Maths NCERT Exemplar Solutions Class 12th Chapter Twelve: Overview, Questions, Preparation

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve 2025 ( Maths NCERT Exemplar Solutions Class 12th Chapter Twelve )

Raj Pandey
Updated on Apr 23, 2025 12:00 IST

By Raj Pandey

Table of content
  • Linear Programming Fill in the blanks type Questions
  • Linear Programming True or False type Questions
  • JEE Mains Solutions 2020, 4th september, Maths, second shift
  • JEE Mains Solutions 2020, 4th september, Maths, First shift
Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Logo

Linear Programming Fill in the blanks type Questions

1. In a LPP, the linear inequalities or restrictions on the variables are called __________.
Constraints
2. In a LPP, the objective function is always __________.
Linear
Q&A Icon
Commonly asked questions
Q:  

In a LPP, the objective function is always __________.

A: 

In a LPP, the objective function is always Linear

Q:  

If the feasible region for a LPP is __________, then the optimal value of the objective function Z=ax+by may or may not exist.

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A: 

If the feasible region for a LPP is o p e n u n b o u n d e d , then the optimal value of the objective function Z=ax+by may or may not exist.

Q:  

In a LPP, if the objective function Z=ax+by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points gives the same __________ value.

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A: 

In a LPP, if the objective function Z=ax+by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points gives the same maximum value.

Q:  

A feasible region of a system of linear inequalities is said to be __________ if it can be enclosed within a circle.

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A: 

A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.

Q:  

A corner point of a feasible region is a point in the region which is the __________ of two boundary lines.

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A: 

A corner point of a feasible region is a point in the region which is the intersection of two boundary lines.

Q:  

The feasible region for an LPP is always a __________ polygon.

A: 

The feasible region for an LPP is always a  convex polygon.

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Logo

Linear Programming True or False type Questions

1. If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = a x + b y  may or may not exist.

True
2. The maximum value of the objective function Z = a x + b y  in a LPP always occurs at only one corner point of the feasible region.
False
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Commonly asked questions
Q:  

If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = a x + b y  may or may not exist.

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A: 

The following statement is true.

Q:  

In a LPP, the minimum value of the objective function Z=ax+by is always 0 if the origin is one of the corner points of the feasible region.

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A: 

The following statement is false.

Q:  

The maximum value of the objective function Z=ax+by in a LPP always occurs at only one corner point of the feasible region.

Read more
A: 

The following statement is false.

Q:  

In a LPP, the maximum value of the objective function  Z=ax+by is always finite.

A: 

The following statement is true.

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Logo

JEE Mains Solutions 2020, 4th september, Maths, second shift

JEE Mains Solutions 2020, 4th september, Maths, second shift

Q&A Icon
Commonly asked questions
Q:  

If a and b are real numbers such that (2+α)4=a+bα , where α=-1+i32 , then a+b is equal to:

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A: 

x2-3x+p=0

α, β, γ, δ in G.P.

α+αr=3

x2-6x+q=0

αr2+αr3=6

(2)÷ (1)r2=2

So,  2q+p2q-p=2r5+r2r5-r=2r4+12r4-1=97

Q:  

Contrapositive of the statement:

If a function f is differentiable at a , then it is also continuous at a ', is:

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A: 

03? g (x)-f (x)=03? ||x-2|-2|dx-03? |x-2|dx

=12×2×2+1+12×1×1-12×2×2+12×1×1

=2+1+12-2+12=1

Q:  

If for some positive integer n , the coefficients of three consecutive terms in the binomial expansion of (1+x)N+5 are in the ratio 5:10:14 , then the largest coefficient in the expansion is:

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A: 

f (2)=8, f' (2)=5, f' (x)1, f'' (x)4, x (1,6)

Using LMVT

f'' (x)=f' (5)-f' (2)5-24f' (5)17

f' (x)=f (5)-f (2)5-21f (5)11

Therefore f' (5)+f (5)28

Q:  

The function f(x)=π4+tan-1x,|x|112(|x|-1),|x|>1 is:

