Eamcet Mathematics - Maxima amd Minima Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 40
A function f such that f' (2) = f" (2) = 0 and f has a local maximum of -17 at 2 is
- (a)
(x-2)4
- (b)
3-(x-2)4
- (c)
-17-(x-2)4
- (d)
none of these
Since, f' (2) = 0 = f" (2)
Let f(x) = A +(x-2)n, n ≥ 3
but f has local maximum at x = 2
So, n may be taken as 4 and since maximum value is -17
Therefore, f (x)=-17-(x-2)4
Let f(x) = a -(x-3)8/9 then maxima of f(x) is
- (a)
3
- (b)
a-3
- (c)
a
- (d)
none of these
f(x)=a-(x-3)8/9
f'(x)=0-\(\frac { 8 }{ 9 } \)(x-3)-1/9
At x = 3, f' (x) is not defined
Hence, x = 3 is the point of extremum
Hence, maximum value of f(x) = a at x = 3.
Let \(f(x)=\begin{cases} \left| { x }^{ 3 }+{ x }^{ 2 }+3x+sinx \right| \left( 3+sin\frac { 1 }{ x } \right) ,x\neq 0 \\ 0,\quad x=0 \end{cases}\)then number of points (where f(x) attains its minimum value) is
- (a)
1
- (b)
2
- (c)
3
- (d)
infinite many
\(f(x)=\begin{cases} \left| { x }^{ 3 }+{ x }^{ 2 }+3x+sinx \right| \left( 3+sin\frac { 1 }{ x } \right) ,x\neq 0 \\ 0,\quad x=0 \end{cases}\)
let g(x)=x3+x2+3x+sinx
g'(x)=3x2+2x+3+cosx
=\(3\left( { x }^{ 2 }+\frac { 2x }{ 3 } +1 \right) +cosx\)
=\(3\left\{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 8 }{ 9 } \right\} +cosx>0\)
and 2<3+sin\(\left( \frac { 1 }{ x } \right) \)<4
Hence, minimum value of f (x) is 0 at x = O.
Hence, number of points = 1.
If \(f(x)={ alog }_{ e }\left| x \right| +{ bx }^{ 2 }+x\) has extremum at x = 1 and x = 3, then
- (a)
a = - 3/4, b = -1/8
- (b)
a = 3/4, b = -1/8
- (c)
a = - 3/4, b = 1/8
- (d)
none of these
For x > 0 or x < 0
f'(x)=\(\frac { a }{ x } \)+2bx+1
f'(1)=0 ⇒ a+2b+1=0...(i)
f'(3)=0 ⇒ \(\frac{a}{3}\)+6b+1=0...(ii)
Solving Eqs. (i) and (ii), we get
a=-3/4, b=-1/8
Let f(x) = 1 + 2x2 + 22 x4 + .... + 210 x20 Then f(x) has
- (a)
more than one minimum
- (b)
exactly one minimum
- (c)
at least one maximum
- (d)
none of the above
f'(x)=4x+16x3+...+210.20.x19
=4x(1+4x2+..+5.210x18) ...(i)
f"(x)=4+48x2+...+210380.x18>0
Minimum at x=0
Let \(f(x)=\begin{cases} { sin }^{ -1 }\alpha +{ x }^{ 2 },\quad 0<x<1 \\ 2x,\quad x\ge 1 \end{cases}\)f(x) can have a minimum at x = 1 is the value of ∝ is
- (a)
1
- (b)
-1
- (c)
0
- (d)
none of these
\(f'(x)=\begin{cases} 2x,\quad 0<x<1 \\ 2,\quad x\ge 1 \end{cases}\)
minimum at x = 1
f(1)=sin-1 ∝+12⇒ 2=sin-1 ∝+1
sin-1 ∝=1 ⇒ ∝=sin 1
A differentiable function f(x) has a relative minimum at x = 0, then the function y = f(x) + ax + b has a relative minimum at x = 0 for
- (a)
all a and all b
- (b)
all b if a = 0
- (c)
all b > 0
- (d)
all a > 0
Given
f'(x)=0 at x=0
f''(0)>0
y = f(x) + ax + b
\(\frac { dy }{ dx } =f'(x)+a=0+a\)
=a
For maximum or minimum
a=0
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =f''(x)>0\)
y is minimum for all b and a = 0
Assuming that the petrol burnt in a motor boat varies as the cube of its velocity, the most economical speed, when going against a current of c km/h is
- (a)
(3c/2) km/h
- (b)
(3c/4) km/h
- (c)
(5c/2) km/h
- (d)
(c/2) km/h
Suppose p is the quantity of petrol burnt in one hour. Then p = k v3 Let the distance of the journey be s km
Duration of the journey=\(\frac { s }{ v-c } \)hours
Total petrol burnt for the whole journey
=\(\frac { sp }{ v-c } =\frac { { skv }^{ 3 } }{ v-c } \)
Take \(u=\frac { { v }^{ 3 } }{ v-c } \)
\(\frac { du }{ dv } =\frac { \left( v-c \right) { 3v }^{ 2 }-{ v }^{ 3 } }{ { \left( v-c \right) }^{ 2 } } =\frac { { v }^{ 2 }\left( 2v-3c \right) }{ { \left( v-c \right) }^{ 2 } } \)
\(\frac { du }{ dv } =0\Rightarrow v=\frac { 3c }{ 2 } ,\frac { 3c }{ 2 } \)is found to give a minimum for u.
