IISER Mathematics - Continuity
Exam Duration: 45 Mins Total Questions : 30
\(f(x)=\begin{cases} |x-a|sin\frac { 1 }{ x-a } ,if\quad x\neq 2 \\ 0,\quad \quad \quad \quad \quad \quad if\quad x=a \end{cases}\)
- (a)
continuous at x=a
- (b)
discontinuous at x=a
- (c)
discontinuous for all \(x \epsilon R\)
- (d)
None of the above
Now, LHL=\(\begin{matrix} lim \\ x\rightarrow { a }^{ - } \end{matrix}f(x)=\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}f(a-h)\)
=\(\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}|a-h-a|sin\left[ \frac { 1 }{ (a-h-a) } \right] \)
=\(\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}|-h|sin\left( \frac { -1 }{ h } \right) \)
=\(\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}h\quad sin1/h=0\times sin\frac { 1 }{ 0 } \)=0 x finite value=0
RHL=\(\begin{matrix} lim \\ x\rightarrow { a }^{ - } \end{matrix}f(x)=\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}(a+h)\)
=\(\begin{matrix} lim \\ h\rightarrow 0 \end{matrix}|a-h-a|sin\left[ \frac { 1 }{ (a-h-a } \right] \)
and f(x)=0
Thus, LHL=RHL=f(a)=0
Hence, the function is continuous at x=a.
If \(f(x)=\frac { { \left( { 3 }^{ x }-1 \right) }^{ 2 } }{ sinx.log_{ e }(1+x) } ,x\neq 0\) is continuous at x=o, then f(o) is
- (a)
loge3
- (b)
2loge3
- (c)
(loge3)2
- (d)
None of the above
Since, f(x) to be continuous at x=0.
\(\therefore \quad \quad \quad \underset { x\rightarrow 0 }{ lim } f(x)=f(0)\Rightarrow \underset { x\rightarrow { 0 } }{ lim } \frac { { \left( { 3 }^{ x }-1 \right) }^{ 2 } }{ sin\quad x.log_{ e }(1+x) } =f(0)\)
\(\Rightarrow f(0)=\underset { x\rightarrow 0 }{ lim } \frac { { \left( \frac { { 3 }^{ x }-1 }{ x } \right) }^{ 2 } }{ \frac { sinx }{ x } .\frac { log_{ e }(1+x) }{ x } } ={ (log_{ e }3) }^{ 2 }\)
If the function \(f(x)=\begin{cases} x+{ a }^{ 2 }\sqrt { 2 } sinx,\quad 0\le x<\frac { \pi }{ 4 } \\ x\quad cot\quad x+\_ b,\quad \frac { \pi }{ 4 } \le x<\frac { \pi }{ 2 } \\ b\quad sin\quad 2x-a\quad cos2x,\quad \frac { \pi }{ 4 } \le x\le \pi \end{cases}\)is continuous in the interval [0,\(\pi\)]then the values of (a, b) are
- (a)
(0,0)
- (b)
\((0,{1\over2})\)
- (c)
(0,1)
- (d)
(-1,1)
At x=\(\frac { \pi }{ 4 } \),
\(\lim _{ x\rightarrow \frac { { \pi }^{ - } }{ 4 } }{ f(x) } =\lim _{ x\rightarrow \frac { { \pi }^{ - } }{ 4 } }{ (x+{ a }^{ 2 }\sqrt { 2 } sinx)={ a }^{ 2 }+\frac { \pi }{ 4 } } \)
and \(\lim _{ x\rightarrow \frac { { \pi }^{ - } }{ 4 } }{ f(x) } =\lim _{ x\rightarrow \frac { { \pi }^{ + } }{ 4 } }{ (x\quad cot\quad x+b)=b+\frac { \pi }{ 4 } } \)
Since, f(x) is continuous at x=\(\frac { \pi }{ 4 } \)
∴ a2+\(\frac { \pi }{ 4 } \)=b+\(\frac { \pi }{ 4 } \) ⇒ a2=b ......(i)
\(\lim _{ x\rightarrow \frac { { \pi }^{ + } }{ 2 } }{ f(x) } =\lim _{ x\rightarrow \frac { { \pi }^{ 4 } }{ 2 } }{ (b\quad sin2x-a\quad cos2x)=a } \)
Since, f(x) is continuous at x=\(\frac { \pi }{ 2 } \)
and \(\lim _{ x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ f(x) } =\lim _{ x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ (x\quad cotx+b) } =b\)
Since, f(x) is continuous at x=\(\frac { \pi }{ 2 } \).
