12th Standard Maths English Medium Applications Of Differential Calculus Reduced Syllabus Important Questions With Answer Key 2021

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Questions:

Generic 3

13 x 3 = 39

The side of a square is equal to the diameter of a circle. If the side and radius change at the same rate then find the ratio of the change of their areas.

A ball is thrown vertically upwards, moves according to the law s = 13.8 t - 4.9 t² where s
is in metres and t is in seconds.
(i) Find the acceleration at t = 1
(ii) Find velocity at t = 1
(iii) Find the maximum height reached by the ball?

Verify LMV theorem for f(x) = x³ - 2x² - x + 3 in [0, 1].

Write the Maclaurin series expansion of the following function
log(1 - x); -1 ≤ x < 1

Suppose that for a function f(x), f'(x) ≤ 1for all 1 ≤ x ≤ 4. Show that f(4) - f(1) ≤ 3.

Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x -axis for the following functions:
f(x) = x² − x, x ∈ [0, 1]

Find the absolute extrema of the following functions on the given closed interval.
\(f(x)=2cosx+sin2x;\left[ 0,\frac { \pi }{ 2 } \right] \)

Prove using the Rolle’s theorem that between any two distinct real zeros of the polynomial \(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\) there is a zero of the polynomial \(na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_{1}\)

Find the points on the curve y² - 4xy = x² + 5 for which the tangent is horizontal.

10.

Find the point on the curve y = x² − 5x + 4 at which the tangent is parallel to the line 3x + y = 7.

11.

A particle moves along a straight line in such a way that after t seconds its distance from the origin is s = 2t² + 3t metres.

12.

A point moves along a straight line in such a way that after t seconds its distance from the origin is s = 2t² + 3t metres.
(i) Find the average velocity of the points between t = 3 and t = 6 seconds.
(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.

13.

A person learnt 100 words for an English test. The number of words the person remembers in t days after learning is given by W(t) = 100 × (1− 0.1t)², 0 ≤ t ≤ 10. What is the rate at which the person forgets the words 2 days after learning?

Generic 5

6 x 5 = 30

14.

A manufacturer can sell x items at a price of rupees \(\left( 5-\frac { x }{ 100 } \right) \) each. The cost price of x items is Rs.\(\left( \frac { x }{ 5 } +500 \right) \) .Find the numbers of items he should sell to earn maximum profit.

15.

missle fired from ground level rises x metres vertically upwards in t seconds and \(x=100t-\frac { 25 }{ 2 } { t }^{ 2 }\). Find the
(i) initial velocity of the missile
(ii) the time when the height of the missile is maximum
(iii) the maximum height reached
(iv) the velocity which the missile strikes the ground.

16.

A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?

17.

A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall.
(i) How fast is the top of the ladder moving down the wall?
(ii) At what rate, the area of the triangle formed by the ladder, wall and the floor is changing?

18.

A conical water tank with vertex down of 12 metres height has a radius of 5 metres at the top. If water flows into the tank at a rate 10 cubic m/min, how fast is the depth of the water increases when the water is 8 metres deep?

19.

A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of s = 16t² in t seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?

Generic 2

8 x 2 = 16

20.

Evaluate the following limits, if necessary using L’Hopitalrule
(i) \(\underset { x\rightarrow 2 }{ lim } \cfrac { sin\pi x }{ 2-x } \)
(ii) \(\cfrac { lim }{ x\rightarrow 2 } \cfrac { { x }^{ n }-{ a }^{ n } }{ x-2 } \)
(iii) \(\underset { x\rightarrow \infty }{ lim } \cfrac { sin\frac { 2 }{ x } }{ \frac { 1 }{ x } } \)
(iv) \(\underset { x\rightarrow \infty }{ lim } \cfrac { { x }^{ 2 } }{ { e }^{ x } } \)

21.

Using Rolle’s theorem find the value of c for f(x) = sin x in[0,2π]

22.

Find the point on the parabola y²=18x at which the ordinate increases at twice the rate of the abscissa.

23.

At what point on the curve y = x² on [-2, 2] is the tangent parallel to X-axis?