Most economical speed=\(\frac{3c}{2}\) km/h
N Characters of information are held on magnetic tape, in batches of x characters each, the batch processing time is ∝+βx2 seconds, ∝ and β are constants. The optical value of x for fast processing is
- (a)
\(\frac { \alpha }{ \beta } \)
- (b)
\(\frac { \beta }{ \alpha } \)
- (c)
\(\sqrt { \frac { \alpha }{ \beta } } \)
- (d)
\(\sqrt { \frac { \beta }{ \alpha } } \)
x batch has characters = N
1 batch has characters = N/x
Let P=(∝+βx2)x\(\frac{N}{x}\)
=\(N\left( \frac { \alpha }{ x } +\beta x \right) \);
\(\frac { dP }{ dx } =N\left( -\frac { \alpha }{ { x }^{ 2 } } +\beta \right) \)
For max or min, \(\frac { dP }{ dx } =0\)
\(x=\pm \sqrt { \alpha /\beta } \)
\(\frac { { d }^{ 2 }P }{ { dx }^{ 2 } } =N\left( \frac { 2\alpha }{ { x }^{ 3 } } \right) \)
For least of P
x=\(\sqrt { \frac { \alpha }{ \beta } } \)
On [1, e], the least and greatest values of f(x) = x2 ln x is
- (a)
e,1
- (b)
1,e
- (c)
0,e2
- (d)
none of these
f(x)=x2 ln x
⇒ f'(x)= x (1 + 2ln x)
and f' (x) > 0 for ∈ [1,e]
f(x) is continuously increasing on [1, e] with the least value zero at x = 1 and the greatest value e2 at x = e.
The minimum value of \(\left( 1+\frac { 1 }{ { sin }^{ n }\alpha } \right) \left( 1+\frac { 1 }{ { cos }^{ n }\alpha } \right) \) is
- (a)
1
- (b)
2
- (c)
(1+2n/2)2
- (d)
none of these
f(x)=\(\left( 1+\frac { 1 }{ { sin }^{ n }\alpha } \right) \left( 1+\frac { 1 }{ { cos }^{ n }\alpha } \right) \)
=\(1+\frac { 1 }{ { sin }^{ n }\alpha } +\frac { 1 }{ { cos }^{ n }\alpha } +\frac { 1 }{ { sin }^{ n }\alpha { cos }^{ n }\alpha } \)
\(f'(\alpha )=-\frac { ncos\alpha }{ { sin }^{ n+1 }\alpha } +\frac { nsin\alpha }{ { cos }^{ n+1 }\alpha } \)
-\(\frac { n }{ { sin }^{ n+1 }\alpha { cos }^{ n+1 }\alpha } \){cos2∝-sin2∝)
f'(∝)=0
cos∝-sin∝=0, f''(π/4)=+ve
∝=π/4
f(π/4)=(1+2n/2)2
The fuel charges for running a train are proportional to the square of the speed generated in mile/h and costs Rs 48 per h at 16 miles/h. The most economical speed if the fixed charges ie, salaries etc amount to Rs 300 per h
- (a)
10 mile/h.