∴ a=b ..........(ii)
From Eqs (i) and (ii), we get (a,b)=(0,0) and (1,1)
If \(f(x) = {-b \pm \sqrt{b^2-4ac} \over 2a}f(x)=\left\{ { log }_{ e }\left( \frac { \sqrt { 9-{ x }^{ 2 } } }{ 2-x } \right) \right\}\) then
- (a)
range of f(x) is y\(\epsilon\)(-1,1)
- (b)
at x=-3, right hand limit of y=(x-3)f(x)=0
- (c)
domain of f(x) is x\(\epsilon\)(-3,2)
- (d)
f(x) is continuous at \(x\neq0\)
f(x) is defined for 9-x2>0 and 2-x>0 ⇒ xદ(-3,2)
∵ Domain of f(x)=(-3,2). Also, range of f(x)=[-1,1]
∵ \(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =f(0)\)
So, f(x) is continous at x=0
Again,
\(\lim _{ x\rightarrow { -3 }^{ + } }{ (x-3)f(x) } =\lim _{ h\rightarrow { 0 } }{ (h-6) } sin\left[ log_{ e }\left\{ \frac { \sqrt { 9-(-3+h)^{ 2 } } }{ 2-(-3+h) } \right\} \right] \)
=\(\lim _{ h\rightarrow { 0 } }{ (h-6)sin\left[ log_{ e }\left\{ \frac { \sqrt { h(6-h) } }{ 5-h } \right\} \right] } \)
=(h-6) x (an oscillating number between -1 and 1)
So , \(\lim _{ x\rightarrow -3^{ + } }{ (x-3) } \) . f(x) does not exist.
The function \(f:R-\left\{ 0 \right\} \rightarrow R\) given by \(f(x)={1\over x}-{2\over e^{2x}-1}\)can be made continuous at x=0 by defining f(0) as
- (a)
2
- (b)
-1
- (c)
0
- (d)
1
Clearly,\(\underset { x\rightarrow 0 }{ lim } \left( \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 } \right) =\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ 2x }-1-2x }{ x({ e }^{ 2x }-1) } \)
\(=\underset { x\rightarrow 0 }{ lim } \frac { 2{ e }^{ 2x }-2 }{ ({ e }^{ 2x }-1)+2xe^{ 2x } } \) [using L'Hospital's rule]
\(=\underset { x\rightarrow 0 }{ lim } \frac { 4{ e }^{ 2x } }{ 4{ e }^{ 2x }+4e^{ 2x } } =1\) [using L'Hospital's rule]
Now, f(x) is continuous at x=0, \(\underset { x\rightarrow 0 }{ lim } \)f(x)=f(0)\(\Rightarrow\)1=f(0)
If \(f(x)=\left| x-1 \right| +\left| x-3 \right| +\left| 5-x \right| ,\quad \forall x\in R\)
f(x) is decreases, then \(x\in \)
- (a)
\((-\infty ,1)\)
- (b)
\((-\infty ,3)\)
- (c)
\((-\infty ,5)\)
- (d)
(3,5)
\(f^{ ' }(x)<0,\quad x\in (-\infty ,3)\)
If \(f(x)=\left| x-1 \right| +\left| x-3 \right| +\left| 5-x \right| ,\quad \forall x\in R\)
The set of values of a such that equation \(\left| f(x) \right| -a=0\) hs no solution
- (a)
(4,5)
- (b)
\((4,\infty )\)
- (c)
\((-\infty ,4)\)
- (d)
(5,6)
\(\because \quad f(x)>0\)
\(\therefore \quad \left| f(x) \right| =f(x)\)
for no.solution, a<4.
Suppose a, b, c, d, be non-zero real numbers and ab>0,
and \(\int _{ 0 }^{ 1 }{ (1+e^{ x^{ 2 } })({ ax }^{ 3 }+{ bx }^{ 2 }+cx+d)dx=\int _{ 0 }^{ 2 }{ (1+e^{ x^{ 2 } })({ ax }^{ 3 }+{ bx }^{ 2 }+cx+d)dx=0 } } \\ \)
Which statement is correct?
