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Given:

Distance between the two cities = 360 km

Formula used:

Average speed = Total distance/Total time

Distance = Speed × Time

Calculation:

Time taken by car A to cover distance from P to Q = 360/40 = 9 hours

Time taken by car A to cover distance from Q to P = 360/50 = 7.2 hours

Average speed = (360 + 360)/(9 + 7.2) = 44.44 km/hr

∴ The average speed of the car is 44.44 km/hr

Correct answer is (a) deer

Given:

Distance between Madhubani and Patna = 300 km

Train starts from Madhubani with 30 km/hr at = 9 am

Train starting from Patna with 20 km/hr at = 10 am

Concept used:

Relative speed = Speed of train A + speed of train B

When both are moving in opposite direction.

Calculation:

Distance travelled by train from Madhubani from 9 am to 10 am = 30 km

Now, distance between trains = 300 - 30 = 270 Km

Time at which the two trains meet = distance/ relative speed

Time of meeting = 270/(30 + 20) = 5 hour 24 min.

That of meeting = 5 hours 24 mins from 10 am

∴ The two trains will meets at 3:24 pm.

Given:

Speed = 8 km/hr

Distance = 24 km

Formula used:

Time = Distance/Speed

Calculation:

Time = 24/8 = 3 hours

As he started his journey at 8 : 00 am 

He will reach his destination at 8 : 00 am + 3 hours 

∴ He will reach his destination at 11 : 00 am

Given:

Speed of first car = 60 km/h

Speed of second car = 30 km/h

Formula used:

Relative Speed =  (Total distance)/Time

Concept used:

Let two cars meet after ‘t’ hours when train start from Patna at 9 A.M

Distance covered in ‘t’ hours at 60 km/h = 60t

Distance covered in ‘(t – 2)’ hours at 30 km/h = 30(t – 2)

According to the Question,

60t + 30(t – 2) = 300

⇒ 60t + 30t – 60 = 300

⇒ 90t = 360

⇒ t = 360/90

⇒ t = 4 hours

Time at he reach = (9 + 4) = 1 P.M

∴ Cars will meet at 1 P.M.

Calculation:

Speed of Tiger = 6 leaps/min = 6 × 10 m/min = 60 m/min

Speed of Deer = 5 leaps/min = 5 × 6 m/min = 30 m/min

Distance b/w Tiger and Deer at start = 60 × 10 = 600 m

Let d be the distance travelled by deer before being caught by tiger.

Thus, (600 + d)/60 = d/30 ⇒  600 + d = 2d ⇒ d = 600 m

So, the number of leaps covered by deer = 600/6 = 100 leaps

∴ The number of leaps covered by deep is 100 leaps.

Time taken by train T1 to complete its journey = 4 hours

Time taken by train T2 to complete its journey = 5 hours

Ratio of speeds of trains T1 and T2 = (Time taken by train T2)/(Time taken by train T1) = 5/4

Ratio of distances travelled by trains T1 and T2 = 5/4

Time taken by trains T1 and T2 to cross each other = (5/9) × 4 = 20/9 = 2 hours 13 minutes

Thus, both the trains T1 and T2 will cross each other at 10:13 am.

(i) A = {1, 2, 4, 6}

B = {3, 7}

And C = {1, 3, 6, 7, 9}

∴ B ∪ C = {1, 3, 6, 7, 9}

Now, A ∩ {B ∪ C} = {1, 2, 4, 6} ∩ {1, 3, 6, 7, 9} = {1, 6}

(ii) A = {2, 4, 6, 8}

B = {3, 10}

C = {2, 3, 8, 9, 10}

∴ B ∩ C = {3, 10}

Now, A ∪ {B ∩ C} = {2, 4, 6, 8} ∪ {3, 10} = {2, 3, 4, 6, 8, 10}

(iii) A = {2, 4, 6, 8}

B = {3, 9}

C = {4, 5, 7, 9, 10}

∴ A ∩ B = ϕ

A ∩ {B ∪ C} = ϕ ∪ C = C = {4, 5, 7, 9, 10}

And B ∪ C = {3, 4, 5, 7, 9, 10}

A ∩ (B ∪ C) = {4}

Thus, (A ∩ B) ∪ C ≠ A ∩ (B ∪ C)

That's why we need to use bracket at right places.

Focal length (f) = + 50 cm

Power of the lens (P) = 100/f = 100/50 =2D.