- (b)
20 mile/h
- (c)
30 mile/h
- (d)
40 rnile/h
Let the speed of the train be v and distance to be covered be s so that total time taken is s/v hours cost of fuel per hour = kv2 (k is constant)
Also, 48 = k .162 by given condition
k = 3/16
Cost of fuel per hour =(3/16) v2
Other charges per hour = 300
Total charge =\(\frac { 3 }{ 16 } { v }^{ 2 }\times \frac { s }{ v } +\frac { 300s }{ v } \)
or C=s\(\left( \frac { 3 }{ 16 } v+\frac { 300 }{ v } \right) \)
\(\frac { dC }{ dv } =s\left( \frac { 3 }{ 16 } -\frac { 300 }{ { v }^{ 2 } } \right) \)
For max or min \(\frac { dC }{ dv } =0\)
v=40mile/h
then \(\frac { { d }^{ 2 }C }{ d{ v }^{ 2 } } >0\)
Most economical speed v =40 mile/h
A cylindrical gas container is closed at the top and open at the bottom; if the iron plate of the top is 5/4 times as thick as the plate forming the cylindrical sides. The ratio of the radius to the height of the cylinder using minimum material for the same capacity is
- (a)
2/3
- (b)
1/2
- (c)
4/5
- (d)
1/3
v=πr2h
If k be the thickness of the sides, then that of the top will be (5/4)k
S=(2πrh)k+(πr2)(5/4)k('s' is volume of material used)
or \(S=2\pi rk.\frac { V }{ { \pi r }^{ 2 } } +\frac { 5 }{ 4 } { \pi r }^{ 2 }k=k\left( \frac { 2V }{ r } +\frac { 5 }{ 4 } { \pi r }^{ 2 } \right) \)
\(\frac { ds }{ dr } =k\left( -\frac { 2V }{ { r }^{ 2 } } +\frac { 5 }{ 2 } \pi r \right) \)
r3=4V/5π
\(\frac { ds }{ dr } =k\left( -\frac { 2V }{ { r }^{ 2 } } +\frac { 5 }{ 2 } \pi r \right) \)=+ ive
when r3=4V/5π or 5πr3=4πr2h
r/h=4/5
Let \(f(x)=\begin{cases} { x }^{ 3 }-{ x }^{ 2 }+10x-5,\quad x\le 1 \\ -2x+{ lo }g_{ 2 }\left( { b }^{ 2 }-2 \right) ,\quad x>1 \end{cases}\)the set of values of b for which f(x) have greatest value at x = 1 is given by
- (a)
1≤b≤2
- (b)
b={1,2}
- (c)
b∈(-∞,-1)
- (d)
none of these
f(1)≥\(\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } \)
\(\Rightarrow f(1)\ge \lim _{ h\rightarrow 0 }{ f(1+h) } \)
1 - 1 + 10 - 5 ≥ \(\lim _{ h\rightarrow 0 }{ -2(1+h) } +{ log }_{ 2 }\left( { b }^{ 2 }-2 \right) \)
5≥-2+log2 (b2-2) and b2-2>0
27≥b2-2 and b2>2
22≤130
b∈[-√130,-√2]U[√2,√130]
Let f:[a,b]⇾R be a function such that for c ∈ (a,b), f'(c)=f''(c)=f'''(c)=fiv(c)=fv(c)=0 then
- (a)
f has local extremum at x = c
- (b)
f has neither local maximum nor local minimum at x=c
- (c)
f is necessarily a constant function
- (d)
it is difficult to say whether (a) or (b)
If even derivative +ive, then minima and if even derivative -ive, then maxima
But here fvi (c) is not given.
Hence, it is difficult to say maximum or minimum ie, a or b.
From the graph we can conclude that the
- (a)
function has some roots
- (b)
function has interval of increase and decrease
- (c)
greatest and the least values of the function exist
- (d)
function is periodic
From the graph of the function we conclude that the function has interval of increase and decrease.
Two towns A and Bare 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school should be built at
- (a)
town B
- (b)
45 km from town A
- (c)
town A
- (d)
45 km from town B
Let the school be at a distance x from A (with 150 students), then total distance travelled by the students is
z =150x + 50 (60 - x)
=3000 + 100x
z will be least when x =0
z =3000
ie, school is built at A.