24.

Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.
\(f(x)=x-2logx, x\in [2,7]\)

25.

Find the slope of the tangent to the following curves at the respective given points
y = x⁴ + 2x²− x at x = 1

26.

A stone is dropped into a pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate at 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?

27.

If the volume of a cube of side length x is v = x³. Find the rate of change of the volume with respect to x when x = 5 units.

Multiple Choice Question

15 x 1 = 15

28.

The statement "If f has a local extremum at c and if f'(c) exists then f'(c) = 0" is ________

(a)

the extreme value theorem

(b)

Fermat's theorem

(c)

Law of mean

(d)

Rolle's theorem

29.

\(\underset { x\rightarrow 0 }{ lim } \frac { x }{ tanx } \) is _________

(a)

(b)

-1

(c)

(d)

∞

30.

If the curves y = 2e^x and y = ae^-x intersect orthogonally, then a = _________

(a)

\(\frac { 1 }{ 2 } \)

(b)

-\(\frac { 1 }{ 2 } \)

(c)

(d)

2e²

31.

In LMV theorem, we have f'(x₁) = \(\frac { f(b)-f(a) }{ b-a } \) then a < x₁ _________

(a)

(b)

≤b

(c)

(d)

≠b

32.

The equation of the tangent to the curve x = t cost, y = t sin t at the origin is __________

(a)

x = 0

(b)

y = 0

(c)

x +y = 0

(d)

x + y = 7

33.

The critical points of the function f(x) = \((x-2)^{ \frac { 2 }{ 3 } }(2x+1)\) are __________

(a)

-1, 2

(b)

1, \(\frac { 1 }{ 2 } \)

(c)

1, 2

(d)

none

34.

Equation of the normal to the curve y = 2x²+3 sin x at x = 0 is __________

(a)

x + y = 0

(b)

3y = 0

(c)

x + 3y = 7

(d)

x + 3y = 0

35.

The point on the curve y = x² is the tangent parallel to X-axis is __________

(a)

(1, 1)

(b)

(2, 2)

(c)

(4, 4)

(d)

(0, 0)

36.

The point of inflection of the curve y = (x - 1)³ is

(a)

(0, 0)

(b)

(0, 1)

(c)

(1, 0)

(d)

(1, 1)

37.

One of the closest points on the curve x² - y² = 4 to the point (6, 0) is

(a)

(2,0)

(b)

\(\left( \sqrt { 5 } ,1 \right) \)

(c)

\(\left( 3,\sqrt { 5 } \right) \)

(d)

\(\left( \sqrt { 13 } ,-\sqrt { 3 } \right) \)

38.

The number given by the Mean value theorem for the function \(\frac { 1 }{ x } \), x ∈ [1, 9] is

(a)

(b)

2.5

(c)

(d)

3.5

39.

The function sin⁴ x + cos⁴x is increasing in the interval

(a)

\(\left[ \frac { 5\pi }{ 8 } ,\frac { 3\pi }{ 4 } \right] \)

(b)

\(\left[ \frac { \pi }{ 2 } ,\frac { 5\pi }{ 8 } \right] \)

(c)

\(\left[ \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right] \)

(d)

\(\left[ 0,\frac { \pi }{ 4 } \right] \)

40.

What is the value of the limit \(\lim _{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right) \text { is }\)

(a)

(b)

(c)

(d)

∞

41.

The tangent to the curve y² - xy + 9 = 0 is vertical when

(a)

y = 0

(b)

\(\\ \\ y=\pm \sqrt { 3 } \)

(c)

\(y=\frac { 1 }{ 2 } \)

(d)

\(y=\pm 3\)

42.