- (a)
ac>0
- (b)
ac<0
- (c)
ad<0
- (d)
None of these
\(\because \quad 1+{ e }^{ x^{ 2 } }>0,\quad { ax }^{ 3 }+{ bx }^{ 2 }+cx+d=0\)
For some values of x between 0 and 1 and 1 and 2
Now, let a2x3+abx2+acx+ad=0 has three real roots
Now, 3a2x2+2abx+ac=0 has two real roots
\(Its\quad roots=\frac { -2ab\pm \sqrt { 4{ a }^{ 2 }{ b }^{ 2 }-12{ a }^{ 2 }(ac) } }{ 6a^{ 2 } } \quad \quad ....(i)\)
obviously both roots are real, when ac<0 and one of there is -ve
Let f and g .be non-increasing and non-decreasing function respectively from [0,\(\infty \)) to [0,\(\infty \)) and h(x)=f(g(x), h(0)=0, then in [0,\(\infty \)), h(x)-h(1) is
- (a)
<0
- (b)
>0
- (c)
=0
- (d)
increasing
\(\because \) f:[0,\(\infty \)), g:[0,\(\infty \))⟶[0,\(\infty \))
and h(x)=f(g(x))
\(\because \) h:[0,\(\infty \))⟶[0,\(\infty \))
\(\therefore \) h:[0,\(\infty \))⟶[0,\(\infty \))
\(\therefore \) h'(x)=f'(g(x))g'(x)\(\le \)0
\(\Rightarrow \) h(x) is monotonically decreasing function
\(\because \quad h(x)\le 0\quad \quad \quad (\because h(0)=0)\)
\(\therefore \quad h(x)\le 0\quad in\quad [0,\infty )\quad ....(i)\)
Also, \(h:[0,\infty )\rightarrow [0,\infty )\)
\(h(x)\ge 0\quad in\quad [0,\infty )\quad \quad ...(ii)\)
From Eqs. (i) and (ii), h(x)=0 in [0,\(\infty \))
\(\therefore \) h(x)-h(1)=0-0=0
In the function f(x)=ax3+bx2+11x-6 satisfies conditions of Rolle's theorem in [1,3] and \(f^{ ' }\left( 2+\frac { 1 }{ \sqrt { 3 } } \right) =0\), then value of a and b are respectively.
- (a)
1,-6
- (b)
-1,6
- (c)
-2,1
- (d)
-1,1/2
∵ f(1)=f(3)
⇒ a+b+11-6=27a+9b+33-6
⇒ 13a+4b=-11 ......(i)
and f'(x)=3ax2+2bx+11
\(\Rightarrow \quad f^{ ' }\left( 2+\frac { 1 }{ \sqrt { 3 } } \right) =3a\left( 2+\frac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }+2b\left( 2+\frac { 1 }{ \sqrt { 3 } } \right) +11=0\)
\(\Rightarrow \quad 3a\left( 4+\frac { 1 }{ 3 } +\frac { 4 }{ \sqrt { 3 } } \right) +2b\left( 2+\frac { 1 }{ \sqrt { 3 } } \right) +11=0\quad \quad ...(ii)\)
From Eqs. (i) and (ii), we get a=1, b=-6
The function \(f(x)=\frac { In(\pi +x) }{ In\quad (e+x) } \) is
- (a)
increasing on [0,∞)
- (b)
decreasing on [0,∞)
- (c)
increasing on [0,\(\pi \)/e) and decreasing on [\(\pi \)/e,∞)
- (d)
decreasing on [0,\(\pi \)/e) and increasing on [\(\pi \)/e,∞)
Since, \(f(x)=\frac { In\quad (\pi +x) }{ In\quad (e+x) } \)
\({ f }^{ ' }(x)=\frac { \frac { In\quad (e+x) }{ (\pi +x) } -\frac { In\quad (\pi +x) }{ (e+x) } }{ \{ In\quad (e+x)\} ^{ 2 } } \)
\(=\frac { (e+x)\quad In\quad (e+x)-(\pi +x)\quad In\quad (\pi +x) }{ (\pi +x)(e+x)[In\quad (e+x)]^{ 2 } } <0\)
\(x\ge 0\)
Hence, f(x) is decreasing function for x\(\in \)[0,∾)
The function f(x)=tanx-x
- (a)
always increases
- (b)
always decreases
- (c)
never decreases