If the function f(x)=2x3-9ax2+12a2x+1 has a local maximum at x=x1 and a local minimum at x=x2 such that x2=x12 then a is equal to
- (a)
0
- (b)
\(\frac{1}{4}\)
- (c)
2
- (d)
either (a) or (c)
f' (x) = 6x2 -18ax + 12a2 = 0 (for max or min)
x = a, 2a
x2=x12
a cannot be negative
If a> 0, then local minimum occurs at x = 2a and local maximum at x = a. Thus, x1 = a and x2 = 2a
2a=a2
a=2
Let \(\begin{cases} { x }^{ 2 }+3x,\quad -1\le x<0 \\ -sin\quad x,\quad 0\le x<\pi /2 \\ -1-cosx,\quad \frac { \pi }{ 2 } \le x\le \pi \end{cases}\) Then global maxima of f(x) and global minima of f(x) are equals
- (a)
-1
- (b)
0
- (c)
-3
- (d)
-2
f(x) increases in the interval -1≤x<0
f(x) decreases in the interval 0≤x<π/2
f(x) increases in the interval π/2≤x≤π
Hence, global maxima occurs at x = 0 and x =π ie, f(0)=f(π)=0 and global minima occurs at x = -1, ie f(-1)=-2
Let f(x)=(x2-1)n (x2+x-1), then f(x) has local extremum at x = 1 when
- (a)
n=2
- (b)
n=3
- (c)
n=4
- (d)
n=6
f'(x)=(x2-1)n {(x2-1)(2x+1)+2nx(x2+x+1)}
=(x+1)n+1 (x-1)n+1 (2x+1)+2nx(x+1)n(x-1)n(x2+x+1)
If power even, then neither max nor min
n = 2,4,6
The critical points of the function f'(x) where f(x)=\(\frac { \left| x-2 \right| }{ { x }^{ 2 } } \)is
- (a)
0
- (b)
2
- (c)
4
- (d)
6
Obviously, at x = 0, f(x)=∞
f'(0) does not exist.
So, x=0 is a critical point
Now, f(x)=\(\begin{cases} \frac { x-2 }{ { x }^{ 2 } } ,\quad x\ge 2 \\ \frac { 2-x }{ { x }^{ 2 } } ,\quad 0<x<2 \end{cases}\)
At x = 2, 4 the function f (x) is not differentiable, so they are critical points.
Let f(x)=\(\int _{ 0 }^{ x }{ \frac { cost }{ t } } dt\) (x>0); then f(x) has
- (a)
maxima, when n = - 2, - 4, - 6, ...
- (b)
maxima, when n = -1, - 3, - 5, ...
- (c)
minima, when n = 0, 2, 4, ...
- (d)
minima, when n = 1, 3, 5, ...
\(f'(x)=\frac { cosx }{ x } \)
For max or min f'(x)=0
cos x=0
⇒ x(2n+1)\(\frac { \pi }{ 2 } \), n∈I
\(f''(x)=\frac { -xsinx-cosx }{ { x }^{ 2 } } =-\frac { 1 }{ { x }^{ 2 } } \left( xsinx+cosx \right) \)
For n = - 2, - 4, - 6, ...
f''(x)>0 (minima)
For n=-1,-3,-5, ...
\(x=-\frac { \pi }{ 2 } ,-\frac { 5\pi }{ 2 } ,-\frac { 9\pi }{ 2 } ,...\)
f''(x)<0 (maxima)
For n=0,2,4...
\(x=\frac { \pi }{ 2 } ,\frac { 5\pi }{ 2 } ,\frac { 9\pi }{ 2 } ,...\)
f''(x)<0, (maxima)
For n=1,3,5,...
