The graph of the curve y = f(x) given below does not represent a function on R, give reasons?

f(x) is not a function as x Î R is not uniquely mapped i.e. for each x there exists 2 y values.

What is principal value of .

sin^{-}^{1}= -
sin^{-}^{1}

= - sin^{-}^{1}= -
sin^{-}^{1}

= - (since principal
range of sin^{-}^{1}x
is ).

If
A = , B = and
if A^{T} = B find x, y.

A^{T}

x + y = 6

2x + y = 1

⇒ x = - 5, y = 11

Find
the point on the curve y = x^{2} + 2x + 17, where the tangent is
parallel to x-axis. .

= 2x + 2 = 0 ( since tangent is parallel to x-axis ⇒ slope = 0)

⇒ x = - 1, if x = - 1, y = 1 - 2 + 17 = 16

\point is (-1, 16).

If |a| = 3, |b| = 2 and find ||.

= |a| |b| cosq⇒ 3 = 3.2. cosq⇒cosq = ⇒q =

= |a| |b| sinq = = .

If are two unit vectors and ‘’ is the angle between them, then find
the value

of .

1 + 1 + 2 cosq =
2 + 2 cosq = 2(1 + cosq) = 2. 2 cos^{2}

⇒= .

If
f(x) = 2x + 3, g(x) = x^{2} find fog (2), gof(2). .

fog(x) = f[g(x)] = f[x^{2}] = 2x^{2} +
3

⇒fog(x) = 2x^{2} + 3

⇒fog(2) = 11

gof(x) = g[f(x)] = g[2x + 3] = (2x+3)^{2}

⇒gof(x) = (2x + 3)^{2}

gof(2) = 49

Prove that

OR

Solve , |x| < 1.

Let

x = cos2q⇒ 2q = cos^{-}^{1}x

1 + cos2q = 2 cos^{2}q

1 - cos2q = 2 sin^{2}q

= = .

Or

⇒

⇒
2x^{2}- 4 = - 3 ⇒ 2x^{2} = 1 ⇒ x
= ±.

Evaluate .

C_{1} + C_{2}
+ C_{3}

= C_{1}- C_{2}

= C_{1}- C_{2}

= C_{3}- C_{1}

= C_{2}- C_{3}

= .

If
y = (sinx)^{x} + x^{sinx} find

y = (sinx)^{x} + x^{sinx},

Let y_{1} = (sinx)^{x }⇒ x
ln sinx = lny_{1}

⇒

⇒

Let y_{2} = x^{sinx} ⇒
lny_{2} = sinx .lnx.

.

A
water tank has the shape of an inverted right circular cone with its axis
vertical and vertex lower most. Its semi-vertical angle is tan^{-}^{1}. Water is poured into it
at a constant rate of 5 cubic meter per minute. Find the rate at which the
level of the water is rising at the instant when depth of water in the tank is
10 m.

Let at any instant radius be ‘r’ and ‘h’ be height of water and volume be V, given = 5 cu. mts

V =

V =

5 =

m/ minutes.

If y = , z = , then find .

y = = 2 tan^{-}^{1}x,

z = cos^{-}^{1}= 2tan^{-}^{1}x ,

⇒

Evaluate dx as limit of sum.

Let us divide the interval [1, 3] in ‘n’ equal intervals

viz. [1 + h], [1 + 2h], [1 + 3h], … [ 1+ (n-1)h, 3].

\sum of area of all ‘n’ rectangle

= h{(2(1)^{2} + 3) + (2(1 + h)^{2}
+ 3) + (2(1 + 2h)^{2} + 3) +...

+ (2(1 +
(n-1)h)^{2}
+ 3)]

=
h[2(1^{2} + 1^{2} + 1^{2} + … + n terms) + 4h (0 + 1 +
2 +

…+
n term) + 2h^{2} (0 + 1^{2} + 2^{2} + 3^{2} + …
+ n terms)

+ (3 + 3 + 3 + … n terms)

=

Where 3 - 1 = nh

⇒ h = and h→ 0 , n →¥.

\ A = .

= = .

Evaluate

Let x- b = t ⇒ dx = dt , x - a = t + b - a

=

= cos(b - a) + sin(b - a)

= t cos(b - a) + sin(b - a) . log|sint| + c .

= (x - b) cos(b - a) + sin(b - a) log|sin(x - b)| + c.

Solve the differential equation

, where y(0) = 1.