A: 

x2a2+y2b2=1 (ab)2b2a=10b2=5a

Now,  ? (t)=512+t-t2=812-t-122
? (t)max=812=23=ee2=1-b2a2=49

a2=81  (From (i) and (ii)

So,  a2+b2=81+45=126

Q:  

If the system of equations

x+y+z=2

2x+4y-z=6

3x+2y+λz=μ

has infinitely many solutions, then:

A: 

x=10

x? =63+a+b8=10

a+b=17

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  σ2=13.5=79+ (a-10)2+ (b-10)28

a2+b2-20 (a+b)=-171

a2+b2=169

From (1) and (2) ; a=12 and b=5

Q:  

Let λ0 be in R. If α and β are the roots of the equation, x2-x+2λ=0 and α and γ are the roots of the equation, 3x2-10x+27λ=0 , then βγλ is equal to:

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A: 

f(x)=a?(b?×c?)=x-23-2x-17-2x

=x3-27x+26

f'(x)=3x2-27=0x=±3 and f''(-3)<0

 local maxima at x=x0=-3

Thus, a?=-3iˆ-2jˆ+3kˆ,b?=2iˆ-3jˆ-kˆ , and c?=7iˆ-2jˆ-3kˆ

a?b?+b?c?+c?a?=9-5-26=-22

Q:  

The circle passing through the intersection of the circles, x2+y2-6x=0 and x2+y2-4y =0 , having its centre on the line, 2x-3y+12=0 , also passes through the point:

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A: 

(a+2bcos?x)(a-2bcos?y)=a2-b2

a2-2abcos?y+2abcos?x-2b2cos?xcos?y=a2-b2

Differentiating both sides:

0-2ab-sin?ydydx+2ab(-sin?x)-2b2cos?x-sin?ydydx+cos?y(-sin?x)=0

At π4,π4 :

abdydx-ab-2b2-12dydx-12=0dxdy=ab+b2ab-b2=a+ba-b;a,b>0

Q:  

The integral π/6π/3?tan3?xsin2?3x2sec2?xsin2?3x+3tan?xsin?6xdx is

A: 

 tan? θ=10x=hx2x2=hx10

tan? ? =15x=hxx1=hx15

Now,  x1+x2=x=hx15+hx10

1=h10+h15h=6

Q:  

The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x -axis and vertices C and D lie on the parabola, y=x2-1 below the x -axis, is:

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A: 

1+1-22.1+1-42.3+.+1-202.19

=α-220β

=11-221+423+..+20219

=11-22r=110? r2 (2r-1)=11-411022-35×11

=11-220 (103)

α=11, β=103

Q:  

Suppose the vectors x1,x2 and x3 are the solutions of the system of linear equations, Ax =b when the vector b on the right side is equal to b1,b2 and b3 respectively. If x1=111,x2=021,x3=001,b1=100,b2=020 and b3=002 , then the determinant of A is equal of A is equal to :

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A: 

Let TV (r) denotes truth value of a statement r .

 Now, if TV (p)=TV (q)=T

TVS1=F

Also, if TV (p)=T and TV (q)=F

TVS2=T

Q:  

The solution of differential equation dydx-y+3xloge?(y+3x)+3=0 is:

(where C is a constant of integration.)

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A: 

u=2z+iz-ki

=2x2+(2y+1)(y-k)x2+(y-k)2+i(x(2y+1)-2x(y-k))x2+(y-k)2

Since Re?(u)+Im?(u)=1

2x2+(2y+1)(y-k)+x(2y+1)-2x(y-k)=x2+(y-k)2

P0,y1Q0,y2y2+y-k-k2=0y1+y2=-1y1y2=-k-k2

?PQ=5

y1-y2=5k2+k-6=0

k=-3,2

So, k=2(k>0)

Q:  

The minimum value of 2 s i n x + 2 c o s x is:

A: 

K-2h-11-23-1=-1K=2h

? [? ABC]=55

12 (5) (h-1)2+ (K-2)2=55

Q:  

Let f:(0,)(0,) be a differentiable function such that f(1)=e and limtxt2f2(x)-x2f2(t)t-x=0 . If f(x)=1 , then x is equal to:

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A: 

Since  (3,3) lies on x2a2-y2b2=1

9a2-9b2=1

Now, normal at  (3,3) is y-3=-a2b2 (x-3) ,

which passes through  (9,0)b2=2a2

So,  e2=1+b2a2=3

Also,  a2=92

(From (i) and (ii)

Thus,  a2, e2=92, 3

Q:  

The distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line x2=y3=z-6 is:

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A: 

xxsin?x+cos?x2dx=xcos?xxcos?xdx(xsin?x+cos?x)2

=xcos?x-1xsin?x+cos?x+cos?x+xsin?xcos2?x1xsin?x+cos?xdx=-xsec?xxsin?x+cos?x+sec2?xdx=-xsec?xxsin?x+cos?x+tan?x+C

Q:  

In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:

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A: 

A2=cos?2θisin?2θisin?2θcos?2θ

Similarly, A5=cos?5θisin?5θisin?5θcos?5θ=abcd

(1) a2+b2=cos2?5θ-sin2?5θ=cos?10θ=cos?75?

(2) a2-d2=cos2?5θ-cos2?5θ=0

(3) a2-b2=cos2?5θ+sin2?5θ=1

(4) a2-c2=cos2?5θ+sin2?5θ=1

Q:  

The angle of elevation of a cloud C from a point P,200m above a still take is 30 . If the angle of depression of the image of C in the lake from the point P is 60 , then PC (in m ) is equal to

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A: 

 r=020?50-rC6=50C6+49C6+48C6+.+30C6

=50C6+49C6+48C6+.+30C6+30C7-30C7=50C6+49C6+48C6+.+31C6+31C7-30C7=50C6+50C7-30C7=51C7-30C7

Q:  

Let x=4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12 . If P(1,β),β>0 is a point on this ellipse, then the equation of the normal to it at P is:

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A: 

[x]2+2 [x+2]-7=0

[x]2+2 [x]+4-7=0 [x]=1, -3x [1,2) [-3, -2)

Q:  

Let i=150Xi=i=1nYi=T , where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi 's and exactly 6 of sets Yi 's then n is equal to:

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A: 

f(x)=13?xdx(1+x)2=13?t.2tdt1+t22 (put x=t )

=-t1+t213+tan-1?t13 [Applying by parts]

=-34-12+π3-π4=12-34+π12

Q:  

If the perpendicular bisector of the line segment joining the points P(1,4) and Q(k,3) has y -intercept equal to -4 , then a value of k is:

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A: 

 max {n (A), n (B)}n (AB)n (U)

7676+63-x100

-63-x-39

63x39

Q:  

Let a1,a2,an be a given A.P. whose common difference is an integer and Sn=a1+a2 ++an . If a1=1,a1=300 and 15n50 , then the ordered pair Sn-4,an-4 is equal to:

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A: 

xdydx-y=x2(xcos?x+sin?x),x>0

dydx-yx=x(xcos?x+sin?x)dydx+Py=Q

So, I.F. =e-1xdx1|x|=1x(x>0)

Thus, yx=1x(x(xcos?x+sin?x))dx

yx=xsin?x+C

?y(π)=πC=1

So, y=x2sin?x+x(y)π/2=π24+π2

Also, dydx=x2cos?x+2xsin?x+1

d2ydx2=-x2sin?x+4xcos?x+2sin?x

Q:  

If a=2iˆ+jˆ+2kˆ , then the value of |iˆ×(a×iˆ)|2+|jˆ×(a×jˆ)|2+|kˆ×(a×kˆ)|2 is equal to

A: 

Given 2x2+3x+410=r=020?arxr

replace x by 2x in above identity :-

2102x2+3x+410x20=r=020??ar2rxr210r=020??arxr=r=020??ar2rx(20-r)( from (i))

now, comparing coefficient of x7 from both sides

(take r=7 in L.H.S. and r=13 in R.H.S.)