\(x=\frac { 3\pi }{ 2 } ,\frac { 7\pi }{ 2 } ,\frac { 11\pi }{ 2 } ,...\)
f''(x)>0 (minima)
The function f(x) = 3 + 2 (a + 1) x + (a2 + 1) x2- x3 has a local minimum at x = x1 and local maximum at x = x2 such that x1<2
- (a)
\(\left( -\infty ,-\frac { 3 }{ 2 } \right) \)
- (b)
\(\left( -\frac { 3 }{ 2 } ,1 \right) \)
- (c)
(0,∞)
- (d)
(1,∞)
2 lies between the roots of f'(x)=0
ie., 2 (a + 1) + 2 (a2 + 1) x - 3x2 = 0
2 (a + 1) + 2 (a2 + 1)(2) - 3 (2)2> 0
\(\Rightarrow a<-\frac { 3 }{ 2 } \)or a>1
If composite function f1(f2(f3(...(fn(x))...) is an increasing function and if r of fi' s are decreasing functions while rest are increasing, then maximum value of r (n - r) is
- (a)
\(\frac { { n }^{ 2 } }{ 4 } \), when n is an even number
- (b)
\(\frac { { n }^{ 2 } }{ 4 } \) when n is an odd number
- (c)
\(\frac { { n }^{ 2 }-1 }{ 4 } \)when n is an odd number
- (d)
\(\frac { { n }^{ 2 }-1 }{ 4 } \) when n is an even number
r must be an even integer because two decreasing are required to make it increasing function
Let y=r(n-r)
when n is odd, \(r=\frac { n-1 }{ 2 } \) or \(\frac { n+1 }{ 2 } \) for maximum value of y
when n is even, r=\(\frac{n}{2}\) for maximum value of y
Maximum (y) =\(\frac { { n }^{ 2 }-1 }{ 4 } \)when n is odd and \(\frac { { n }^{ 2 } }{ 4 } \)when n is even.
Four points A, B, C and D lie in that order on the parabola y = ax2+bx+c and the coordinates of A, Band D are known A(- 2, 3); B(- 1, 1); D(2, 7).
The value of a is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
y = ax2 + bx + c...(i)
passing through A (- 2, 3), B(-1, 1), D (2,7) then
3=4a-2b+c...(ii)
1=a-b+c...(iii)
and 7 = 4a + 2b + c...(iv)
Solving Eqs. (ii), (iii) and (iv) we get
a=1, b = 1, c = 1
Four points A, B, C and D lie in that order on the parabola y = ax2+bx+c and the coordinates of A, Band D are known A(- 2, 3); B(- 1, 1); D(2, 7).
The value of b is
- (a)
-3
- (b)
-1
- (c)
0
- (d)
1
y = ax2 + bx + c...(i)
passing through A (- 2, 3), B(-1, 1), D (2,7) then
3=4a-2b+c...(ii)
1=a-b+c...(iii)
and 7 = 4a + 2b + c...(iv)
Solving Eqs. (ii), (iii) and (iv) we get
a=1, b = 1, c = 1
Four points A, B, C and D lie in that order on the parabola y = ax2+bx+c and the coordinates of A, Band D are known A(- 2, 3); B(- 1, 1); D(2, 7).
The value of c is
- (a)
3
- (b)
2
- (c)
0
- (d)
none of these
y = ax2 + bx + c...(i)
passing through A (- 2, 3), B(-1, 1), D (2,7) then
3=4a-2b+c...(ii)
1=a-b+c...(iii)
and 7 = 4a + 2b + c...(iv)
Solving Eqs. (ii), (iii) and (iv) we get
a=1, b = 1, c = 1
Four points A, B, C and D lie in that order on the parabola y = ax2+bx+c and the coordinates of A, Band D are known A(- 2, 3); B(- 1, 1); D(2, 7).
If roots of ax2+bx+c=0 are ∝ and β, then equation whose roots are ∝19 and β7 is
- (a)
x2-x+1=0
- (b)
x2-2x+3=0
- (c)
x2+x+1=0
- (d)
x2+2x+3=0
x2+x+1=0
\(x=\omega ,{ \omega }^{ 2 }\)
∝=ω,β=ω2
∝19 = ω19
ω=∝
β7=ω14
ω2=β
Equation x2+x+1=0
Four points A, B, C and D lie in that order on the parabola y = ax2+bx+c and the coordinates of A, Band D are known A(- 2, 3); B(- 1, 1); D(2, 7).
