x<−11,x≥−1\begin{aligned} & \text{ Sol. } f(x)=\left\{\begin{array}{c} x+1, x<0 \\\ 1-x, 0 \leq x<1 \\ x-1,1 \leq x \end{array}\right. \\\\ & g(x)=\left\{\begin{array}{c} x+1, x<0 \\ 1, x \geq 0 \end{array}\right. \\\\ & g(f(x))=\left\{\begin{array}{c} x+2, x<-1 \\ 1, x \geq-1 \end{array}\right. \end{aligned}

∴g(f(x))\therefore \mathrm{g}(\mathrm{f}(\mathrm{x})) is continuous everywhere g(f(x))\mathrm{g}(\mathrm{f}(\mathrm{x})) is not differentiable at x=−1\mathrm{x}=-1 Differentiable everywhere else.

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