Mathematics

CBSE Class 12

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3.

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and Rs. z respectively.

Since total cash award is Rs. 6,000.

Three times the award money for hard work and honesty amounts to Rs.11,000.

Award money for honesty and hard work is double that given for regularity.

The above system of equations can be written in matrix form AX = B as:

Here,

Thus, A is non-singular. Hence, it is invertible.

Hence, x = 500, y = 2000, and z = 3500.

Thus, award money given for honesty, regularity and hardwork is Rs. 500, Rs.2000 and Rs. 3500 respectively.

The school can include awards for obedience.

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4.

Show that the function is continuous but not differentiable at x=3.

Let c be a real number.

Since f is continuous at x = 3.

**Case III: **C>3. Then f(c) = c - 3

Since,

Therefore, f is a continuous function.

Now, we need to show that

Consider the left hand limit of f at x = 3

Consider the right hand limit of f at x = 3

Since the left and right hand limits are not equal, f is not differentiable at x = 3.

Since,

Therefore, f is a continuous function.

Now, we need to show that

Consider the left hand limit of f at x = 3

Consider the right hand limit of f at x = 3

Since the left and right hand limits are not equal, f is not differentiable at x = 3.

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6.

Find the value of a if

Equating the corresponding elements, we get,

Now consider the equations,

a - b = -1 and 2a - b = 0

Subtracting first equation from second, we get: a = 1

a - b = -1 and 2a - b = 0

Subtracting first equation from second, we get: a = 1

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Write the principal value of

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8.

Using properties of determinants prove the following:

Applying we have

Expanding along R_{1, }we have:_{}

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