If area of quadrilateral ABCD is greatest, then the coordinates of Care
- (a)
\(\left( \frac { 7 }{ 4 } ,\frac { 1 }{ 2 } \right) \)
- (b)
\(\left( \frac { 1 }{ 2 } ,\frac { 7 }{ 4 } \right) \)
- (c)
\(\left( \frac { 1 }{ 2 } ,-\frac { 7 }{ 4 } \right) \)
- (d)
\(\left(- \frac { 1 }{ 2 } ,\frac { 7 }{ 4 } \right) \)
Let the point C(x, x2+x+1)
Area of quadrilateral ABCD
A(say)=\(\frac { 1 }{ 2 } \left| \begin{matrix} -2 & 3 \\ -1 & 1 \\ x & { x }^{ 2 }+x+1 \\ 2 & 7 \\ -2 & 3 \end{matrix} \right| \)
=\(\frac{3}{2}\)(-x2+x+6)
\(\frac { dA }{ dx } =\frac { 3 }{ 2 } (-2x+1)\)and \(\frac { { d }^{ 2 }A }{ d{ x }^{ 2 } } =-3<0\)
For max. or min, of A, \(\frac { dA }{ dx } \)=0
x=\(\frac{1}{2}\)
Then, \(y={ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+\left( \frac { 1 }{ 2 } \right) +1\)
\(\frac { 1+2+4 }{ 4 } =\frac { 7 }{ 4 } \)
Then, C≡\(\left( \frac { 1 }{ 2 } ,\frac { 7 }{ 4 } \right) \)
A cubic f(x)=ax3+bx2+cx+d vanishes at x=-2 and has relative minimum/maximum at x = - 1 and x=\(\frac{1}{3}\) and if \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
The value of c is
- (a)
-2
- (b)
-1
- (c)
0
- (d)
2
f'(-1)=f'\(\left( \frac { 1 }{ 3 } \right) \)=0
f'(x)=a'(x+1)\(\left( x-\frac { 1 }{ 3 } \right) \)
\(a'\left( { x }^{ 2 }+\frac { 2 }{ 3 } x-\frac { 1 }{ 3 } \right) \)
Integrating W.r.t. x, we get
\(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x \right) +\lambda \)...(i)
where λ is constant of integration.
and f(- 2) = 0
⇒ \(\frac { a' }{ 3 } (-8+4+2)+\lambda =0\)
\(\lambda =\frac { 2a' }{ 3 } \)
From Eq. (i), \(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) \)
Also, \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
⇒ \(\frac { a' }{ 3 } \int _{ -1 }^{ 1 }{ \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) } \)dx=\(\frac{14}{3}\)
⇒ \(0+\frac { 2a' }{ 3 } \int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }+2 \right) } dx=\frac { 14 }{ 3 } \)
Then, f(x) = x3 + x2 - x + 2
Comparing with f(x) = ax3+bx2+cx+d
a = 1, b = 1, C = - 1, d = 2
A cubic f(x)=ax3+bx2+cx+d vanishes at x=-2 and has relative minimum/maximum at x = - 1 and x=\(\frac{1}{3}\) and if \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
The value of d is
- (a)
5
- (b)
2
- (c)
0
- (d)
-4
f'(-1)=f'\(\left( \frac { 1 }{ 3 } \right) \)=0
f'(x)=a'(x+1)\(\left( x-\frac { 1 }{ 3 } \right) \)
\(a'\left( { x }^{ 2 }+\frac { 2 }{ 3 } x-\frac { 1 }{ 3 } \right) \)
Integrating W.r.t. x, we get
\(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x \right) +\lambda \)...(i)
where λ is constant of integration.
and f(- 2) = 0
⇒ \(\frac { a' }{ 3 } (-8+4+2)+\lambda =0\)
\(\lambda =\frac { 2a' }{ 3 } \)
From Eq. (i), \(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) \)
Also, \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
⇒ \(\frac { a' }{ 3 } \int _{ -1 }^{ 1 }{ \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) } \)dx=\(\frac{14}{3}\)
⇒ \(0+\frac { 2a' }{ 3 } \int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }+2 \right) } dx=\frac { 14 }{ 3 } \)
Then, f(x) = x3 + x2 - x + 2
Comparing with f(x) = ax3+bx2+cx+d
a = 1, b = 1, C = - 1, d = 2
A cubic f(x)=ax3+bx2+cx+d vanishes at x=-2 and has relative minimum/maximum at x = - 1 and x=\(\frac{1}{3}\) and if \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
The function f(x) is
- (a)
x3+x2+x-2
- (b)
x3-x2+x-2
- (c)
x3-x2-x-2
- (d)
x3+x2-x-2
f'(-1)=f'\(\left( \frac { 1 }{ 3 } \right) \)=0
f'(x)=a'(x+1)\(\left( x-\frac { 1 }{ 3 } \right) \)
\(a'\left( { x }^{ 2 }+\frac { 2 }{ 3 } x-\frac { 1 }{ 3 } \right) \)
Integrating W.r.t. x, we get
\(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x \right) +\lambda \)...(i)
where λ is constant of integration.
