GATE Computer Science Engineering(CSE) 2027 Test: Calculus- 2 Free Online

For the integrand 

This gives a new lower bound  and upper bound  Now, our integral becomes:

Since the antiderivative of cos(u) is sin (u) , applying the fundamental theorem of calculus, we get:

 is a continuously increasing function, and for a continuously increasing function f(x)

But in question, summation of L.H.S. above, a = 3 and in R.H.S, a = 2, so we don't know whether S > T. So we compute some initial values :

and since we already know that

Use Integration by Parts
ln(x) dx
set
u = ln(x),    dv = dx
then we find
du = (1/x) dx,    v = x
substitute

substitute u=ln(x), v=x, and du=(1/x)dx

f''(1) < 0, so x = 1 is point of local maxima, f''(3) > 0, so x = 3 is point of local minima.
Also the end points 0 and 6 are critical points. 0 is point of local minima, because it is to the left of x = 1 (which is point of maxima). Similarly x = 6 is point of local maxima.

(b) Since xcosx is an odd function, by the properties of definite integration, answer is 0.

There is a mod term in the given integral. So, first we have to remove that. We know that x is always positive here and sin x is positive from 0 to π. From π to 2π,  x is positive while sin x changes sign. So, we can write

Take the series as 1/x
and use  Riemann integral to evaluate the series
integrate 1/ x dx from x=1001 to 3001 then we get 1.0979 which is in between 1 and 3/2
So C is correct option. 1<X<3/2

Integral of a multiplied by b equals
a multiplied by integral of b
minus
integral of derivative of a multiplied by integral of b

Since substituting x=1 we get 0/0 which is indeterminate
after applying L'H we get ((7x⁶)-(10x⁴))/((3x²)-(6x))
now substituting x=1 we get -3/-3
=1
hence answer is 1

Since we have a 0/0 form, we can apply the L'Hôpital's rule.

Hence, option c is correct.

substitute h=x-4..so it becomes lim_{h->0 (sinh)/h ...} which is a standard limit.. Ans would be 1.

i will solve by two methods 
method 1
y=lim_n->inf (1-1/n)²ⁿ
taking log 

log y =lim_n->inf 2n log(1-1/n)        =lim_n->inf log (1-1/n)/(1/2n)---------------(converted this so as to have form 0/0)
apply l hospital rule
log y=lim_n->inf (1/1-1/n)1/n²  / (-1/2n²) log y=^-2
y=e^-2 method two
it  takes 1 to power infinity form  

lim_x->inff(x)g(x)
=elim_x->inf(f(x)-1)g(x) (f(x)-1)*g(x)=-1/n*2n=-2 ie -2
constant so we get  final ans e^-2

Use LH rule:
First Derivative: [x(e^x) + (e^x-1) - 2(sinx)]/[xsinx + (1 - cosx)]
Second Derivative: [xe^x + e^x + e^x - 2cosx]/[{xcosx + sinx + sinx]
Third Derivative: [xe^x + e^x + e^x + e^x + 2sinx]/[-xsinx + cosx + cosx + cosx]
Put x = 0: [0+1+1+1+0]/[0+1+1+1] = 3/3 = 1.

Apply an exponential of a logarithm to the expression.

Since the exponential function is continuous, we may factor it out of the limit.

The numerator of grows asymptotically slower than its denominator as x approaches ∞.
Since  grows asymptotically slower than e^z as x approaches
e⁰

now to calculate values we use Squeezing Theorem.


Hence,

Now we can see that after the 10/10 term, all subsequent terms are < 1, and keep decreasing. As we increase the value of n it the product will get close to 0.

Apply an exponential of a logarithm to the expression. 

Since the exponential function is continuous, we may factor it out of the limit.

Logarithmic functions grow asymptotically slower than polynomials.
since log(x) grows asymptotically slower than the polynomial x as  x approaches ∞,

at  it means that x = 0  is local minima.
but at  so we can't apply second derivative test. So, we can apply first derivative test.
 is not changing sign on either side of 2. So,x = 2 is neither maxima nor minima.

So, only one extremum i.e. x=0.

Given ,              x + y + z = 3
==>    x + (y/2) + (y/2) + (z/3) + (z/3) + (z/3) = 3
Now using A.M. G.M inequality , we have :  
[x + (y/2) + (y/2) + (z/3) + (z/3) + (z/3) ] / 6  >=   (x . (y/2) .(y/2) . (z/3) . (z/3) .(z/3))^(1/6)
==>    (x . (y/2) .(y/2) . (z/3) . (z/3) .(z/3))^(1/6)    <=   1/2
==>    (x . (y/2) .(y/2) . (z/3) . (z/3) .(z/3))     <=  1 / 64
==>     x . y² . z³    <=  108 / 64  =  27 / 16

is an extremum (either maximum or minimum).