- (d)
some times increases and some times decreases
f(x)=tan x-x
∴ f'(x)=sec2 x-1≥0
If f''(x)>0, ∀x\(\in \)R, f'(3)=0 and g(x)=f(tan2x-2 tan x+4), 0<x<\(\frac { \pi }{ 2 } \), then g(x) is increasing in
- (a)
\(\left( 0,\frac { \pi }{ 4 } \right) \)
- (b)
\(\left( \frac { \pi }{ 6 } ,\frac { \pi }{ 3 } \right) \)
- (c)
\(\left( 0,\frac { \pi }{ 3 } \right) \)
- (d)
\(\left( \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right) \)
g'(x)=f'(tan2 x-2 tan x+4).(2 tan x-2). sec2 x......(i)
∵ f''(x)>0 ⇒ f'(x) is increasing function
so f'(tan2 x-2 tan x+4)=f'(tan x-1)2+3)>f'(3)=0
\(\forall x\in \left( 0,\frac { \pi }{ 4 } \right) \cup \left( \frac { \pi }{ 4 } '\frac { \pi }{ 2 } \right) \)
Also, (tan x-1)>0, for \(x\in \left( \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right) \)
So, g(x) is increasing function in \(\left( \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right) \)
By LMVT, which of the following is true for x>1
- (a)
1+ x In x<1+ln x
- (b)
1+ ln x<x<1+x ln x
- (c)
x,1+x ln x<1+ ln x
- (d)
1+ ln x<1+x ln x<x
Let f(x)=1+x ln x in (1,x)
∴ f'(x)=(1+ln x)>0 (∵ x>1)
By LMVT, for 1<c<x
\(f^{ ' }(c)=\frac { f(x)-f(1) }{ x-1 } =\frac { (1+x\quad lln\quad x)-1 }{ x-1 } =\frac { x\quad lm\quad x }{ x-1 } >0\)
or x ln x>x-1 or x ln x+1>x
Also, similarly by LMVT
Let g(x)=1+ ln x in (1,x)
we get, 1+ ln x<x
Hence, 1+ln x<x<1+x ln x
For all x\(\in \)(0,1)
- (a)
ex<1+x
- (b)
loge(1+x)<x
- (c)
sin x>x
- (d)
loge x>x
Let f(x)=ex-(1+x)
∴ f'(x)=ex-1>0 (∵ 0<x<1)
∴ f(x) is increasing
f(x)>f(0) ⇒ ex-(1+x)>0 ⇒ ex>1+x (∵ x>0)
Now, let g(x)=loge(1+x)-x
\(g^{ ' }(x)=\frac { 1 }{ 1+x } -1=\frac { x }{ 1+x } <0\)
∴ g'(x) is decreasing
⇒ x>0 ⇒ g(x)<g(0) ⇒loge(1+x)-x<0
or loge(1+x)<x
Again, let h(x)=sin x-x ⇒ h'(x)=cos x-1<0
h(x) is decreasing
∵ x>0 ⇒ h(x)<h(0) ⇒ sin x-x<0 ⇒ sin x<x
In last, let t(x)=loge x-x
∴ t'(x)=\(\frac { 1 }{ x } -1=\frac { 1-x }{ x } >0\)
⇒ t(x) is increasing
x<1 ⇒ t(x)<t(1) ⇒ loge x-x<-1
⇒ loge x<x-1
If \(f(x)=2x+cot^{ -1 }x+ln(\sqrt { 1+x^{ 2 } } -x),\)then f(x)
- (a)
increases in (0,∾)
- (b)
decreases in [0,∾)
- (c)
neither increases nor decreases in (0,∾)
- (d)
increases in (-∾,∾)
\(f^{ ' }(x)=2-\frac { 1 }{ 1+{ x }^{ 2 } } +\frac { 1 }{ (\sqrt { 1+x^{ 2 } } -x } \times \left( \frac { x }{ \sqrt { 1+x^{ 2 } } } -1 \right) \)
\(=2-\frac { 1 }{ 1+{ x }^{ 2 } } -\frac { 1 }{ (\sqrt { 1+x^{ 2 } } } \)
\(=\frac { 2+2x^{ 2 }-1-\sqrt { 1+x^{ 2 } } }{ (1+x^{ 2 }) } \)
\(=\frac { x^{ 2 }+\sqrt { (1+x^{ 2 }) } (\sqrt { 1+x^{ 2 } } -1 }{ (1+x^{ 2 }) } >0\)
for all x
f(x) is increasing in (-∾,∾)
and in particular (0,∾)
The interval to which b may belong so that the function
