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1/2.

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2.

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A team of 8 couples(husband and wife) attended a lucky draw in which 4 persons were picked up for a prize. The
probability that at least one couple is selected for the prize is

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A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is
selected at random from the bag and tossed. The probability of getting a head is

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The objective function of a linear programming problem is a

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linear equation.

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The vertices of a triangle ABC are A(–1, 2, –3), B(5, 0, –6) and C(0, 4, –1). The direction cosines of the bisector of the angle BAC are

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The angle between the line and the plane x + y + 4 = 0 is

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45°

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Let A = {2, 3, 4}, B = {5, 6, 7, 9} and let f = {(1, 4), (3, 6), (4, 7)} be a function from A to B, then function is ______________.

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Since, every element of set A has only one image in set B i.e., 2→5, 3→6 and 4→7. Therefore, f is one-one function.

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If A and B are two square matrices and K is a scalar quantity then K(A+B) = ______.

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K(A+B) = KA+KB

Q:

$The function f given by f (x) = x^{3} - {6x}^{2} + 14x, x \in R is________on R .$

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\begin{array}{l} The given function, \\ f (x) = x^{3} - 6 x^{2} + 14 x \\ Differentiating w .r .t . x, we get \\ f' (x) = 3 x^{2} - 12 x + 14 \\ = 3 x^{2} - 12 x + 12 + 2 \\ = 3 (x^{2} - 4 x + 4) + 2 \\ = 3 {(x - 2)}^{2} + 2 > 0 in every interval of R . \\ Therefore, the function f is \underline{strictly increasing} on R . \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@F369@

Q:

If x = 4at and y = 3at², then the equation of tangent at t = 2 is _______________.

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\begin{array}{l} Given, x = 4at and y = {3at}^{2} \\ Differentiating w .r .t . x, we get \\ \frac{d x}{d t} = 4 a and \frac{d y}{d t} = 6 a t \\ \frac{d y}{d x} = \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} \\ = \frac{6 a t}{4 a} \\ = \frac{3 t}{2} \\ {(\frac{d y}{d x})}_{t = 2} = \frac{3 (2)}{2} \\ = 3 \\ At t = 2, \\ x = 4 a (2) and y = 3 a {(2)}^{2} \\ = 8 a = 12 a \\ The equation of tangent is, \\ y - y_{1} = {(\frac{d y}{d x})}_{t = 2} (x - x_{1}) \\ y - 12 a = 3 (x - 8 a) \\ y - 12 a = 3 x - 24 a \\ \Rightarrow 3 x - y = 12 a \\ Therefore, the required equation of tangent is \underline{3 x - y = 12 a} . \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7E82@

Q:

$\begin{array}{l} If \hat{a} is a unit vector and (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15, then the value of \\ | \vec{x} | is ____. \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@7F17@$

A:

\begin{array}{l} Since \hat{a} is a unit vector, so | \hat{a} | = 1. \\ ∴ (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15 \\ \vec{x} . \vec{x} + \vec{x} . \hat{a} - \hat{a} . \vec{x} - \hat{a} . \hat{a} = 15 \\ \Rightarrow {| \vec{x} |}^{2} - 1 = 15 \\ [∵ \vec{x} . \hat{a} = \hat{a} . \vec{x} and \hat{a} . \hat{a} = 1] \\ \Rightarrow {| \vec{x} |}^{2} = 16 \\ \Rightarrow | \vec{x} | = 4 \\ ∴ If \hat{a} is a unit vector and (\vec{x} - \hat{a}) (\vec{x} + \hat{a}) = 15, then the \\ value of | \vec{x} | is \underline{4} . \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@64EA@

Q:

$\begin{array}{l} The scalar projection of the vector \vec{a} = 4 \hat{i} - 3 \hat{j} + 2 \hat{k} on the \\ vector \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k} is _______. \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8D0C@$

A:

\begin{array}{l} Scalar projection of \vec{a} on \vec{b} is given by \\ \frac{\vec{a} . \vec{b}}{| \vec{b} |} = \frac{(4 \hat{i} - 3 \hat{j} + 2 \hat{k}) . (4 \hat{i} - 4 \hat{j} + 7 \hat{k})}{| 4 \hat{i} - 4 \hat{j} + 7 \hat{k} |} \\ = \frac{16 + 12 + 14}{\sqrt{4^{2} + 4^{2} + 7^{2}}} \\ = \frac{42}{\sqrt{81}} \\ = \frac{42}{9} \\ = \frac{14}{3} \\ ∴ The scalar projection of the vector \vec{a} = 4 \hat{i} - 3 \hat{j} + 2 \hat{k} on \\ the vector \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k} is \underline{\frac{14}{3}} . \end{array} MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2Daebbfv3ySLgzGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@2731@

Q:

The value of determinant will _______ if its rows and columns are interchanged.