Objective function is Max z = 2x₁ + x₂

By changing given inequalities into equations

3x₁ + x₂ + x₃ = 3

x₁ + 3x₂ + x₄ = 3

x₁, x₂, x₃, x₄ \(\geq\) 0

First simplex table

Second simplex table

Third simplex table

\(\because\) All Z_j - C_j \(\geq\) 0

Correct answer is 79

f(x) = 4|2x + 3| + 9[x +1/2] -12[x + 20]

x ∈ (-20, 20)

f(x) is not Diff. at x = I∈{-19, -18, ....0,...19} = 39

at x = I + 1/2 , f(x) Non diff. at 39 points

Check at x = -3/2 Discount at x = -3/2 

∴ N.R(1)

No. of point of non-differentiabilty

= 39 + 39 + 1 = 79

 tan²(sec^-12) = sec²(sec^-12) - 1 

cot²(cosec^-13) = cosec²(cosec^-13) - 1   

Therefore, 

tan²(sec^-12) + cot²(cosec^-13) 

= sec²(sec^-12) - 1 + cosec²(cosec^-13) - 1 

= [sec(sec^-12)]² + [cosec(cosec^-13)]² - 2 

= (2)² + (3)² - 2 

= 4 + 9 - 2 

= 11

Concept:

The area of the region bounded by the curve \(\rm f(x)\) , \(\rm x \in [-a, a]\) and the x-axis is given by

A = \(\rm \displaystyle\int_{-a}^a f(x) \;dx \)

Calculations:

We know that The area of the region bounded by the curve \(\rm f(x)\) , \(\rm x \in [-a, a]\) and the x-axis is given by

A = \(\rm \displaystyle\int_{-a}^a f(x) \;dx \)

The area of the region bounded by the curve \(\rm f(x)=1 - \dfrac{x^2}{4}, x \in [-2, 2]\) and the x-axis

A = \(\rm \displaystyle\int_{-2}^2 (1 - \dfrac {x^2}{4})dx \)

⇒A = \(\rm \displaystyle \left[x- \dfrac {x^3}{12}\right]^2_{-2}\)

⇒A = \(\rm [2 - (-2)]- \left[\frac {8}{12} -\frac {-8}{12} \right]\)

⇒A = 4 \(-\dfrac 4 3\)

⇒A = \(\dfrac 8 3\)

Hence, the area of the region bounded by the curve \(\rm f(x)=1 - \dfrac{x^2}{4}, x \in [-2, 2]\) and the x-axis is  \(\dfrac 8 3\) square units

Concept:

For a function y = f(x):

• At the points of local maxima or minima, f'(x) = 0.
• In the regions where f(x) is increasing, f'(x) > 0.
• In the regions where f(x) is decreasing, f'(x) < 0.

Calculation:

f(x) = x3 - 5x2 + 7x + 4

⇒ f'(x) = 3x² - 10x + 7

For f(x) to be increasing, f'(x) > 0.

⇒ 3x2 - 10x + 7 > 0

⇒ 3x2 - 7x - 3x + 7 > 0

⇒ x(3x - 7) - (3x - 7) > 0

⇒ (3x - 7)(x - 1) > 0

⇒ [3x - 7 > 0 AND x - 1 > 0] OR [3x - 7 < 0 AND x - 1 < 0]

⇒ [x > 7/3 AND x > 1] OR x < 7/3 AND x < 1

⇒ x > 7/3 OR x < 1

⇒ x ∈ (7/3, ∞) ∪ (-∞, 1)

Concept:

The nth number in A.P. series = a + (n-1)d

The sum of the n numbers in A.P. series = \(\rm {n\over2}\left[2a + (n-1)d\right]\)

Where 'a' is the first number of the series and 'd' is the common difference

Calculation:

Let the first element of A.P. is 'a' and common difference is 'd'

The 5th number of A.P. (a₅) = a + 4d

The 13th number of A.P. (a₁₃) = a + 12d

Given a₅ + a₁₃ = 0

a + 4d + a + 12d = 0

2a + 16d = 0

a + 8d = 0

a₉ = 0

sin θ + cosec θ = 2

= sin θ + \(\cfrac1{sin\,θ}\) = 2

sin² θ + 1 = 2 sin θ

= sin² θ - 2sin θ + 1 = 0

= (sin θ - 1)² = 0

= sin θ - 1 = 0

= sin θ = 1

\(\therefore\) cosec θ = \(\cfrac1{sin\,θ}\) = \(\cfrac1{1}\) = 1

sin¹⁰ θ + cosec¹⁰ θ = 1¹⁰ + 1¹⁰ = 1 + 1 = 2

Given:

y = x^{n – 1} log x       ----(1)

Diff. (1) w.r.t. ‘x’ we get;

\({y_1} = \left( {n - 1} \right){x^{n - 2}}\log x + {x^{n - 1}}.\frac{1}{x}\)

xy₁ = (n - 1) x^{n - 1} log x + x^{n – 1}

xy₁ = (n - 1) y + x^{n – 1}       ----(2)

∵ y = x^{n – 1} log x

Diff. (2) (n - 1) times by Leibinitz’s theorem,

Xy_n + (n - 1) y_{n – 1} = (n - 1) y_{n – 1} + (n - 1)!

xy_n = (n - 1)!

\({y_n} = \frac{{\left( {n - 1} \right)!}}{x}\)

n^th derivative with respect to \(x = \frac{1}{2}\) is;

y_n = 2(n - 1)!