210a7=a13213a7a13=23=8

Q:  

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is

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A: 

D1=-74-181515b6=0b=-3D=14-1315ab6=021a-8b-66=0P:2x-3y+6z=15

so required distance =217=3

Q:  

Let PQ be a diameter of the circle x2+y2=9 . If α and β are the lengths of the perpendiculars from P and Q on the straight line, x+y=2 respectively, then the maximum value of αβ is......

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A: 

D=1-232111-7a=0a=8

also,  D1=9-23b1124-78=0b=3

hence,  a-b=8-3=5

Q:  

If the variance of the following frequency distribution :

 Class :10-2020-3030-40 Frequency :2x2 is 50 , then x is equal to:

A: 

 Since,  limx0? f (x)x exist f (0)=0

Now,  f' (x)=limh0? f (x+h)-f (x)h=limh0? f (h)+xh2+x2hh ( take y=h)

=limh0? f (h)h+limh.0? (xh)+x2

f' (x)=1+0+x2

f' (3)=10

Q:  

Let {x} and [x] denote the fractional part of x and the greatest integer x respectively of a real number x . if 0n{x}dx,0n[x]dx and 10n2-n,(nN,n>1) are three consecutive terms of a G.P. then n is equal to(Integration)

Read more
A: 

We have,  1-  (probability of all shots result in failure)  > 1 4

1 - 9 10 n > 1 4 3 4 > 9 10 n n 3

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Logo

JEE Mains Solutions 2020, 4th september, Maths, First shift

JEE Mains Solutions 2020, 4th september, Maths, First shift

Q&A Icon
Commonly asked questions
Q:  

Let α and β be roots of x2-3x+p=0 and γ and δ be the roots of x2-6x+q=0 . If α,β , γ,δ , form a geometric progression. Then ratio (2q+p):(2q-p) is:

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A: 

 x2-3x+p=0

α, β, γ, δ in G.P.

α+αr=3

x2-6x+q=0

αr2+αr3=6

(2)÷ (1)r2=2

So,  2q+p2q-p=2r5+r2r5-r=2r4+12r4-1=97

Q:  

Let f(x)=|x-2| and g(x)=f(f(x)),x[0,4] . Then 03?(g(x)-f(x))dx is equal to:

A: 

03? g (x)-f (x)=03? ||x-2|-2|dx-03? |x-2|dx

=12×2×2+1+12×1×1-12×2×2+12×1×1

=2+1+12-2+12=1

Q:  

Let f be a twice differentiable function on (1,6) . If f(2)=8,f'(2)=5,f'(x)1 and f " (x)4 , for all x(1,6) , then:

Read more
A: 

 f (2)=8, f' (2)=5, f' (x)1, f'' (x)4, x (1,6)

Using LMVT

f'' (x)=f' (5)-f' (2)5-24f' (5)17

f' (x)=f (5)-f (2)5-21f (5)11

Therefore f' (5)+f (5)28

Q:  

Let x2a2+y2b2=1(ab) be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, ?(t)=512+t-t2 , then a2+b2 is equal to:

Read more
A: 

 x2a2+y2b2=1(ab);2b2a=10b2=5a

Now, ?(t)=512+t-t2=812-t-122
?(t)max=812=23=ee2=1-b2a2=49

a2=81 (From (i) and (ii)

So, a2+b2=81+45=126

Q:  

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5,7,10,12,14,15 , then the absolute difference of the remaining two observations is:

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A: 

x=10

x? =63+a+b8=10

a+b=17

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  σ2=13.5=79+ (a-10)2+ (b-10)28

a2+b2-20 (a+b)=-171

a2+b2=169

From (1) and (2) ; a=12 and b=5

Q:  

Let x0 be the point of local maxima of f(x)=a(b×c) where a=xiˆ-2jˆ+3kˆ,b=-2iˆ+xjˆ-kˆ and c=7iˆ-2jˆ+xkˆ . Then the value of ab+bc+ca at x=x0 is

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A: 

f(x)=a?(b?×c?)=x-23-2x-17-2x

=x3-27x+26

f'(x)=3x2-27=0x=±3 and f''(-3)<0

 local maxima at x=x0=-3

Thus, a?=-3iˆ-2jˆ+3kˆ,b?=2iˆ-3jˆ-kˆ , and c?=7iˆ-2jˆ-3kˆ

a?b?+b?c?+c?a?=9-5-26=-22

Q:  

