**Question 6.** Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

test

**Q 15)**

Given function is
Given function is satisfies for the all real values of x
case (i) k < 0
Hence, function is continuous for all values of x < 0
case (ii) x = 0
L.H.L at x= 0
R.H.L. at x = 0
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0
case (iii) k > 0
Hence , function is continuous for all values of x > 0
case (iv) k < 1
Hence , function is...

**Q18. **The general solution of the differential equation is

(A)

(B)

(C)

(D)

Given equation is
we can rewrite it as
It is type of equation where
Now,
Now, the general solution is
Therefore, (C) is the correct answer

**Q16**. The general solution of the differential equation is

(A)

(B)

(C)

(D)

Given equation is
we can rewrite it as
Integrate both the sides
we will get
Therefore, answer is (C)

**Q17. **The general solution of a differential equation of the type is

(A)

(B)

(C)

(D)

Given equation is
and we know that the general equation of such type of differential equation is
Therefore, the correct answer is (C)

**Q15.**** **The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Let n be the population of the village at any time t.
According to question,
Now, at t=0, n = 20000 (Year 1999)
Again, at t=5, n= 25000 (Year 2004)
Using these values, at t =10 (Year 2009)
Therefore, the population of the village in 2009 will be 31250.

**Q14. **Find a particular solution of the differential equation , given that when

Given equation is
we can rewrite it as
Integrate both the sides
Put
put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
Now, put the value of C
Therefore, the particular solution is

**Q13. **Find a particular solution of the differential equation , given that .

Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when
at
Now, put the value of C
Therefore, the particular solution is

**Q12. **Solve the differential equation .

Given,
This is equation is in the form of
p = and Q =
Now, I.F. =
We know that the solution of the given differential equation is:

**Q11. **Find a particular solution of the differential equation , given that, when . (Hint: put )

Given equation is
Now, integrate both the sides
Put
Now, given equation become
Now, integrate both the sides
Put again
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is

**Q10. **Solve the differential equation

**Q8. **Find the equation of the curve passing through the point whose differential equation is

Given equation is
we can rewrite it as
Integrate both the sides
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point
So,
Now,
Therefore, the equation of the curve passing through the point whose differential equation is is

**Q9. **Find the particular solution of the differential equation , given that when .

Given equation is
we can rewrite it as
Now, integrate both the sides
Put
Put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is

**Q7. **Show that the general solution of the differential equation is given by , where *A* is parameter.

**Q6. **Find the general solution of the differential equation

Given equation is
we can rewrite it as
Now, integrate on both the sides
Therefore, the general solution of the differential equation is

**Q5. **Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Now, equation of the circle with center at (x,y) and radius r is
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
-(i)
Differentiate it w.r.t x
we will get
-(ii)
Put value from equation (ii) in equation (i)
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is

**Q4. **Prove that is the general solution of differential equation, where c is a parameter.

Given,
Now, let y = vx
Substituting the values of y and y' in the equation,
Integrating both sides we get,
Now,
Let
Now,
Let v2 = p
Now, substituting the values of I1 and I2 in the above equation, we get,
Thus,

**Q3. **Form the differential equation representing the family of curves given by , where *a* is an arbitrary constant.

Given equation is
we can rewrite it as
-(i)
Differentiate both the sides w.r.t x
-(ii)
Put value from equation (ii) in (i)
Therefore, the required differential equation is

**Q2. **Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv)

Given,
Now, differentiating both sides w.r.t. x,
Putting values in LHS
Therefore, the given function is the solution of the corresponding differential equation.

**Q2. **Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii)

Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.

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