JEE Main Mathematics - Limits
Exam Duration: 60 Mins Total Questions : 30
The value of \(\underset{x\rightarrow {\pi\over4}}{lim}{2-cot x-cot^3x\over 1-cot^3x}\)is
- (a)
\(45\over3\)
- (b)
1
- (c)
0
- (d)
None of the above
\(\underset { x\rightarrow 2 }{ lim } \left[ \frac { \sqrt { 1-\{ cos2(X-2)\} } }{ X-2 } \right] \)is equal to
- (a)
\(\sqrt{2}\)
- (b)
\(-\sqrt{2}\)
- (c)
\(1\over\sqrt{2}\)
- (d)
does not exist
Let \(\alpha\) and \(\beta\) be the distinct roots of ax2+bx+c=0, then \(\underset { x\rightarrow \alpha }{ lim } \frac { 1-cos(ax^{ 2 }+bx+c) }{ (x-\alpha )^{ 2 } } \) is equal to
- (a)
\({a^2\over2}(\alpha-\beta)^2\)
- (b)
\(-{a^2\over2}(\alpha-\beta)^2\)
- (c)
\(-{a^2\over2}(\alpha+\beta)^2\)
- (d)
None of these
\(\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } \frac { \left[ 1-tan\left( \frac { x }{ 2 } \right) \right] (1-sin\quad X) }{ \left[ 1+tan\left( \frac { x }{ 2 } \right) \right] (\pi -2x)^{ 3 } } \)is
- (a)
1/8
- (b)
0
- (c)
\(1\over2\)
- (d)
\(\infty\)
\(\underset { x\rightarrow 0 }{ lim } \frac { \sqrt { 1-cos2x } }{ \sqrt { 2x } } =k\) is equal to
- (a)
\(\lambda \)
- (b)
-1
- (c)
zero
- (d)
does not exist
\(\underset { x\rightarrow 0 }{ lim } { \left( \frac { x^{ 2 }+5x+3 }{ { x }^{ 2 }+x+2 } \right) }^{ x }\) is equal to
- (a)
e4
- (b)
e2
- (c)
e3
- (d)
e
For \(x \epsilon R,\) \(\underset { x\rightarrow 0 }{ lim } { \left( \frac { x-3 }{ { x }+2 } \right) }^{ x }\)is equal to
- (a)
e
- (b)
e-1
- (c)
e-5
- (d)
e5
if \(f(x)=\begin{cases}x+2, x\le -1 \\ cx^2, x>-1\end{cases},\)then find c, if \(\underset { x\rightarrow -1 }{ lim } f(x)\) exists.
- (a)
-1
- (b)
1
- (c)
0
- (d)
2
Let \(f(x)=\lim _{ n\rightarrow \infty }{ n({ x }^{ 1/n } } -1),x>0\) then f(xy) is equal to
- (a)
f(x)+f(y)
- (b)
f(x).f(y)
- (c)
x f(y)+y f(x)
- (d)
x f (y)
ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is altitude from A to BC,then \(\lim _{ h\rightarrow 0 }{ \frac { \triangle }{ { p }^{ 3 } } } \) is equal to (where \(\triangle \) is the area and P is the perimeter of the
triangle ABC)
- (a)
\(\frac { 1 }{ 32r } \)
- (b)
\(\frac { 1 }{ 64r } \)
- (c)
\(\frac { 1 }{ 128r } \)
- (d)
\(\frac { 1 }{ 216r } \)
Let\(f(x)={ 3x }^{ 10 }={ 7x }^{ 8 }+{ 5x }^{ 6 }-{ 21x }^{ 3 }+{ 3x }^{ 2 }-7\) The value of \(\lim _{ h\rightarrow 0 }{ \frac { f(1-h)-f(1) }{ { h }^{ 3 }+3h } } \) is equal to
- (a)
22/3
- (b)
50/3
- (c)
25/3
- (d)
53/3
The graph of the function y = f(x) has a unique tangent at the point (a, 0) through which the graph passes, Then \(\lim _{ x\rightarrow a }{ \frac { \log _{ e }{ \{ 1+6f(x)\} } }{ 3f(x) } } \) is
- (a)
0
- (b)
1
- (c)
2
- (d)
none of these
If α and β be the roots of ax2+bx+c=0 , then \(\lim _{ x\rightarrow \infty }{ \frac { 1-cos({ ax }^{ 2 }+bx+c) }{ (x-\infty )^{ 2 } } } \) is equal to
- (a)
0
- (b)
\(\frac { 1 }{ 2 } \left( \alpha -\beta \right) ^{ 2 }\)
- (c)
\(\frac { { a }^{ 2 } }{ 2 } \left( \alpha -\beta \right) ^{ 2 }\)
- (d)
\(-\frac { { a }^{ 2 } }{ 2 } \left( \alpha -\beta \right) ^{ 2 }\)
\(\lim _{ x\rightarrow 0 }{ \frac { \tan { ([-\pi ^{ 2 }]{ x }^{ 2 })- } \tan { ([-\pi ^{ 2 }]{ )x }^{ 2 } } }{ { sin }^{ 2 }x } } \) equals where[.] denotes the greatest integer function)
- (a)
0
- (b)
1
- (c)
tan 10-10
- (d)
∝
Evaluate of the following limits.