If (a+2bcosx)(a-2bcosy)=a2-b2 , where a>b>0 , then dxdy at π4,π4 is:

A: 

(a+2bcos?x)(a-2bcos?y)=a2-b2

a2-2abcos?y+2abcos?x-2b2cos?xcos?y=a2-b2

Differentiating both sides:

0-2ab-sin?ydydx+2ab(-sin?x)-2b2cos?x-sin?ydydx+cos?y(-sin?x)=0

At π4,π4 :

abdydx-ab-2b2-12dydx-12=0dxdy=ab+b2ab-b2=a+ba-b;a,b>0

Q:  

Two vertical poles AB=15m and CD=10m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD , then the height of P (in m ) above the line AC is:

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A: 

tan? θ=10x=hx2x2=hx10

tan? ? =15x=hxx1=hx15

Now,  x1+x2=x=hx15+hx10

1=h10+h15h=6

Q:  

If 1+1-221+1-423+1-625+.+1-20219=α-220β , then an ordered pair (α,β) is equal to:

A: 

1+1-22.1+1-42.3+.+1-202.19

=α-220β

=11-221+423+..+20219

=11-22r=110? r2 (2r-1)=11-411022-35×11

=11-220 (103)

α=11, β=103

Q:  

Given the following two statements:

S1:(qp)(pq) is a tautology.

S2:q(pq) is a fallacy. Then:

A: 

Let TV (r) denotes truth value of a statement r .

 Now, if TV (p)=TV (q)=T

TVS1=F

Also, if TV (p)=T and TV (q)=F

TVS2=T

Q:  

Let u=2z+iz-ki,z=x+iy and k>0 . If the curve represented by Re(u)+Im(u)=1 intersects the y -axis at the points P and Q where PQ=5 , then the value of k is:

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A: 

u=2z+iz-ki

=2x2+(2y+1)(y-k)x2+(y-k)2+i(x(2y+1)-2x(y-k))x2+(y-k)2

Since Re?(u)+Im?(u)=1

2x2+(2y+1)(y-k)+x(2y+1)-2x(y-k)=x2+(y-k)2

P0,y1Q0,y2y2+y-k-k2=0y1+y2=-1y1y2=-k-k2

?PQ=5

y1-y2=5k2+k-6=0

k=-3,2

So, k=2(k>0)

Q:  

A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1) . If BAC=90 , and ar(ABC)=55 sq. units, then the abscissa of the vertex C is:

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A: 

K-2h-11-23-1=-1K=2h

? [? ABC]=55

12 (5) (h-1)2+ (K-2)2=55

Q:  

Let P(3,3) be a point on the hyperbola, x2a2-y2b2=1 . If the normal to it at P intersects the x -axis at (9,0) and e is its eccentricity, then the ordered pair a2,e2 is equal to:

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A: 

Since  (3,3) lies on x2a2-y2b2=1

9a2-9b2=1

Now, normal at  (3,3) is y-3=-a2b2 (x-3) ,

which passes through  (9,0)b2=2a2

So,  e2=1+b2a2=3

Also,  a2=92

(From (i) and (ii)

Thus,  a2, e2=92, 3

Q:  

The integral xxsinx+cosx2dx is equal to (where C is constant of integration):

A: 

 xxsin?x+cos?x2dx=xcos?xxcos?xdx(xsin?x+cos?x)2

=xcos?x-1xsin?x+cos?x+cos?x+xsin?xcos2?x1xsin?x+cos?xdx=-xsec?xxsin?x+cos?x+sec2?xdx=-xsec?xxsin?x+cos?x+tan?x+C

Q:  

If A=cos?θisin?θisin?θcos?θ,θ=π24 and A5=abcd , where i=-1 , then, which one of the following is not true?

A: 

A2=cos?2θisin?2θisin?2θcos?2θ

Similarly, A5=cos?5θisin?5θisin?5θcos?5θ=abcd

(1) a2+b2=cos2?5θ-sin2?5θ=cos?10θ=cos?75?