A stone is thrown up vertically. The height it reaches at time t seconds is given by x = 80t -16t². The stone reaches the maximum height in time t seconds is given by

(a)

(b)

2.5

(c)

(d)

3.5

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Answers Key:

Generic 3

13 x 3 = 39

2:π

s = 13.8 t - 4.9 t²
v = \(\frac { ds }{ dt } \) =13.8 - 4.9 (2t)
=13.8-9.8t
When t = 1, v = 13.8 - 9.8(1)
4 m/sec.
Acceleration = \(\frac { d^{ 2 }x }{ { dt }^{ 2 } } \) = -9.8 m/sec²
At maximum height, v = 0
∴ 13.8 - 9.8 t = 0
⇒ 13.8 = 9.8 t
⇒ t = \(\frac { 13.8 }{ 9.8 } \) = 1.40 sec
At t = 1.4 sec,
distance (s) = 13.8 (1.40) - 4.9 (1.40)²
= 19.32 - 9.604 = 9.716 m

f(x) = x³-2x²-x+3
f'(x) = 3x² - 4x - 1
f'(c) = 3c² - 4c- 1
f(a) = f(0) = 3
f(b) = f(1) = 1³-2-1+3 = 1
Then, if atleast one C \(\in \) (0, 1) such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ 3c²-4c-1 = \(\frac { 1-3 }{ 1-0 } \)
⇒ 3c²-4c+1 = 0
⇒ c = \(\frac { 4\pm \sqrt { 16-4(3) } }{ 2(3) } \)
⇒ c = \(\frac { 4\pm 2 }{ 6 } =\frac { 6 }{ 6 } \) or \(\frac { 2 }{ 6 } \)
⇒ 1 or \(\frac { 1 }{ 3 } \).

Function and its derivatives	log (1-x) cos x and its derivatives	Value at x = 0
f(x)	log (1-x)	log 1 = 0
f^I(x)	\(\frac{-1}{1-x}\) = -1(1 - x)^-1	\(\frac{-1}{1}\) = -1
f^Il(x)	-1(1-x)^-2	-1
f^IIl(x)	-2(1-x)^-3	-2
f^IV(x)	-6 (1 - x)^-4	-6
f^V(x)	-24 (1 - x)^-5	-24

Meclaurin's expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\).................
log(1-x) = \(0-\frac { 1 }{ 1! } x-\frac { { x }^{ 2 } }{ 2! }- \frac { 2 }{ 3! } { x }^{ 3 }-\frac { { 6x }^{ 4 } }{ 4! } +.....\)
= \(-x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } \)+ .....
log(1 - x) = \(-\left( x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +.... \right) \)

f'(x) = ≤1 for all 1 ≤ x ≤ 4
Using Lagrange's mean value theorem,
f'(x) = \(\frac { f(b)-f(a) }{ b-a } \) [∵ f(x) is continuous in [1, 4] and differentiable in (1, 4)]
f'(x) = \(\frac{f(4)-f(1)}{4-1}\)
f'(x) = \(\frac { f(4)-f(1) }{ 3} \)
⇒ \(\frac { f(4)-f(1) }{ 3} \) = f'(s)
⇒ \(\frac { f(4)-f(1) }{ 3} \) ≤ 1[∵ f'(x) ≤ 1]
⇒ f(4) - f(1) ≤ 3
Hence proved

Given f(x) = x² − x, x ∈ [0, 1]
(i) f(x) is continuous in [0, 1]
(ii) f(x) is differentiable in (0, 1)
(iii) f(0) = 0² - 0 = 0
f(1) = 12-1 = 1-1 = 0
∴ f(0) = f(1)
By Rolle's theorem, there exists C ∈ [0, 1] such that
f'(c) = 0
⇒ 2c - 1 = 0
⇒ 2c = 1
⇒ c = \(\frac12\) ∈ [0, 1]

