Q.1. Find the derivative of sin (x^{2} + 1) with respect to x from first principle.
Ans. 2 x cos (x^{2} + 1)
Solution. Let f (x) = sin (x^{2} + 1) then
Q.2. Find the derivative of
Ans. 2/9
Solution.
Q.3.
Ans.
Solution. We have,
(Clearly y is not defined at x = 1)
Q. 4.
Ans.
Solution. We are given
Here y is the sum of two functions and in the second function base as well as power are functions of x. Therefore we will use logarithmic differentiation here.
Let y = u + v
Differentiating (1) with respect to x, we get
Taking log on both sides on equn (2), we get
log v = x log tan x
Differentiating the above with respect to x, we get
Substituting the value of
Q. 5. Let f be a twice differentiable function such that
f "(x) = – f (x), and f '(x)= g (x), h(x) = [f (x)]^{2} + [g (x)]^{2}
Find h(10) if h(5) = 11
Ans. 11
Solution. Given that f is twice differentiable such that
f " (x) = – f (x) and f '(x) = g(x)
h (x) = [f (x)]^{2} + [g (x)]^{2}
To find h (10) when h (5) = 11.
Consider h'(x) = 2f f ' + 2gg' = 2f (x) g(x) + 2g(x)f " (x)
[∴ g(x) = f '(x) ⇒ g' (x) = f ''(x)]
= 2f (x) g (x) + 2g(x) (– f (x))
= 2f (x) g (x) – 2f (x) g (x) = 0
⇒ h is a constant function
∴ h (5) = 11 ⇒ h (10) = 11.
Q. 6. If a be a repeated root of a quadratic equation f(x) = 0 and A(x), B(x) and C(x) be polynomials of degree 3, 4 and 5 respectively, then show that divisible by f(x), where prime denotes the derivatives.
Solution.
Given that α is a repeated root of quadratic equation f (x) = 0
∴ We must have f (x) = k (x – α)^{2} ; where k is a non zero real no.
If we put x = α on both sides of eq. (1); we get
Hence (x – α) is a factor of F(x) Differentiating eq. (1) w.r. to x, we get
Putting x = α on both sides, we get
[as R_{1} and R_{3} are identical]
⇒ (x – α) is a factor of F'(x) al so. Or we can say (x – α)^{2} is a factor of F(x).
⇒ F (x) is divisible by f (x).
Q. 7. If x = sec θ – cos θ and y = sec^{n} θ  cos^{n} θ , then show that
Solution. We have, x = sec θ – cos θ, y = sec^{n} θ – cos^{n} θ
sec θ tan θ + tan θ cos θ = tan θ (sec θ + cos θ)
= n sec^{n} θ tan θ +n tan θ cos^{n} θ = n tan θ (sec^{n} θ + cos^{n} θ)
...(1)
Also x^{2} + 4 = (sec θ – cos θ)^{2} + 4
= sec^{2} θ + cos^{2} θ – 2 sec θ cos θ + 4
= sec2 θ + cos2 θ + 2
= (sec θ + cos θ)^{2} ...(2)
and y^{2} + 4 = secn θ – cosn θ)2 + 4
= sec^{2n} θ + cos^{2n} θ – 2 secn θ cosn θ + 4
= sec ^{2n} θ + cos ^{2n} θ + 2
= (secn θ + cosn θ)^{2} ...(3)
Now we have to prove
[Using (1) and (2)]
= n^{2} (sec^{n} q + cos^{n} q)^{2}
= n^{2} (y^{2} + 4) [From eq. (3)]
= RHS
Q. 8. Find dy/dx at x = – 1, when
Ans. 0
Solution. We have given the function
For x = – 1, we have
....(2)
Differentiating both sides with respect to x, we get
Now differentiating eq. (1), we get
Q. 9.
Solution.
⇒ log y = 3 log x – log (x – a) – log (x – b) – log (x – c)
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