Question Bank
UNIT -1
Question Bank
UNIT -1
FUNCTIONS OF ONE VARIABLE
Q 1:
Find the nth derivative of sin3 x
Solution:
We know that sin 3x= 3sin x 4sin3 x = sin3x = 
Differentiate n times w.r.t x,
dn/dxn ( sin3 x) = 1/4 dn/dxn (3 sinx- sin3x)
=1/4 ( -3n. Sin( 3x+ nπ/2) + 3 sin (x+ nπ/2)) nϵz
Q 2:
Find the nth derivative of sin 5x. Sin 3x.?
Sol: let y = sin 5x.sin 3x= 1/ 2( sin 5x.sin 3x)
⇒y= ½( cos 2x - cos8x)
⇒ y= ½ ( cos 2x- cos8x )
Differentiate n times w.r.t x,
Yn = ½ (dn/dxn) ( cos 2x - cos8x )
⇒ yn = ½ ( 2 n (cos( 2x+ nπ/2)- 8n .cos (8x + nπ/2)) nϵz.
Successive n th derivative of nth elementary function ie., exponential
If y = ae n x + be –nx , then show that y2= n2y
Sol: Y= aenx + be-nx
y 1 = a.n.enx - b.n.e-nx
y2 = an2 enx – bn2 e-nx = n2 (ae nx+ be –nx)
y2= n2y.
Q 3:
If y= e-kx/2(a cosnx+ b sinnx) then show that.,y2+ ky1+(n2+ k2/4)y =0
Sol : y= e-kx/2(a cosnx+ b sinnx)
Differentiating w.r.to. x.,
Y1 = e-kx/2( -an sin nx + bn cos nx) - k/2.y
Y1+ k/2.y = ne-kx/2 ( -an sin nx + bn cos nx) (1)
Differentiating w.r.to x.,
Y2+ k/2.y1 = ne-kx/2 (-k/2) ( -an sin nx + bn cos nx) + n e-kx/2(-an cosnx- bn sinnx).
= -(k/2) (y1+ k/2 y)- n2 y = - (k/2 y1)- ( k2/4)y- n2y.
y2 + ky1 +(n2+ k2/4)y = 0.
Successive differentiation of nth derivative of elementary functions ie., logarithmic
Q 4:
If Y
2) = log( x + 
2) ) then show that (1 + x 2) y1 +xy =1
Sol: Y
2) = log( x + 
2) )
Differentiating w.r.to x.,
y.1/ 2
2 .2x+ 
2. Y1.
=(1+x ) y +xy = 
. 1/x+ 
2 .
2) +x/
2 =1
Q5:
Solution:
It is an alternating series ,with
here (-1)n defines only the sign
for all n
Now,
=
=0 as 
Hence by Leibnitz’s rule of alternating series 
is converging.
Q 6:
Q = 9R2- 
to find 
Solution:
Q = 9R2-15R-3
Now,
= 18R+45R-2 = 18R +
Power series:
A power series can be written in the form of
It can be written in the form of a function as.,
f(x) =
If we let 
=1 for all n then the power series tends to geometric series
f(x) = 1+
The above series converges if -1<x<1
The above series diverges if x≥ 1 and x
Q 7:
n
L= 
= 
1/n
L= 
n =
= 
<1
Hence the series converges.
Q 8:
Use lagranges mean value thorem to determnine a point P on the curve y=
,where the tangent is parallel to the chord joining (2,0) and (3,1).
Solution:
Consider y=
in [2,3]....(1)
(i)Function is continuous in[2,3] as algebraic expression with positive exponent is continuous
(ii)y’=
,y’ exists in (2,3) hence the function is derivable in(2,3)
Hence the conditions of LMV theorem are satisfied.
Hence thereexists, at least one c
,such that 
For x=
,tangent is parallel to the chor joining(2,0) and (3,1)
By substituting in equation (1) we get ,
=
= 
Q 9:
Find the slope point on the graph y=
on[1,2] at which the tangent to the graph has the same slope as the line that passes through the end points (1,2) and (3,12)of the closed interval.
Solution:
Step 1: find the slope of the points that passes through the endpoints at
the required slope is given by 
=
=9
Step 2:
Find the point on the curve where the slope is 9
Y’=6x
6x=9 
the instantaneous rate of change of slope is 9
Q 10:
Show that sinx<x for all x>0
Solution:
If x>2
,then sinx
1<2
<x. If 0<x
2
,then by the mean value theorem,there-exists c in open interval(0,2
) such that
x=c = cosc<1
Thus, sinx<x in this case also.
Q 11:
Find the values of the c,which satisfy the equation 
as stated in mean value theorem for the function f(x) =
in the interval [1,3]
Solution:
First let us check the mean value theorem conditions,
F(x) is continuous in its domain[0,
and hence the given interval [1,3]
F(x) is also differentiable in the given interval .plugging a=1 and b=3 in the expression on the left side of the equation.
Now, the derivative of the function can be found using chain rule as
f’(x) =
Hence, the equation can be formed as
f’(c)= 
Cross multiplying and squaring the equation reduces to,
2(c-1)=1 which gives the solution as c=3/2 which lies in the interval [1,3].
Q 12:
In a geometric progression,the second term is 12 and the sixth term is 192.find the eleventh term
Solution:
The required geometric progression is,
a+ar+ar2+ar3+.....
Given the second term is ar=12...(1)
Sixth term is ar5 = 192.....(2)
Dividing eq(2) by eq(1) we get ,
r4 = 16
Substituting r in eq(1) we get ar=12/2=6
eleventh term is given by ar10 = 6
210 =6144
Q 13:
If sn denotes the sum of first n terms of an A.P and [s3n-sn-1]/[s2n-s2n-1]=31,then the value of n is
Solution:
S 3n = [3n / 2] [2a + (3n − 1) d]
Sn−1 = ([n − 1] / 2) * [2a + (n − 2) d]
⇒ S3n − Sn−1 = [1 / 2] [2a (3n − n + 1)] + [d / 2] [3n (3n − 1) − (n − 1) (n − 2)]
= [1 / 2] [2a (2n + 1) + d (8n2 − 2)]
= a (2n + 1) + d (4n2 − 1)
= (2n + 1) [a + (2n − 1) d]
S2n − S2n−1 = T2n = a +(2n − 1) d
⇒ S3n − Sn − S2n −S2n−1 = (2n + 1)
Given, ⇒ [S3n − Sn−1] / [S2n − S2n−1] = 31
⇒ n = 15
Q 14:
If Sn denotes the sum of first n terms of an A.P ,whose first term is 
is independent of x,then SP =?
Solution:
= 
For 
to be independent
2a-d=0 or 2a=d
Now,
Sp = 
Q 15:
Find the taylor series for the following:
= 
<1
(x/10)<1 and (x/10) > -1
Therefore radius of convergence is (-10,10)
ROC =10
Q 16:
f(n)5 = 
Here the ROC is 4
Q 17:
f(x)=
= f(0)+f’(0)x+
x2 + 
x3 +......
= 1+x+x2 +x3 + .....
= 
Q 18:
Solve the following improper integral 
dx
Solution:
By definition
Here we consider at c=0
= 
Here the integral converges at 
Q 19:
Find 
Solution:
The denominator of 
is 0 when x=2, so the function is not defined even when x=2.so,
=
-ln(4) =-
So the integral diverges