\(\lim _{ x\rightarrow -1 }{ [1+x+{ x }^{ 2 }+...+{ x }^{ 10 }] } \)is
- (a)
0
- (b)
1
- (c)
2
- (d)
does not exit
Evaluate of the following limits.
\(\lim _{ x\rightarrow 1 }{ \left[ \frac { x^{ 2 }+1 }{ x+100 } \right] } \)=
- (a)
1
- (b)
\(\frac{2}{101}\)
- (c)
0
- (d)
\(\frac{101}{2}\)
Evaluate of the following limits.
\(\lim _{ x\rightarrow 1 }{ \frac { (2x-3)(\sqrt { x } -1) }{ 2x^{ 2 }+x-3 } } \)=
- (a)
\(\frac{1}{10}\)
- (b)
\(\frac{2}{15}\)
- (c)
-\(\frac{1}{10}\)
- (d)
-\(\frac{2}{15}\)
\(\lim _{ k\rightarrow \infty }{ \left( \frac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ k }^{ 3 } }{ { k }^{ 4 } } \right) } \)is equal to
- (a)
0
- (b)
2
- (c)
\(\frac{1}{3}\)
- (d)
\(\frac{1}{4}\)
Find the positive integer n so that \(\lim _{ x\rightarrow 3 }{ \frac { { x }^{ n }-3^{ n } }{ x-3 } } \)=108.
- (a)
1
- (b)
2
- (c)
3
- (d)
4
The value of \(\lim _{ x\rightarrow \infty }{ \frac { 1-cos(ax^{ 2 }+bx+c) }{ (x-\propto )^{ 2 } } } \), where α and β are roots of the equation ax2+bx+c=0, is
- (a)
\(\frac { (\alpha -\beta )^{ 2 } }{ 2 } \)
- (b)
(a-b)2
- (c)
\(\frac { { \alpha }^{ 2 }(a-b)^{ 2 } }{ 2 } \)
- (d)
\(\frac { { \alpha }^{ 2 }(\alpha -\beta )^{ 2 } }{ 2 } \)
\(\lim _{ x\rightarrow y }{ \left( \frac { { x }^{ y }-{ y }^{ x } }{ { x }^{ x }-{ y }^{ y } } \right) } \), is equal to
- (a)
\(\frac { 1+logy }{ logy } \)
- (b)
\(\frac { 1-logy }{ 1+logy } \)
- (c)
\(\frac { 1-logy }{ log(ey) } \)
- (d)
log(ex)
Find the derivative of the following function.
f(x)=2x2+3x-5 at x=-1
- (a)
0
- (b)
1
- (c)
-1
- (d)
-2
Find the derivative of f(x)=\(\frac { x+1 }{ x } \)
- (a)
\(\frac { 1 }{ x } \)
- (b)
\(\frac { 1 }{ x^{ 2 } } \)
- (c)
0
- (d)
-\(\frac { 1 }{ x^{ 2 } } \)
If \(f(x)=1-\sqrt { x } +(1+\sqrt { x } )^{ 2 },f'(1)\) is equal to
- (a)
\(\cfrac { 1 }{ 2 } \)
- (b)
1
- (c)
\(\cfrac { 3 }{ 2 } \)
- (d)
2
If y=x tanx, then dy/dx is equal to
- (a)
\(\cfrac { tan\quad x }{ x-{ x }^{ 2 }-{ y }^{ 2 } } \)
- (b)
\(\cfrac { y }{ x-{ x }^{ 2 }-{ y }^{ 2 } } \)
- (c)
\(\cfrac { tan\quad x }{ y-x } \)
- (d)
\(\cfrac { cosxsinx+x }{ { cos }^{ 2 }x } \)
If siny=xsin(a+y), then dy/dx is
- (a)
sin(a+y)
- (b)
sin2(a+y)
- (c)
\(\cfrac { { sin }^{ 2 }(a+y) }{ sina } \)
- (d)
\(\cfrac { sin\quad (a+y) }{ sina } \)
Which of the following is CORRECT?
- (a)
If f(x)=x4(5sinx-3cosx)then
f'(x)=x3sinx(20+3x)-x3cosx(5x-1). - (b)
If f(x)=(x+secx)(x-tanx) then
f'(x)=(1+secxtanx)(x-tanx) - (c)
If f(x)=(ax2+cotx)(p+qcosx) then
f'(x)=-qsinx(ax2+cotx) - (d)
None of these
\(\underset { x\rightarrow 0 }{ lim } \cfrac { sinx }{ \sqrt { x+1 } -\sqrt { 1-x } } \) is
- (a)
2
- (b)
0
- (c)
1
- (d)
-1
If y =\(\frac { sin(x+9) }{ cosx } ,\quad then\quad \frac { dy }{ dx } \)at x = 0 is
- (a)
cos 9
- (b)
sin 9
- (c)
0
- (d)
1