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Solution :
Let
Solution :
Let
Solution :
Let,
Solution :
Let,
Solution :
Let y =
Solution :
Let ,y =
Solution :
Let, y =
Question
9.
Solution :
Let, y =
Question
10.
, for some fixed a> 0 and x > 0
Solution :
Let y =
Since a is constant, aa is also a constant.
∴ ds/dx = 0 .....(5)
From (1), (2), (3), (4), and (5), we obtain
Question
11.
, for x > 3
Solution :
Question
12. Find dy/dx , if
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Solution :
Question 1
3. Find
Solution :
Question
14. If
Solution :
It is given that,
Differentiating both sides with respect to x, we obtain
Hence, proved.
Question
15.
Solution :
It is given that,
Differentiating both sides with respect to x, we obtain
= - c
which is constant and independent of a and b
Hence, proved.
Question
16. If cos y = x cos (a + y), with cos a ≠ ± 1, prove that prove that
Solution :
It is given cos y = x cos (a + y)
Hence, proved.
Question
17. If x = a (cos t + t sin t) and y = a (sin t – t cos t), find
Solution :
It is given that, x = a(cost + tsin t) and y = a (sin t - t cost)
Question
18. If f (x) = |x|
3
show that f''(x) exists for all real x, and find it.
Solution :
It is known that,
Therefore, when x ≥ 0, f(x) = |x|
3
= x
3
In this case, f'(x) = 3x
2
and hence, f''(x) = 6x
When x < 0, f(x) = |x|
3
= (-x
3
) = -x
3
In this case, f'(x) = -3x
2
and hence, f''(x) = -6x
Thus, for f(x) = |x|
3
, f''(x) exists for all real x and is given by,
Question
19. Using mathematical induction prove that
for all positive integers n.
Solution :
For n = 1,
∴P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is,
It has to be proved that P(k + 1) is also true.
Thus, P(k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, proved.
Question
20. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution :
sin (A + B) = sin A cos B + cos A sin B
Differentiating both sides with respect to x, we obtain
Question
21. Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?
Solution :
y={|x| −∞< x ≤ 1
2−x 1≤ x ≤ ∞
It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.
Question
22. If
, prove that
Solution :
Question
23. If
, show that
Solution :
It is given that,
