JEE Advanced (Subjective Type Questions): Differentiation | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Find the derivative of sin (x² + 1) with respect to x from first principle.

Ans. 2 x cos (x² + 1)

Solution. Let f (x) = sin (x² + 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 R₁ and R₃ 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 = secⁿ θ - cosⁿ θ , then show that 

Solution. We have, x = sec θ – cos θ, y = secⁿ θ – cosⁿ θ

 sec θ tan θ + tan θ cos θ = tan θ (sec θ + cos θ) 

= n secⁿ θ tan θ +n tan θ cosⁿ θ = n tan θ (secⁿ θ + cosⁿ θ)

     ...(1)

Also x² + 4  = (sec θ – cos θ)² + 4    
= sec² θ + cos² θ – 2 sec θ cos θ  + 4    
= sec2 θ + cos2 θ + 2    
= (sec θ + cos θ)²          ...(2)
and y² + 4 = secn θ – cosn θ)2 + 4  
= sec²ⁿ θ + cos²ⁿ 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)]

= n² (secⁿ q + cosⁿ q)²
= n² (y² + 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)