\(f(x)=\left( 1-\frac { \sqrt { 21-4b-b^{ 2 } } }{ b+1 } \right) x^{ 3 }+5x+\sqrt { 6 } \) is increasing at every point of its domain is
- (a)
[-7,-1]
- (b)
[-6,-2]
- (c)
[2,2.5]
- (d)
[2,3]
\(f^{ ' }(x)=3\left( 1-\frac { \sqrt { 21-4b-b^{ 2 } } }{ b+1 } \right) \)x2+5>0(given)
\(\therefore \quad \left( 1-\frac { \sqrt { 21-4b-b^{ 2 } } }{ b+1 } \right) \ge 0\)
\(\Rightarrow \quad \frac { \sqrt { 21-4b-b^{ 2 } } }{ b+1 } \le 1\quad ...(i)\)
and 21-4b-b2≥0 ⇒ -7≤b≤3 ...(ii)
Case I: If b+1<0 ie, b<-1 ...(iii)
Then, inequality (ii) is always true, then from (ii) and (iii) we get,
-7≤b<-1 ....(iv)
Case II: If b+1>0 ie, b>-1 ....(v)
From Eq (i), \(\sqrt { 21-4b-b^{ 2 } } \le (b+1)\)
⇒ 21-4b-b2≤+2b+1
⇒ 2b2+6b-20≥0
⇒ b2+3b-10≥0
⇒ (b+5)(b-2)≥0
\(b\in (-\infty ,-5)\cup [2,\infty )\)
From Eqs (v) and (vi), we get
b≥2 ...(vii)
Combining Eqs (iv) and (vii), we get
\(b\in [-7,-1)\cup [2,\infty )\)
Let f be a function satisfying f(x + y) = f(x) + f(y) and f(x) = X2 g(x) for all x and y, where g(x) is a continuous function, then f' (x) is equal to
- (a)
g'(x)
- (b)
g(0)
- (c)
g(0)+g'(x)
- (d)
0
\(f'(x)=\underset{h\rightarrow 0}{lim}{f(x+h)-f(x)\over h}\)
\(=\underset{h\rightarrow 0}{lim}{f(x)+f(h)-f(x)\over h}\)
\(=\underset{h\rightarrow 0}{lim}{f(h)\over h}=\underset{h\rightarrow 0}{lim}{h^2g(h)\over h}=0\)
Let f: R ⟶ R be a differentiable function and f(1) = 4.then the value of \(\underset{h\rightarrow 0}{lim}\int_4^{f(x)}{2t\over x-1}dt\ is\)
- (a)
8f' (1)
- (b)
4f' (1)
- (c)
2f' (1)
- (d)
f' (1)
\(\underset{x\rightarrow 0}{lim}{{\{ \ t^2\}}_4^{f(x)}\over (x-1)}=\underset{x\rightarrow 0}{lim}{\{ f(x)\}^2-16\over (x-1)}\)
\(=\underset{x\rightarrow 0}{lim}{2f(x)f'(x)-0\over 1}=2f(1)f'(1)\)
= 2 x 4 x f' (1)
= 8f' (1)
If the derivative of the function \(f(x)=\begin{cases} bx^2 + ax + 4; x \ge-1\\ ax^2 + b ; x < - 1' \end{cases}\)is everywhere continuous then
- (a)
a = 2, b = 3
- (b)
a = 3, b = 2
- (c)
a = - 2, b = - 3
- (d)
a = - 3, b = - 2
We have \(f(x)=\begin{cases} bx^2 + ax + 4; x \ge-1\\ ax^2 + b ; x < - 1' \end{cases}\)
\(f'(x)=\begin{cases}2ax, x<-1 \\2bx+a, x\ge-1 \end{cases}\)
Since, f(x) is difterentiable at x = - 1, therefore it is continuous at x = - 1and hence,
\(\underset{x\rightarrow-1-}{lim}f(x)=\underset{x\rightarrow-1+}{lim}f(x)\)
a+b=b-a+4
a= 2
and also \(\underset{x\rightarrow-1-}{lim}f(x)=\underset{x\rightarrow-1+}{lim}f(x)\)
-2a=-2b+a
3a=2b ⇒ b=3
Hence, a=2,b=3
If \(f(x)-\int_{-1}^x|t|dt, x\ge-1,\) then
- (a)
f and f' are continuous for x + 1>
- (b)
f is continuous but f' is not so for x + 1> a
- (c)
f and f' are continuous at x = a
- (d)
f is continuous at x = a but f' is not so
We have
\(f(x)=\int_{-1}^{x}|t|dt,x\ge -1\)
f'(x)=|X|
f' (x) is a linear function hence f(x) and I' (x) are continuous for x + 1 > O.