So, the maximum value is at x = 2 which is 10 as there are no other extremum for the given function.

Since a polynomial is defined and continuous everywhere, we only need to check the critical point and the boundaries.

Critical point:  which is the minimum.
Boundaries: 
Since p(x) increases as x goes farther away from the 1, but 
p(x)  is defined on an open interval,
p(x) never attains a maximum!

Hence, e. None of the above is the correct answer.

Digvijay is right, f(x) will be minimum at x = 2,
Here is another approach.
Since log and exponent are monotonically increasing functions, the problem of minimizing f(x) can be reduced to just minimizing the quadratic expression
x² - 4x + 5,
this  quadratic expression can be written as (x² - 4x + 4) + 1 which is equal to
(x - 2)² + 1.
now since (x - 2)^{^}2 can not be less than 0, so (x - 2)^{^}2 + 1 can not be less than 1.
Also (x - 2)² + 1 will be at its minimum value (= 1), when x = 2.
so value of f(x) will be minimum at x = 2.

This is a repeating question on continuity. Let me solve it a non-standard way -- which should be useful in GATE.
From the question f is a function mapping the set of real (or rational) numbers between [0,1] to [0.4,0.6]. So, clearly the co-domain here is smaller than the domain set. The function is not given as onto and so, there is no requirement that all elements in co-domain set be mapped to by the domain set. We are half done now. Lets see the options:

A False, as we can have f(0.5) = 0.4, continuity does not imply anything other than all points being mapped being continuous.
C. Again false, we can have  for all .
D. False, same reason as for A.

D. False, same reason as for A.
Only B option left- which needs to be proved as correct now since we also have E option. We know that for a function all elements in domain set must have a mapping. All these can map to either 1 or more elements but at least one element must be there in the range set. i.e., f(x) = y is true for some y which is in [0.4,0.6]. In the minimal case this is a single element say c. Now for x = 1/0.8, option B is true. In the other case, say the minimal value of f(x) = a and the maximum value be f(x) = b. Now,

as per Intermediate Value theorem between  and  are also in the range set as f is continuous. Now, we need to consider x in the range [0.5, 0.75] as then only f(x) can be 0.8x and be in [0.4,0.6]. In our case we have

 Lets assume  Now, for all other points in must be between a and b and all points between a and b must be mapped by some x.

Moreover,for  aand for  So, if we plot this line should cross

 must be above or equal to the line 0.8x (shown below) and for x = 0.75 it must be belowor equal which means an intersection must be there.

This shows there exist some  a stronger case than option B. So, B option is true. Now please try for and see if it is true.

x>=4 and y>=4 , So we can take both x =5 & y = 5
x+y >= 10 => Satisfied , 5+5 = 10
2x + 3y >= 20. Satisfied.
This is infact minimum value.
Other options =>
4,4 => x+y constraint fail
4,5 => x+y fail
6,4 => Still giving 52 as sum which is more than 50 !,  This can not  be answer.
7,3 => 49+9 > 58 > 50.

Minimum value of function occurs at end points or critical points
f'(x)=1+logx
Equate it to 0
x=1/e
f''(x)=1/x
Put x=1/e f''(x)=e so minima at 1/e But 1/e=0.36
But x∈[1/2,infinity)
So min occurs at 1/2
So min value=1/2 log 1/2
So ans is c

For unstable equilibrium point, dx/dt >0
At x = 0, dx/dt =  (1-0)(2-0)(3-0) = 6 > 0
Hence x = 0 is point of unstable equilibrium.
We can understand equilibrium in terms of radioactive decay.
Let dN/dt = -KN ;K>0 its significance is that an element is loosing energy so it is getting stability because we know more energy an element gets,more de-stability it gains and vice versa

which lie between given domain in question
it means it is local maxima and at which is local minima and since it at  is local maxima so before it graph is strictly increasing so  is also local minima
so there is two local minima  

Sine function increases till π/2 and so for the considered interval π/4 would be a local minimum. From π/2, value of sine keeps on decresing till 3π/2 and hence 3π/2 would be another local minima. So, (D) is the correct answer here.

Since the polynomial has highest degree 7. So there are 7 roots possible for it
now suppose if an imaginary number a+bi is also root of this polynomial then a-bi will also be the root of this polynomial
That means there must be even number of complex root possible becoz they occur in pair.
Now we will solve this question option wise
A) All complex root
This is not possible. The polynomial has 7 roots and as I mention a polynomial should have even number of complex root and  7 is not even. So this option is wrong
B) At least one real root
This is possible. Since polynomial has 7 roots and only even number of complex root is possible, that means this polynomial has max 6 complex roots and Hence minimum one real root. So this option is correct
C) Four pairs of imaginary roots 4 pair means 8 complex root. But this polynomial can have atmost 7 roots. So this option is also wrong