f'(x) = -2 sin x + 2 cos 2x
f'(x) = 0
\(\Rightarrow\) 2 sin x + 2 cos 2x = 0
\(\Rightarrow\) 2 sin x + 2(1 - 2 sin²x) = 0
\(\Rightarrow\) 4 sin²x + 2 sin x - 2 = 0
\(\Rightarrow\) 4 sin² x + 2 sin x - 2 = 0
\(\Rightarrow\) 2 sin² x + sin x-1 = 0
\(\Rightarrow\) (sin x+1) (2sin x-1) = 0
\(\Rightarrow\) \(sinx=-1\ or\ sinx=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(sinx=-sin\frac { \pi }{ 2 } \) or
\(sinx=sin\frac { \pi }{ 6 } \)
\(\Rightarrow sinx=sin\left( -\frac { \pi }{ 2 } \right) \)
\(sinx=sin\frac { \pi }{ 6 } \)
\(\Rightarrow x=-\frac { \pi }{ 2 } or \ x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=\frac { \pi }{ 6 } \)
\(\\ \\ \\ \\ \\ \\ \left[ \because x=-\frac { \pi }{ 2 }∉ \left[ 0.\frac { \pi }{ 2 } \right] \right] \)
\(\therefore\) The critical number is \(x=\frac { \pi }{ 6 } \)
Evaluating f(x) at the end points x = 0, \(x=\frac { \pi }{ 2 } \) and at the critical number \(x=\frac { \pi }{ 6 } \) we get.
f(0) = 2 cos0 + sin0 = 2
\(f(\frac { \pi }{ 2 } )=2cos\frac { \pi }{ 2 } +sin\pi =0\)
\(f\left( \frac { \pi }{ 6 } \right) =2cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 3 } \)
\(2\left( \frac { \sqrt { 3 } }{ 2 } \right) +\frac { \sqrt { 3 } }{ 2 } =\frac { 3\sqrt { 3 } }{ 2 } \)
From these values, the absolute maximum is \(\frac { 3\sqrt { 3 } }{ 2 } \) which occurs at \(x=\frac { \pi }{ 6 } \) and the absolute minimum is 0 which occurs at \(x=\frac { \pi }{ 2 } \)

Let P(x) = \(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\). Let \(\alpha<\beta \) be two real zeros of P(x). Therefore, \(P(\alpha)=P(\beta)=0.\) Since P(x) is continuous in \([\alpha, \beta]\) and differentiable in \((\alpha, \beta)\) by an application of Rolle’s theorem there exists \(\gamma \in (\alpha,\beta)\) such that \(P'(\gamma)=0\). Since,
\(P'(x)=na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_{1}\) which completes the proof.

Equation of the given curve is y² - 4xy = x² + 5..(1)
Differentiating with respect to 'x' we get,
\(2y\frac { dy }{ dx } -4\left[ x\frac { dy }{ dx } +y(1) \right] =2x\)
⇒ \(2y\frac { dy }{ dx } -4x\frac { dy }{ dx } -4y=2x\)
⇒ \(\frac { dy }{ dx } \) (2y -4x) = 2x + 4y
⇒ \(\frac { dy }{ dx } \) = \(\frac { x+2y }{ y-2x } \)
Since the tangent to the curve is horizontal, \(\frac { dy }{ dx } \) = 0
ஃ \(\frac { x+2y }{ y-2x } \) = 0
⇒ x+ 2y = 0
⇒ x = -2y .....(2)
Substituting (2) in (1) we get,
y² - 4(-2y)y = (-1y)² + 5
⇒ y² + 8y² = 4y² + 5
⇒ y² = 4y²+ 5
⇒ 5y² = 5
⇒ y² = 1
⇒ y = 土 1
From (2), When y = 1, x = - 2
When y = -1, x = 2
∴ The required points are (2, -1) and (-2,1)

10.

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

11.

12.

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec
Instantaneous Velocity at t = 6
= V(6) = 27m/sec

13.

We have,
\(\frac{d}{dt}W(t)=-20\times(1-0.1t)\)
Therefore at t = 2, \(\frac{d}{dt}W(t)=-16\)
That is, the person forgets at the rate of 16 words after 2 days of studying.

Generic 5

6 x 5 = 30

14.

240

15.

100 m / s, t = 4 sec, 200 m / s, −100 m / s

16.