The function \(f(x)={e^{tan\ x-1}-1\over e^{tan\ x}+1}\) discontinuous at x is equal to
- (a)
nπ+π, n ∈ I
- (b)
nπ+π/2,n∈I
- (c)
nπ+π/4,n∈I
- (d)
nπ+π/8,n∈I
For x =nπ+π/2,n∈I f(x)=∞
\(f(x)={tan([x]\pi)\over [1+|In(sin^2 \ x+1 )|]}\)where [.] denotes the greatest integer function, then f(x) is
- (a)
continuous ∀ x ∈ R
- (b)
discontinuous ∀ x ∈ R
- (c)
non-differentiable ∀ x ∈ R
- (d)
a periodic function with fundamental period not defined
f(x)=0 [∴tan [x] n = 0]
is a constant function
Let \(f(x)=\begin{cases} {a(1-x\ sin\ x)+b\ cos\ x+5\over x^2}, x<0\\ 3, x=0\\ \left\{1+\left(cx+dx^3\over x^2\right)\right\}^{1/x}, x>0 \end{cases}\) If is continuous at x = 0
The value of b is
- (a)
-1
- (b)
In 3
- (c)
0
- (d)
-4
\(Let\ \underset{x\rightarrow0-}{lim} f(x) = \underset{h\rightarrow0}{lim} f(0 - h)\)
\(= \underset{h\rightarrow0}{lim} {a(1-h\ sin\ h)+b\ cos\ h+5\over h^2}\)
f(x) is continuous
at h ➝ 0, Numerator must be = 0
ie, a+b+5=0 ... (i)
Now, from Eq. (i), -1 + b + 5 = 0
⇒ b=-4
Let \(f(x)=\begin{cases}x+a, x<0\\ |x-1|, x\ge0 \end{cases}\)and \(g(x)=\begin{cases}x+1, x<0\\ (x1)^2+b, x\ge0 \end{cases}\) Where, n is odd natural number> 1and f' (0) > 0 where a and b are non-negative real numbers.
The value of b, if (gof) x is continuous for all real x, is
- (a)
-1
- (b)
0
- (c)
1
- (d)
2
(gof)x is continuous
g(x) and f(x) are also continuous.
Then, \(f(0) =\underset{x\rightarrow0-} {lim} f(x) = \underset{h\rightarrow0}{lim} f(0 - h)\)
\(1=\underset{h\rightarrow0}{lim} (0-h+a)=a\)
a=1
and \(g(0) =\underset{x\rightarrow0-} {lim} g(x) = \underset{h\rightarrow0}{lim} g(0 - h)= \underset{h\rightarrow0}{lim}(0+h+1)\)
1+b=1
b=0
For these values of a and b, (gof)x ∀ x ∈ ( - 1, 1) is
- (a)
even
- (b)
odd
- (c)
neither even nor odd
- (d)
none of these
From above,
(gof)x = X2 ∀ X ∊ (-1,1)
which is even
Let f(x) = \(\underset{n\rightarrow\infty}{lim}\) (sin x)2n, then f is
- (a)
continuous at x = π / 2
- (b)
discontinuous at x = π / 2
- (c)
discontinuous at x = -π / 2
- (d)
discontinuous at an infinite number of points
Since, \(\underset{n\rightarrow\infty}{lim}x^{2m}=\begin{cases}0, if |x| <1\\ 1,\ if |x|=1\end{cases}\)
\(f(x)=\underset{n\rightarrow\infty}{lim}x^{2m}(sin\ x)^{2n}\)
\(\underset{n\rightarrow\infty}{lim}x^{2m}=\begin{cases}0, if |sin\ x| <1\\ 1,\ if |sin\ x|=1\end{cases}\)
This shows that f(x) is continuous ∀ x, except possibility when |sin x | = 1 ie, when
\(x=(2k+1){\pi\over 2}(k\epsilon I)\)
For these points
\(\underset{x\rightarrow(2k+1){\pi\over 2}}{lim\ f(x)}=0\neq1=f\left[(2k+1){\pi\over 2}\right]\)
⇒ f(x) is discontinuous for these points.
