If $$\alpha > \beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\

Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$ Replace $x \rightarrow \frac{1}{x}$ $$ \frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0 $$ So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots. Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}$ $$ =\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}} $$ $$ \begin{aligned} &

Limits, Continuity and Differentiability — Page 14 | JEE Mathematics

Q131

Let

f(x)=\left[x^{2}-x\right]+|-x+[x]|

, where

x \in \mathbb{R}

and

[t]

denotes the greatest integer less than or equal to

t

. Then,

f

is :

A continuous at

x=0

, but not continuous at

x=1

B continuous at

x=0

and

x=1

C continuous at

x=1

, but not continuous at

x=0

D not continuous at

x=0

and

x=1

Correct Answer

Option C

Solution

We have,

\begin{aligned} f(x) & =\left[x^2-x\right]+|-x+[x]| \\\\ & =[x(x-1)]+\{x\} \end{aligned}

f(x)=\left\{\begin{array}{ccc} x+1 & ; & -0.5 < x < 0 \\ 0 & ; & x=0 \\ -1+x & ; & 0 < x <1 \\ 0 & ; & x=1 \\ x-1 & ; & 1 < x < 1.5 \end{array}\right.

At $x=0$ ,

\begin{array}{r} \text{ LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \\\\ \text{ RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \\\\ f(0)=0 \end{array}

$\therefore f(x)$ is not continuous at $x=0$ At $x=1$

\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\\\ \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\\\ f(1) & =0 \end{aligned}

$\therefore f(x)$ is continuous at $x=1$ Hence, $f(x)$ is continuous at $x=1$ , but not continuous at $x=0$ .

Q132

\lim \limits_{n \rightarrow \infty} \dfrac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}

is equal to :

A

\dfrac{2}{3}

B

\dfrac{1}{2}

C

\dfrac{3}{4}

D

\dfrac{1}{3}

Correct Answer

Option D

Solution

\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}

\begin{aligned} & \text{ Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\ & =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\ & =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n \end{aligned}

We will take term with the greatest power of

n

=\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4

\begin{aligned} & \text{ Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \\ & =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right) \end{aligned}

Greatest power of

n

is

\frac{n^4}{4}

\lim \limits_{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}

Q133

If

\alpha > \beta > 0

are the roots of the equation

a x^{2}+b x+1=0

, and

\lim\limits_{x \rightarrow \dfrac{1}{\alpha}}\left(\dfrac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\dfrac{1}{2}}=\dfrac{1}{k}\left(\dfrac{1}{\beta}-\dfrac{1}{\alpha}\right), \text{ then } \mathrm{k} \text{ is equal to }

:

A

2 \beta

B

\beta

C

\alpha

D

2 \alpha

Correct Answer

Option D

Solution

Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$ Replace $x \rightarrow \dfrac{1}{x}$

\frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0

So, $\dfrac{1}{\alpha}, \dfrac{1}{\beta}$ are the roots.

Now, $\lim\limits_{x \rightarrow \dfrac{1}{\alpha}}\left[\dfrac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right]^{\dfrac{1}{2}}$

=\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^2}\right]^{\frac{1}{2}}

\begin{aligned} & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[\frac{2 \sin ^2 \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{4 \times 2 \alpha^2 \frac{\left(x-\frac{1}{\alpha}\right)^2\left(x-\frac{1}{\beta}\right)^2}{4}\left(x-\frac{1}{\beta}\right)^2}\right]^{\frac{1}{2}} \\\\ & =\lim _{x \rightarrow \frac{1}{\alpha}}\left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}\left(x-\frac{1}{\beta}\right)\right] \\\\ & =\frac{1}{2 \alpha}\left[\frac{-1}{\alpha}+\frac{1}{\beta}\right) \\\\ & \Rightarrow \frac{1}{k}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right]=\frac{1}{2 \alpha}\left[\frac{1}{\beta}-\frac{1}{\alpha}\right] \\\\ & \Rightarrow k=2 \alpha \end{aligned}

Q134

\lim\limits_{x \rightarrow 0}\left(\left(\dfrac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\dfrac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)

is equal to _____________.