A:

remain unchanged

Q:

Find the area of a triangle whose vertices are given by (x₁, y₁), (x₂, y₂) and (x₃, y₃).

A:

Q:

Evaluate

A:

Q:

Evaluate:

A:

Q:

Evaluate

A:

Q:

Evaluate

A:

Q:

A:

Q:

Evaluate

A:

Q:

A:

Q:

Construct a 2 2 matrix A = [ a_ij] whose elements are given by

A:

Q:

A:

Q:

A:

Q:

Write in the simplest form.

A:

Q:

Find the value of .

A:

Q:

If A, B, C have position vectors (2, 0, 0), (0, 1, 0) and (0, 0, 2), show that ABC is isosceles.

A:

Q:

Prove that the angle in a semi-circle is a right angle.

A:

Let O be the centre of the semi-circle and r be its radius.

Let C be any point on the semi-circle.

Now,

Q:

A:

Q:

Evaluate .

A:

Q:

A:

Q:

A:

Q:

A:

Q:

Use differential to approximate 0.037.

A:

Let y = x, x = 0.040 and x +x = 0.037.

Then x = 0.037 - 0.047 = - 0.003

For x = 0.040, y = 0.040 = 0.2

Let dx = x = - 0.003

Now, y = x

i.e., dy/dx = 1/2x = 1/0.4.

Therefore, dy = dy/dx x dx

i.e., dy = (1/0.4) ( - 0.003) = -3/100

y = -3/400

Hence, 0.037 = y + y = 0.2 - (3/400) = 0.1925.

Q:

Solve the following linear programing problem graphically:
Maximise Z = 60x + 15y

Subject to constraints

x + y 50
3x + y 90
x, y 0

A:

Maximise Z = 60x + 15y (1)

Subject to constraints

x + y 50 (2)
3x + y 90 (3)
x, y 0 (4)

Graph the inequalities (2), (3) and (4). The feasible region (shaded) is shown below:

Corner Point	Z = 60x +15y
A(0, 50)	750
B(20, 30)	1650
C(30, 0)	1800Maximum

The maximum value of Z on the feasible region occurs at x = 30 and y = 0.

Q:

Find the area of the region {x, y : 0 y x² + 1, 0 y y x +1, 0 x 2}.

A:

Q:

Using the method of integration, find the area of the region bounded by inequation 2x+y8, x+2y8, x 0 and y0.

A:

Q:

In a class of 60 students, 32 like Maths, 30 like Biology and 24 like both Maths and Biology. If one of these students is selected at random, find the probability that the selected student

(a) likes Maths or Biology
(b) likes neither Maths nor Biology
(c) likes Maths but not Biology.

A:

Let the probability that a student likes Maths be denoted by P(M) and Biology be denoted by P(B).It is given that
P(M) = 32/60 = 8/15, P(B) = 30/60 = 1/2, P(MB) = 24/60 = 2/5

(a) Probability that the selected student likes Maths or Biology
P(MB) = P(M) + P(B) – P(MB)

= 8/15 + 1/2 - 2/5 = 19/30.

(b) Probability that the student likes neither Maths nor Biology = 1 - P(MB)

= 1 - 19/30 = 11/30

(c) Probability that a student likes Maths but not Biology = P(M) – P(MB)

= 8/15 - 2/5 = 2/15.

Q:

P,Q,R shot to hit a target. If P hits it 3 times in 4 trials, Q hits it 2 times in 3 trials and R hits it 4 times in 5 trials, what is the probability that the target is hit by at least two persons?

A:

Let E₁, E₂ and E₃be independent events that P, Q and R hit the target respectively. Then
P(E₁) = 3/4 , P(E₂) = 2/3 and P(E₃) = 4/5
P(E₁^c) = 1 - 3/4 = 1/4 , P(E₂^c) = 1 - 2/3 = 1/3 and P(E₃^c) = 1 - 4/5 = 1/5 .