and f(- 2) = 0
⇒ \(\frac { a' }{ 3 } (-8+4+2)+\lambda =0\)
\(\lambda =\frac { 2a' }{ 3 } \)
From Eq. (i), \(f(x)=\frac { a' }{ 3 } \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) \)
Also, \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
⇒ \(\frac { a' }{ 3 } \int _{ -1 }^{ 1 }{ \left( { x }^{ 3 }+{ x }^{ 2 }-x+2 \right) } \)dx=\(\frac{14}{3}\)
⇒ \(0+\frac { 2a' }{ 3 } \int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }+2 \right) } dx=\frac { 14 }{ 3 } \)
Then, f(x) = x3 + x2 - x + 2
Comparing with f(x) = ax3+bx2+cx+d
a = 1, b = 1, C = - 1, d = 2
A cubic f(x)=ax3+bx2+cx+d vanishes at x=-2 and has relative minimum/maximum at x = - 1 and x=\(\frac{1}{3}\) and if \(\int _{ -1 }^{ 1 }{ f(x)dx=\frac { 14 }{ 3 } } \)
f(x) decreases in the interval
- (a)
\(\left( -\frac { 1 }{ 3 } ,1 \right) \)
- (b)
\(\left( -\frac { 1 }{ 3 } ,-1 \right) \)
- (c)
\(\left( -1,\frac { 1 }{ 3 } \right) \)
- (d)
\(\left( 1\frac { 3 }{ 2 } \right) \)
f'(x)=3(x+1)\(\left( x-\frac { 1 }{ 3 } \right) \)<0
-1
x∈\(\left( -1,\frac { 1 }{ 3 } \right) \)
The point (0,5) is closest to the curve x2=2y at
- (a)
(2√2,0)
- (b)
(2,2)
- (c)
(-2√2,0)
- (d)
(2√2,4)
Let P\(\left( t,\frac { { t }^{ 2 } }{ 2 } \right) \) be a point on x2=2y and A be (0,5)
If AP=d
z=d2=t2+\({ \left( \frac { { t }^{ 2 } }{ 2 } -5 \right) }^{ 2 }\)
\(\frac { dz }{ dt } =2t+2\left( \frac { { t }^{ 2 } }{ 2 } -5 \right) .t\)
=t3-8t
=t(t2-8)
\(\frac { { d }^{ 2 }z }{ d{ t }^{ 2 } } ={ 3t }^{ 2 }-8\)
\(\frac { dz }{ dt } =0\Rightarrow t=0\)or t=±2√2
At t=0 \(\frac { { d }^{ 2 }z }{ { dt }^{ 2 } } \) is -ive
Hence, the closest point is (2√2,4)
The least value of 'a' for which \(\frac { 4 }{ sinx } +\frac { 1 }{ 1-sinx } \)=a has atleast one solution in the interval (0,π/2)
- (a)
9
- (b)
8
- (c)
4
- (d)
1
f(x)=\(\frac { 4 }{ sinx } +\frac { 1 }{ 1-sinx } =a\)
f'(x)=-\(\frac { 4cosx }{ { sin }^{ 2 }x } +\frac { cosx }{ { \left( 1-sinx \right) }^{ 2 } } \)
=\(cosx\left( \frac { 1 }{ { \left( 1-sin\quad x \right) }^{ 2 } } -\frac { 4 }{ { sin }^{ 2 }x } \right) \)
f'(x)=0
⇒ \(\frac { 1 }{ { \left( 1-sin\quad x \right) }^{ 2 } } -\frac { 4 }{ { sin }^{ 2 }x } =0\) as cos x≠0 in \(\left( 0,\frac { \pi }{ 2 } \right) \)
This gives sin x=\(\frac{2}{3}\)
Substituting this in Eq. (i), we get
a=9
The global maxima of f(x)=[2{-x2+x+1}] is (where {x} denotes fractional part of x and [.] denotes greatest integer function)
- (a)
2
- (b)
1
- (c)
0
- (d)
none of these
0≤{-x2+x+1}<1
⇒ 0≤ 2{-x2+x+1}<2
[2{-x2+x+1}]=0,1
Global maximum of f(x) is 1.
