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Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - JEE MCQ


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17 Questions MCQ Test - Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths

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*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 1

Let P(x) = a0 + a1x2 + a2x4 + ...... + anx2n be a polynomial in a real variable x with 0 < a0 < a1 < a2 < ..... < an. . The function P(x) has

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 1

∴ P (x) has only one minimum at x = 0.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 2

If the line ax + by + c = 0 is a normal to the curve xy = 1, then

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 2

Let the line ax + by + c = 0 be normal to the curve xy = 1 at the point (x',y'), then
x ' y ' = 1¼(1)  pt(x ',y') lies on the curve]
Also differentiating the curve xy = 1 with respect to x

Also equation of normal suggests, slope of normal 
∴ We must have,

            … (2)

Now from eq. (1),  are of same sign

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*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 3

The smallest positive root of the equation, tan x – x = 0 lies in

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 3

It is clear from the graph that the curves y = tan x and y = x intersect at P in (π, 3π / 2) .
Thus the smallest +ve root of tan x – x = 0 lies in (π, 3π / 2) .

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 4

Let f and g be in creasin g and decreasing function s, respectively from [0, ∞) to [0, ∞). Let h(x) = f (g(x)). If h(0) = 0, then h(x) – h (1) is

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 4

Since g is decreasing in [0, ∞)

      ........ (1)
Also g(x), g(y) ∈ [0, ∞) and f is increasing from [0, ∞) to [0, ∞).


⇒ h is decreasing function from [0, ∞) to [0, ∞)

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 5

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 5

We are given that



Also f (x) being polynomial for x ∈[-1, 2) U (2, 3]
f (x) is cont. on [– 1, 3] except possibly at
At x = 2,

⇒ f (x) is continuous at x = 2
Hence f (x) is continuous on [– 1, 3] Again at x = 2

∴ f '(2) does not exist. Hence f (x) can not have max. value at x = 2.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 6

Let h(x) = f(x) – (f(x))2 + (f(x))3 for every real number x. Then

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 6

We have

Note that h'(x) <0 whenever f'(x) <0 and h'(x) >0 whenever f'(x) >0 , thus, h (x) increases (decreases) whenever f (x) increases (decreases).

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 7

 for every real number x, then the minimum value of f

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 7

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 8

The number of values of x where the function f(x) = cos x + cos (√2 x) attains its maximum is

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 8

The maximum value of f (x) = cos x + cos (√2 x) is 2 which occurs at x = 0. Also, there is no value of x for which this value will be attained again.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 9

The function  dt has a local minimum at x =

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 9

Critical points are 0, 1, 2, 3. Consider change of sign of

Change is from –ve to  +ve, hence minimum at x = 3.
Again minimum and maximum occur alternately.
∴ 2nd minimum is at x = 1

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 10

f(x) is cubic polynomial with f(2) = 18 and f(1) = –1. Also f(x) has local maxima at x = –1 and f '(x) has local minima at x = 0, then

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 10

Let f (x) = ax3 + bx2 + cx + d
Then, f (2) = 18  ⇒ 8a + 4b + 2c + d = 18 … (1)

f (1) = – 1 ⇒ a + b + c + d = – 1 … (2)
f (x) has local max. at x = – 1
⇒ 3a – 2b + c = 0 … (3)
f '(x) has local min. at x = 0 ⇒ b = 0 … (4)
Solving (1), (2), (3) and (4), we get

f ''(-1) <0, f ''1) > 0 ⇒ x =-1 is a point of local max. and x = 1 is a point of local min. Distance between (– 1, 2) and (1, f (1)), i.e. (1, – 1) is 

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 11

then g(x) has

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 11




∴ g ''(1 + ln 2) =-2 and g ''(e) = 1⇒ g (x) has local max. at x = 1 + ln 2 and local min. at x = e.

Also graph of g '(x) suggests, g (x) has local max. at x = 1 and local min. at x = 2

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 12

For the function

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 12


Þ f '(x) is strictly decreasing in [1, ∞)

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 13

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 13

∴ x = 2 is a point of local maxima and x = 3 is a point of local minima
Also or x ∈ (2, 3) f ’(x)< 0
⇒ f is decreasing on (2, 3)
Also we observe  f ''(0) < 0 and f ''(1) > 0
∴ There exists some C Î(0, 1) such that f'' (C) = 0
∴ All the options are correct.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 14

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume.

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 14

Let L = 8x, B = 15x and y be the length of square cut off from each corner. Then volume of box
= (8x – 2y) (15x – 2y)y
V = 120x2y – 46xy2 + 4y3


*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 15

 Then

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 15



∴ f is monotonically increasing on [1, ∞) (a) is correct.
For x ∈ ( 0,1) , f ' (x)>0
∴ (b) is not correct

∴ f (2x) is an odd function.
∴ (d) is correct.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 16

be continuous functions which are twice differentiable on the interval (–1, 2). Let the values of f and g at the points –1, 0 and 2 be as given in the following table:

In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct statement(s) is(are)

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 16

Let h(x) = f(x) – 3g(x) h(–1) = h(0) = h(2) = 3
∴ By Rolle’s theorem h'(x) = 0 has atleast one solution in (–1, 0) and atleast one solution in (0, 2) But h''(x) never vanishes in (–1, 0) and (0, 2) therefore h'(x) = 0 should have exactly one solution in each interval.

*Multiple options can be correct
Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 17

 be twice differen tiable functions such that f" and g" are continuous functions on = g(2)= 0, f"(2) ≠ 0 and g'(2) ≠ 0. If 

Detailed Solution for Multiple Correct MCQ Of Applications Of Derivatives, Past Year Questions JEE Advance, Class 12, Maths - Question 17


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