IF =

Solution is tanx = t, sec^{2}x
dx = dt

y.⇒ y
e^{tanx} = [t e^{t}-e^{t} ] + c

ye^{tanx} = [tanx- 1] e^{tanx}
+ c

⇒ y = (tanx- 1) + c e^{-}^{tanx} .

If are 3 mutually perpendicular vectors with magnitudes |a| = 2, |b| = 3, |c| = 4 find .

= 4.4 + 9.9 + 16.16 since

= 16 + 81 + 256 = 353

= .

Find the equation of the plane passing through (1, 2, 3); (-1, 4, 7), (0, 0, 2). .

OR

Find the vector, Cartesian equation of line passing though (1, 2, -7); (0, 3, 8). .

Equation of plane passing (a_{1}, b_{1},
c_{1}), (a_{2}, b_{2}, c_{2}), (a_{3},
b_{3}, c_{3})

i.e.

or

and also Cartesian form

A person speaks truth in 60% of cases and his wife speaks false in 80% of cases if both speak on same issue find the probability they agree upon each other. .

P(H) = , P() =

, P(w) =

= 44%

Evaluate

⇒

=

=

Show that the volume of largest cone that can be inscribed in a sphere is of the volume of sphere.

r =

h = R + x

=> volume of cone = V_{max}
=

= (let)

f’(x) =

=

x = (point of maxima) (Check using double derivative test)

V_{max} = =

Also volume of sphere =

V_{max} = V

Hence Proved.

Find the Area common to y^{2} =
4ax, x^{2} = 4ay (a > 0). .

OR

Find the area of the region {(x, y) |
0 ≤ y ≥ x^{2} + 1, 0< y < x + 1, 0 < x < 2}

Required Area =

=

=

= = sq. units.

OR

(y -
1) = x^{2} and

y = x + 1

x -y = - 1;

Solve

y =
x^{2 }+ 1

y = x + 1

to
get point of intersections x^{2} + 1 = x + 1

= x(x - 1) = 0 , x = 0, 1

Required area =

= sq. units.

A variance plane is at constant distance P from the origin and meets the co-ordinates axes at A, B and C. Show that the locus of centroid of triangle ABC .

Or

Find the vector equation of the plane passing through the point of intersection of planes

, and passing through (1, 1, 1).

Let the plane meets area at

A(a, 0, 0), B(0, b, 0),

C(0, 0, c). Equation of plane .

= (x_{1}, y_{1},
z_{1})

perpendicular distance from (0, 0, 0)

= ==

=>= .

Evaluate .

=

= tan^{2}x
= 1 ⇒ 2
tanx sec^{2}x dx = dt

I = .

A fair die is tossed 3 times

(i) Find the probability of getting a number greater than 4 at last once. .

(ii) Find the probability getting an even number of odd number of times. .

(i) Let getting number > 4 is success

P(success) = , P(failure) = q = , n = 3

P(number greater than 4 at Least once) = 1 - P(not even once)

= 1
-^{3}C_{0} a^{3}p^{0} = 1 -^{3}C_{0}=

(ii) Let getting even is success p = , q =

odd no. of time P(x = 1) or P(x = 3) .

P(x
= 1) + P(x - 3) = ^{3}C_{1} a^{2} p^{1}
+ ^{3}C_{3} p^{3} = 3 =

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour an machine to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit? If he operates his machines for a most 12 hours a day.

Let us construct the table with given data

Item |
No of packages |
Time on machine A |
Time on machine B |
Profit |

Nut bolt |
x Y |
1 x hrs 3 y hrs |
3 x hrs 1 y hr |
(35/2) x Rs 7 y Rs |

Total |
x+3y hrs |
3x + y hrs |
(35/2)x +y hrs |

P =

x + 3y ≤ 12

3x + y ≤ 12 x , y ≥ 0

Hence max. is P at (3, 3) i.e. 73.50 when x = 2, y = 3.

Find the local maximum and local minimum values of the function

f(x) = sin2x - x,

f¢(x) = 2 cos2x - 1

f¢(x) = 0 ⇒ cos2x = ⇒ 2x = , 2x = -

x = , x = -

f¢¢= - 4 sin= - 4. = - 2< 0.

f¢¢= - 4 sin= 4. = 2> 0.

⇒f(x) has local max. at x = + ,
f_{max} = =

f(x) has local min. at x = -, f_{min} = = .