\(f(xd)={2-(256-7x)^{1/8}\over (5x+32)^{1/5}}(x\neq0);\) then for f to be continuous every where, f(0) is equal to
- (a)
-1
- (b)
1
- (c)
26
- (d)
none of htese
f(0) = RHL of f(x) at x = 0
\(=\underset{x\rightarrow0^+}{lilm}f(x)\)
\(=\underset{h\rightarrow0}{lilm}f(0+h)\)
\(=\underset{h\rightarrow0}{lilm}{2-(256-7h)^{1/8}\over (5h+32)^{1/5}-2}\)
\(=\underset{h\rightarrow0}{lilm}{(256)^{1/8}-(256-7h)^{1/8}\over (5h+32)^{1/5}-(32)^{1/5}}\)
\(=\underset{h\rightarrow0}{lilm}={{(256)^{1/8}-(256-7h)^{1/8}\over256-(256-7h)}\times7h}\over{(5h+32)^{1/5}-(32)^{1/5}\over (5h+32)-32}\times5h\)
\(={7\over 5}{{1\over 8}(256)^{-7/8}\over {1\over 5}(32)^{-4/5}}\)
\(={7\over 8}.2^{-3}={7\over 64}\)
f \(g(x)=\begin{cases}[f(x)], x\epsilon\left(0,{\pi\over2}\right)\cup\left({\pi\over 2}\pi\right)\\ 3, x=\pi/2 \end{cases}\)where [x] denotes the greatest integer function and \(f(x)={2(sin\ x-sin^n\ x)+|sin\ x-sin^n\ xl\over 2 (sin\ x - sin^n x) -|sin\ x-sin^n x|},n\epsilon R\) then
- (a)
g(x) is continuous and differentiable at
- (b)
g(x) is continuous and differentiable at x =π/2, when a < n < 1
- (c)
g(x) is continuous but not differentiable at x =π/2, when a < n < 1
- (d)
g(x) is continuous but not differentiable, at x =π/2, when a < n < 1
For 0 <n <1, sin x <sinn x
and for n > 1, sin x > sinn x
Now, for 0 < n < 1
\(f(x)={2 (sin\ x - sin^n x) - (sin\ x - sin^n x)\over 2 (sin\ x - sin^n\ x) + (sin\ x - sin^n\ x)}={1\over 3}\)
and for n > 1,
\(f(x)={2 (sin\ x - sin^n x) - (sin\ x - sin^n x)\over 2 (sin\ x - sin^n\ x) + (sin\ x - sin^n\ x)}={3}\)
For n > 1, g(x) = 3, x ∈ (0, π)
g(x) is continuous and differentiable at
\(x={\pi\over 2}\)and for 0 < n < 1,
\(g(x)=\begin{cases} \left[1\over 3\right]=0,x\epsilon\left(0,{\pi\over 2}\right)\cup\left({\pi\over 2},\pi\right)\\ 3, x={\pi\over 2} \end{cases}\)
g(x) is not continuous at \(x={\pi\over 2}\)Hence, g(x) is also not differentiable at \(x={\pi\over 2}\)
If f(x) = {\(\begin{matrix} \frac { 1-cos4x }{ { x }^{ 2 } } , & x<0 \\ a & ,x=0 \\ \frac { \sqrt { x } }{ \sqrt { 16+\sqrt { x } -4 } } , & x>0 \end{matrix}\)
is continuous at x =0, then a =
- (a)
4
- (b)
6
- (c)
8
- (d)
none of these
since f(x) is continuous at x = 0
\(\therefore \) L.H.L. at x = 0 = f(0)= R.H.L. at x = 0
\(\underset { x\longrightarrow 0 }{ lim } \frac { 1-cos4x }{ { x }^{ 2 } } =a=\underset { x\longrightarrow 0 }{ lim } \frac { \sqrt { x } }{ \sqrt { 16+\sqrt { x } -4 } } \)
\(\underset { x\longrightarrow 0 }{ lim } \frac { 4cos4x }{ { 2x } } =a=\underset { x\longrightarrow 0 }{ lim } \sqrt { 16+\sqrt { x } +4 } \)
\(\underset { x\longrightarrow 0 }{ lim } \frac { 4cos4x }{ { 4x } } \times \frac { 4x }{ 2x } =a=\underset { x\longrightarrow 0 }{ lim } \sqrt { 16+\sqrt { x } +4 } \)
\(\therefore \) 8 = a = 8