A 15

B 18

C 9

D 24

Correct Answer

Option B

Solution

\begin{aligned} & \lim _{x \rightarrow 0}\left[\left(\frac{1-\cos ^2(3 x)}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)\right)^5}\right)\right] \\\\ & =\lim _{x \rightarrow 0}\left[\frac{1-\cos ^2(3 x)}{9 x^2} \times \frac{9 x^2}{\cos ^3(4 x)}\right] \times \frac{\frac{\sin ^3 4 x}{(4 x)^3} \times 64 x^3}{\left[\frac{\log _e(2 x+1)}{2 x}\right]^5 \times(2 x)^5} \\\\ & =\left[\frac{1 \times 9 \times 1}{(1)}\right] \times\left[\frac{1 \times 64}{1 \times 32}\right] \\\\ & =9 \times 2=18 \end{aligned}

Q135

Let

a_{1}, a_{2}, a_{3}, \ldots, a_{\mathrm{n}}

be

\mathrm{n}

positive consecutive terms of an arithmetic progression. If

\mathrm{d} > 0

is its common difference, then

\lim\limits_{n \rightarrow \infty} \sqrt{\dfrac{d}{n}}\left(\dfrac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\dfrac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)

is

A

\dfrac{1}{\sqrt{d}}

B 1

C 0

D

\sqrt{d}

Correct Answer

Option B

Solution

\lim\limits_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)

Now, $\begin{aligned} & \dfrac{1}{\sqrt{a_1}+\sqrt{a_2}}+\dfrac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\\\ = & \dfrac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\dfrac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\ldots+\dfrac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}} \\\\ = & \dfrac{\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+. .+\sqrt{a_n}-\sqrt{a_{n-1}}}{d}\end{aligned}$ $\left(\because a_2-a_1=a_3-a_2=\ldots a_n-a_{n-1}=d\right)$ $\begin{aligned} & =\dfrac{\sqrt{a_n}-\sqrt{a_1}}{d} \\\\ & =\dfrac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\end{aligned}$ $\therefore \lim\limits_{n \rightarrow \infty} \sqrt{\dfrac{d}{n}}\left(\dfrac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\right)$ $\begin{aligned} & =\lim _{n \rightarrow \infty}\left[\dfrac{1}{\sqrt{d}}\left(\dfrac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{n}}\right)\right] \\\\ & =\lim _{n \rightarrow \infty}\left[\dfrac{1}{\sqrt{d}}\left(\sqrt{\dfrac{a_1}{n}+\left(d-\dfrac{d}{n}\right)}-\sqrt{\dfrac{a_1}{n}}\right)\right] \\\\ & =\dfrac{1}{\sqrt{d}}(\sqrt{0+d-0}-\sqrt{0})=\dfrac{\sqrt{d}}{\sqrt{d}}=1\end{aligned}$

Q136

Let

f(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}

. If

\mathrm{m}

and

\mathrm{n}

denote the number of points where

f

is not continuous and not differentiable respectively, then

\mathrm{m}+\mathrm{n}

is equal to :

A 5

B 3

C 2

D 0

Correct Answer

Option B

Solution

$\begin{aligned} & f(x)=\left|2 x^2+5\right| x|-3| \\\\ & \text{ Graph of } y=\left|2 x^2+5 x-3\right|\end{aligned}$ Now $f(x)$ is continuous $\forall x \in R$ but non- differentiable at $x=\dfrac{-1}{2}, \dfrac{1}{2}, 0$

\begin{aligned} \therefore m & =0 \\\\ n & =3 \\\\ m+n & =3 \end{aligned}

Q137

Let

f(x)=\left\{\begin{array}{l}x-1, x \text{ is even, } \\ 2 x, \quad x \text{ is odd, }\end{array} x \in \mathbf{N}\right.

. If for some

\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21

, then

\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\dfrac{|x|^3}{\mathrm{a}}-\left[\dfrac{x}{\mathrm{a}}\right]\right\}

, where

[t]

denotes the greatest integer less than or equal to

t

, is equal to :

A 169

B 121

C 225

D 144

Correct Answer

Option D

Solution

$f(x)=\left\{\begin{array}{l}x-1, x \text{ is even, } \\ 2 x, \quad x \text{ is odd, }\end{array} x \in \mathbf{N}\right.$ Let $a$ is odd

\begin{aligned} & \Rightarrow f(a)=2 a \\\\ & \Rightarrow f(f(a))=2 a-1 \\\\ & \Rightarrow f(f(f(a)))=2(2 a-1) \end{aligned}