Let x represent the distance covered by the car, y represent the distance covered by the police jeep, and s represent the distance between the car and jeep.
ஃ Given = x = 0.8 km, y = 0.6 km,

\(\frac { dy }{ dt } \) = -60km/hr,
\(\frac { ds }{ dt } \) = 20 km/hr,
In ΔABC, S² = x² + y² ......(1)
⇒ S² = (0.8)² + (0.6)²
= 0.64 + 0.36
⇒ S² = 1
⇒ s = 1 ....(2)
Differentiating (1) with respect to 't' we get,
\(2s\frac { ds }{ dt } =2x\frac { dx }{ dt } +2y\frac { dy }{ dt } \)
⇒ \(s\frac { ds }{ dt } =x\frac { dx }{ dt } +y\frac { dy }{ dt } \) [Divided by 2]
⇒\(1\left( \frac { ds }{ dt } \right) =(0.8)\left( \frac { dx }{ dt } \right) +(0.6)(-60)\)
⇒ 1(20) = (0.8) \(\left( \frac { dx }{ dt } \right) \) + (0.6)(-60)
⇒ 20 = (0.8) \(\left( \frac { dx }{ dt } \right) \) - 36
⇒ 20 + 36 = (0.8) \(\frac { dx }{ dt } \)
⇒ \(\frac { dx }{ dt } =\frac { 56 }{ 0.8 } \) = 70km/hr.
⇒Speed of the car is 70 km/hr.

17.

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB²= AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
⇒ y² = 289 - 64 = 225
⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

18.

Given h = 12 m
r = 5 m
∴ tan α = \(\frac { OA }{ VO } =\frac { 5 }{ 12 } =\frac { r }{ h } \)
⇒ 12r = 5h
⇒ r = \(\frac{5h}{12}\) ...(1)
Given rate of changeof volume = \(\frac{dv}{dt}=10\)
Volume of cone V = \(\frac13\) πr²h
= \(\frac { 1 }{ 3 } \)π\({ \left( \frac { 5h }{ 12 } \right) }^{ 2 }.h\) [From (1)]
\(v=\frac { 1 }{ 3 } \pi { \left( \frac { 25h }{ 144 } \right) }^{ 2 }=\frac { { 25\pi h }^{ 3 } }{ 3\times 144 } \)
Differentiating with respect to 't' we get,
\(\frac { dv }{ dt } =\frac { 25\pi }{ 3\times 144 } 3{ h }^{ 2 }\frac { dh }{ dt } \)
= \(\frac { 25\pi }{ 144 } { h }^{ 2 }\frac { dh }{ dt } \)
[∵ Given h = 8, \(\frac { dh }{ dt } \) = 10m³/min]
\(\Rightarrow \frac { 10\times 144 }{ 25\pi \times 64 } =\frac { dh }{ dt } \)
\(\Rightarrow \frac { dh }{ dt } =\frac { 9 }{ 10\pi } \)
The depth of the water increasing at the rate of \(=\frac { 9 }{ 10\pi } \)m / min.

19.

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

Generic 2

8 x 2 = 16

20.

(i) π
(ii) πX2^n-1
(iii) 2
(iv) 0

21.

\(\theta =\frac { \pi }{ 2 } ,\frac { 2\pi }{ 2 } \varepsilon \left( 0,2\pi \right) \)

22.

\(\left( \frac { 9 }{ 8 } ,\frac { 9 }{ 2 } \right) \)

23.

Y = x² is continuous on [-2, 2] and differentiable on [-2, 2]
f(a) = f(-2) = (-2)² = 4
f(b) = f(2) = 2² = 4
∴ f(a) = f(b)
Since the tangent is parallel to X - axis, f'(c) = 0
⇒ 2c = 0
⇒ c = 0
∴ When c = 0, y = 0
∴ AE (0, 0) the tangent is parallel to X- axis.

24.

Given \(f(x)=x-2logx, x\in [2,7]\)
(i) f(x) is continuous in [2, 7]
(ii) f(x) is differentiable in (2, 7)
f(2) = 2 - 2 log 2
= 2 - log 2² = 2 - log 4
f (7) = 7 - 2 log 7
= 7 - log 7² = 7 - log 49
Since f(2) ≠ f (7), Rolle's theorem is not applicable.

25.

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

26.

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.

27.

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.

Multiple Choice Question

15 x 1 = 15

28.