If f(x)=\(\begin{cases} 1+{ x }^{ 2 }-3x,\quad x<0 \\ cosx+2x,\quad x\ge 0 \end{cases}\) , then the global maximum and local minimum values of f(x) for x∈[-2,2] are respectively.
- (a)
4+cos 2,1
- (b)
11,1
- (c)
11, not exist
- (d)
none of these
Since, the given function is continuous but not differentiable
at x = 0 and Lf" (0) = - 3, Rf' (0) = 2
⇒ x = 0 is point of local minima and f(0) = 1
Also, there is no other critical point
Now, f(-2) = 11, f(2) = 4 + cos 2
On comparing values of function at x = - 2, 0, 2
global maximum occurs at x = - 2 ⇒ f( -2) = 11
Hence, global maximum and local minimum values are 11, 1.
Let f(x)=\(\begin{cases} \left| { x }^{ 2 }-3x \right| +a,\quad 0\le x<32 \\ -2x+3,\quad x\ge \frac { 3 }{ 2 } \end{cases}\)
If f(x) has a local maxima at x=\(\frac{3}{2}\) , then
- (a)
a≤0
- (b)
a≤\(-\frac{9}{4}\)
- (c)
a≥\(\frac{9}{4}\)
- (d)
none of these
f(\(\frac{3}{2}\))=0
\(\lim _{ x\rightarrow \frac { 3 }{ 2 } }{ \left| { x }^{ 2 }-3x \right| } +a\le 0\)
⇒ \(\frac { 9 }{ 4 } +a\le 0\Rightarrow a\le -\frac { 9 }{ 4 } \)
The greatest value of the function f(x)=2 sin x+ sin 2x on the interval \(\left[ 0,\frac { 3\pi }{ 2 } \right] \) is
- (a)
\(\frac { 3\sqrt { 3 } }{ 2 } \)
- (b)
3
- (c)
\(\frac{3}{2}\)
- (d)
none of these
f(x)= 2sin x + sin 2x
f' (x) = 2 cos x + 2 cos 2x
= 2 (cos x + cos 2x)
= 4 cos\(\left( \frac { 3x }{ 2 } \right) \)cos\(\left( \frac { x }{ 2 } \right) \)
f'(x)=0
Then, x=\(\frac { \pi }{ 3 } ,\pi \)
Now, f(0)=0
\(f\left( \frac { \pi }{ 3 } \right) =\frac { 2\sqrt { 3 } }{ 2 } +\frac { \sqrt { 3 } }{ 2 } =\frac { 3 }{ 2 } \sqrt { 3 } \)
\(f\left( \frac { 3\pi }{ 2 } \right) \)=2(-1)+0=-2
⇒ Hence, greatest value =\(\frac { 3\sqrt { 3 } }{ 2 } \)
The difference between the greatest and the least values of the function \(f(x)=\int _{ 0 }^{ x }{ \left( { at }^{ 2 }+1+cos\quad t \right) dt, } \)a>0 for x∈ [2,3] is
- (a)
\(\frac { 19 }{ 3 } a+1+(sin3-sin2)\)
- (b)
\(\frac { 18 }{ 3 } a+1+2\quad sin3\)
- (c)
\(\frac { 18 }{ 3 } a-1+2\quad sin3\)
- (d)
none of these
f' (x) = ax2 + 1 + cos x > 0, for a > 0 and ∀ x∈R
⇒ f(x) is an increasing function
⇒ max f(x) = f(3) and min f(x) = f(2)
⇒ Difference = f(3) - f(2) =\(\int _{ 2 }^{ 3 }{ \left( { at }^{ 2 }+1+cost \right) } dt\)
=\(\frac{19}{3}\)a + 1 + (sin 3 - sin 2)