(2) a2-d2=cos2?5θ-cos2?5θ=0

(3) a2-b2=cos2?5θ+sin2?5θ=1

(4) a2-c2=cos2?5θ+sin2?5θ=1

Q:  

The value of r=02050-rC6 is equal to:

A: 

r=020?50-rC6=50C6+49C6+48C6+.+30C6

=50C6+49C6+48C6+.+30C6+30C7-30C7=50C6+49C6+48C6+.+31C6+31C7-30C7=50C6+50C7-30C7=51C7-30C7

Q:  

Let [t] denote the greatest integer t . Then the equation in x,[x]2+2[x+2]-7=0 has:

A: 

  [x]2+2 [x+2]-7=0

[x]2+2 [x]+4-7=0 [x]=1, -3x [1,2) [-3, -2)

Q:  

Let f(x)=x(1+x)2dx(x0) . Then f(3)-f(1) is equal to:

A: 

f(x)=13?xdx(1+x)2=13?t.2tdt1+t22 (put x=t )

=-t1+t213+tan-1?t13 [Applying by parts]

=-34-12+π3-π4=12-34+π12

Q:  

A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B . if x% of the people read both the newspapers, then a possible value of x can be:

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A: 

max {n (A), n (B)}n (AB)n (U)

7676+63-x100

-63-x-39

63x39

Q:  

Let y=y(x) be the solution of the differential equation, xy'-y=x2(xcosx+sinx),x>0 . If y(π)=π , then y''π2+yπ2 is equal to:

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A: 

xdydx-y=x2(xcos?x+sin?x),x>0

dydx-yx=x(xcos?x+sin?x)dydx+Py=Q

So, I.F. =e-1xdx1|x|=1x(x>0)

Thus, yx=1x(x(xcos?x+sin?x))dx

yx=xsin?x+C

?y(π)=πC=1

So, y=x2sin?x+x(y)π/2=π24+π2

Also, dydx=x2cos?x+2xsin?x+1

d2ydx2=-x2sin?x+4xcos?x+2sin?x

Q:  

Let 2x2+3x+410=r=020?arxr . Then a7a13 is equal to

A: 

Given 2x2+3x+410=r=020?arxr

replace x by 2x in above identity :-

2102x2+3x+410x20=r=020??ar2rxr210r=020??arxr=r=020??ar2rx(20-r)( from (i))

now, comparing coefficient of x7 from both sides

(take r=7 in L.H.S. and r=13 in R.H.S.)

210a7=a13213a7a13=23=8

Q:  

If the equation of a plane P , passing through the intersection of the planes, x+4y-z+7 =0 and 3x+y+5z=8 is ax+by+6z=15 for some a,bR , then the distance of the point (3,2,-1) from the plane P is(

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A: 

D=14-1315ab6=021a-8b-66=0P:2x-3y+6z=15

so required distance =217=3

Q:  

If the system of equations

x-2y+3z=9

2x+y+z=b

x-7y+az=24

has infinitely many solutions, then a-b is equal to

A: 

D=1-232111-7a=0a=8

also,  D1=9-23b1124-78=0b=3

hence,  a-b=8-3=5

Q:  

Suppose a differentiable function f(x) satisfies the(Functions)

identity f(x+y)=f(x)+f(y)+xy2+x2y , for all real x and y . If limx0f(x)x=1 then f'(3) is equal to

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A: 

Since,  limx0? f (x)x exist f (0)=0

Now,  f' (x)=limh0? f (x+h)-f (x)h=limh0? f (h)+xh2+x2hh ( take y=h)

=limh0? f (h)h+limh.0? (xh)+x2

f' (x)=1+0+x2

f' (3)=10

Q:  

The probability of a man hitting a target is 110 . The least number of shots required, so that the probability of his hitting the target at least once is greater than 14 , is(Probability)

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A: 

We have,  1-  (probability of all shots result in failure)  > 1 4

1 - 9 10 n > 1 4 3 4 > 9 10 n n 3

qna

Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Exam

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