$2(2 a-1)=21$ Not possible for any $a \in N$ Let $a$ is even

\begin{aligned} & \Rightarrow f(a)=a-1 \\\\ & \Rightarrow f(f(a))=2(a-1) \\\\ & \Rightarrow f(f(f(a)))=2(a-1)-1=2 a-3 \\\\ & 2 a-3=21 \quad \Rightarrow a=12 \end{aligned}

Now

\begin{aligned} & \lim _{x \rightarrow 12^{-}}\left(\frac{|x|^3}{2}-\left[\frac{x}{12}\right]\right) \\\\ & =\lim _{x \rightarrow 12^{-}} \frac{|x|^3}{12}-\lim _{x \rightarrow 12^{-}}\left[\frac{x}{12}\right] \\\\ & =144-0=144 . \end{aligned}

Q138

The value of

\lim \limits_{n \rightarrow \infty}\left(\sum\limits_{k=1}^n \dfrac{k^3+6 k^2+11 k+5}{(k+3)!}\right)

is :

A 5/3

B 2

C 4/3

D 7/3

Correct Answer

Option A

Solution

\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right) \end{aligned}

\begin{aligned} & =\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{\mathrm{n!}}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots-\frac{1}{(\mathrm{n}+3)!}\right) \\ & \quad=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3} \end{aligned}

Q139

Let

f: \mathbf{R} \rightarrow \mathbf{R}

be defined as :

f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x1\end{cases}

If

f

is continuous everywhere in

\mathbf{R}

and

m

is the number of points where

f

is NOT differential then

\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}

equals :

A 1

B 4

C 3

D 2

Correct Answer

Option D

Solution

At $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ is continuous therefore, $\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$

f(1)=3+c

.........(1)

\begin{aligned} & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\\\ & \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3 .........(2) \end{aligned}

from (1) and (2)

\mathrm{c}=0

at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$ is continuous therefore,

\begin{aligned} & \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right) ........(3) \\\\ & \mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2 ...........(4) \end{aligned}

$\mathrm{f}\left(0^{-}\right)$ has to be equal to 2 $\lim\limits_{h \rightarrow 0} \dfrac{a-b \cos (2 h)}{h^2}$ = $\lim\limits_{h \rightarrow 0} \dfrac{a-b\left\{1-\dfrac{4 h^2}{2 !}+\dfrac{16 h^4}{4 !}+\ldots\right\}}{h^2}$ = $\lim\limits_{h \rightarrow 0} \dfrac{a-b+b\left\{2 h^2-\dfrac{2}{3} h^4 \ldots\right\}}{h^2}$ for limit to exist $a-b=0$ and limit is $2 b$ from (3), (4) and (5)

\mathrm{a}=\mathrm{b}=1

checking differentiability at $\mathrm{x}=0$

\begin{aligned} & \text{ LHD : } \lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^2}-2}{-h} \\\\ & \lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !} \ldots\right)-2 h^2}{-h^3}=0 \\\\ & \text{ RHD : } \lim _{h \rightarrow 0} \frac{(0+h)^2+2-2}{h}=0 \end{aligned}

Function is differentiable at every point in its domain

\therefore \mathrm{m}=0

m + a + b + c = 0 + 1 + 1 + 0 = 2

Q140

Consider the function.

f(x)=\left\{\begin{array}{cc} \frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x3 \\\\ \mathrm{~b} & , x=3, \end{array}\right.

where

[x]

denotes the greatest integer less than or equal to

x

. If

\mathrm{S}

denotes the set of all ordered pairs (a, b) such that

f(x)

is continuous at

x=3

, then the number of elements in

\mathrm{S}

is :

A Infinitely many

B 4

C 2

D 1

Correct Answer

Option D

Solution

f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|} \quad

(for

f(x)

to be cont.)

\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x} Hence

f\left(3^{-}\right)=\frac{-a}{b}

Then

f\left(3^{+}\right)=2^{\lim _\limits{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2

and

f(3)=b

. Hence

\mathrm{f}(3)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3^{-}\right)

\begin{aligned} & \Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}} \\ & \mathrm{b}=2, \mathrm{a}=-4 \end{aligned}

Hence only 1 ordered pair

(-4,2)$$.