The target is hit by atleast two persons in the following mutually exclusive ways:

(a) P hits, Q hits and R does not hit i.e., E₁E₂E₃^c
(b) P hits, Q does not hit and R hits i.e., E₁E₂^cE₃
(c) P does not hit, Q hits and R hits i.e., E₁^cE₂E₃
(d) P hits, Q hits and R hits i.e., E₁E₂E₃
Required Probability = P[(a) (b) (c) (d)] = P(a) + P(b) + P(c) + P(d)

= (3/4 x 2/3 x 1/5) + (1/4 x 2/3 x 4/5) + (3/4 x 2/3 x 4/5)

= 25/30 = 5/6.

Q:

A:

Q:

A:

Solved Board Paper For CBSE Class 12 Mathematics

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Solved Board Paper For CBSE Class 12 Mathematics

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A team of 8 couples(husband and wife) attended a lucky draw in which 4 persons were picked up for a prize. The probability that at least one couple is selected for the prize is

A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head is

The objective function of a linear programming problem is a

The vertices of a triangle ABC are A(–1, 2, –3), B(5, 0, –6) and C(0, 4, –1). The direction cosines of the bisector of the angle BAC are

The angle between the line and the plane x + y + 4 = 0 is

Let A = {2, 3, 4}, B = {5, 6, 7, 9} and let f = {(1, 4), (3, 6), (4, 7)} be a function from A to B, then function is ______________.

If A and B are two square matrices and K is a scalar quantity then K(A+B) = ______.

The function f given by f( x ) = x 3 – 6x 2 + 14x, x∈R is ________on R.

If x = 4at and y = 3at2, then the equation of tangent at t = 2 is _______________.

The value of determinant will _______ if its rows and columns are interchanged.

Find the area of a triangle whose vertices are given by (x1, y1), (x2, y2) and (x3, y3).

Evaluate

Evaluate:

Evaluate

Evaluate

Evaluate

Construct a 2 2 matrix A = [ aij ] whose elements are given by

Write in the simplest form.

Find the value of .

If A, B, C have position vectors (2, 0, 0), (0, 1, 0) and (0, 0, 2), show that ABC is isosceles.

Prove that the angle in a semi-circle is a right angle.

Evaluate .

Use differential to approximate 0.037.

Solve the following linear programing problem graphically: Maximise Z = 60x + 15y Subject to constraints x + y 50 3x + y 90 x, y 0

Find the area of the region {x, y : 0 y x2 + 1, 0 y y x +1, 0 x 2}.

Using the method of integration, find the area of the region bounded by inequation 2x+y8, x+2y8, x 0 and y0.

In a class of 60 students, 32 like Maths, 30 like Biology and 24 like both Maths and Biology. If one of these students is selected at random, find the probability that the selected student (a) likes Maths or Biology (b) likes neither Maths nor Biology (c) likes Maths but not Biology.

P,Q,R shot to hit a target. If P hits it 3 times in 4 trials, Q hits it 2 times in 3 trials and R hits it 4 times in 5 trials, what is the probability that the target is hit by at least two persons?

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Board Paper Solution 2020

Exam Weightage

A team of 8 couples(husband and wife) attended a lucky draw in which 4 persons were picked up for a prize. The
probability that at least one couple is selected for the prize is

A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is
selected at random from the bag and tossed. The probability of getting a head is

$The function f given by f (x) = x^{3} - {6x}^{2} + 14x, x \in R is________on R .$

If x = 4at and y = 3at², then the equation of tangent at t = 2 is _______________.

Find the area of a triangle whose vertices are given by (x₁, y₁), (x₂, y₂) and (x₃, y₃).

Construct a 2 2 matrix A = [ a_ij] whose elements are given by

Solve the following linear programing problem graphically:
Maximise Z = 60x + 15y

Subject to constraints

x + y 50
3x + y 90
x, y 0

Find the area of the region {x, y : 0 y x² + 1, 0 y y x +1, 0 x 2}.

In a class of 60 students, 32 like Maths, 30 like Biology and 24 like both Maths and Biology. If one of these students is selected at random, find the probability that the selected student

(a) likes Maths or Biology
(b) likes neither Maths nor Biology
(c) likes Maths but not Biology.