Concept:

• \(\rm d\over dx\)xn = nxn-1
• \(\rm d\over dx\)sin x = cos x
• \(\rm d\over dx\)cos x = -sin x
• \(\rm d\over dx\)ex = ex
• \(\rm d\over dx\)ln x = \(\rm1\over x\)
• \(\rm d\over dx\)(ax + b) = a
• \(\rm d\over dx\)tan x = sec2 x
• \(\rm d\over dx\)f(x)g(x) = f'(x)g(x) + f(x)g'(x)

Calculation:

Let y = \(\rm x^{e^{x}}\)

Taking log both sides, we get

ln y = ln \(\rm x^{e^{x}}\)

ln y = e^x (ln x)                                 (∵ log mn = n log m)

Differentiating with respect to x, we get

\(\rm {1\over y}{dy\over dx} = e^x(\ln x)+e^x\left({1\over x}\right)\)

\(\rm {dy\over dx} = y\left[e^x(\ln x)+{e^x\over x}\right]\)

\(\boldsymbol{\rm {dy\over dx} = x^{e^{x}}e^x\left[\ln x+{1\over x}\right]}\)

Concept:

Let u = f(x)  and v = g(x)

The derivative of u with respect to v is \(\rm \dfrac {du}{dv}\)

By chain Rule 

\(\rm \frac {du}{dv} = \dfrac {\frac {du}{dx}}{\frac {dv}{dx}}\)

Calculations:

Let u = sin2 x  and v = cos x

The derivative of sin2 x with respect to cos x is \(\rm \dfrac {du}{dv}\)

By chain Rule 

\(\rm \frac {du}{dv} = \dfrac {\frac {du}{dx}}{\frac {dv}{dx}}\)         ....(1)

On differentiating w.r.t x respectively, we get

⇒\(\rm \dfrac {du}{dx} = 2\sin x\; \cos x\) and \(\rm \dfrac {dv}{dx} = -\sin x\)

Equation (1) becomes,

\(\rm \dfrac {du}{dv} = \dfrac {2\sin x\cos x}{-\sin x}\)

⇒\(\rm \dfrac {du}{dv} = -2\cos x\)

Hence, the derivative of sin2 x with respect to cos x is - 2 cos x.

Concept:

Integration by Parts:

• ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ f'(x) [∫ g(x) dx] dx.

Integration by substitution:

• If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.

Definite Integral:

• If ∫ f(x)dx = g(x) + C, then \(\rm \displaystyle \int_a^b f(x)dx = [ g(x)]_a^b\) = g(b) - g(a).

Derivatives of Trigonometric Functions:

• \(\rm \dfrac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cos x=-\sin x\\ \dfrac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \dfrac{d}{dx}\cot x=-\csc^2 x\\ \dfrac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \dfrac{d}{dx}\csc x=-\cot x\csc\)

Trigonometric identities:

• sin² θ + cos² θ = 1.
• tan2 θ + 1 = sec2 θ.

Calculation:

Let us first consider \(\rm \displaystyle\int \tan^8 x \ dx\).

= \(\rm \displaystyle\int \tan^6 x \tan^2 x \ dx\)

= \(\rm \displaystyle\int \tan^6 x (\sec^2 x-1) \ dx\)

= \(\rm \displaystyle\int \tan^6 x \sec^2 x\ dx-\int \tan^6 x \ dx\)

Now, let's consider \(\rm \displaystyle\int \tan^6 x \sec^2 x \ dx\).

Substitute tan x = u ⇒ sec² x dx = du.

∴ \(\rm \displaystyle\int \tan^6 x \sec^2 x \ dx=\int u^6\ du=\dfrac{u^7}{7}+C=\dfrac{\tan^7x}{7}+C\)

And, \(\rm \displaystyle\int \tan^8 x \ dx=\dfrac{\tan^7x}{7}+C-\int \tan^6 x \ dx\)

⇒ \(\rm \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^8 x \ dx=\left [\dfrac{\tan^7x}{7} \right ]_0^{\tfrac{\pi}{4}}-\int_0^{\tfrac{\pi}{4}} \tan^6 x \ dx\)

⇒ \(\rm \displaystyle\int_0^{\tfrac{\pi}{4}} \tan^8 x \ dx+\int_0^{\tfrac{\pi}{4}} \tan^6 x \ dx=\left [\dfrac{\tan^7 \tfrac{\pi}{4}}{7} - \dfrac{\tan^7 0}{7}\right ]\)

Using \(\rm \tan\left( \dfrac{\pi}{4}\right) = 1\) and tan 0 = 0, we get:

⇒ \(\rm \displaystyle I_8+I_6=\dfrac{1}{7}\).

Concept:

Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times \int vdx\right)dx \) + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Formula:

\(\rm \int \frac{1}{x^2 -a^2}dx = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + c\)

Calculation:

Let I = \(\rm \displaystyle∫\dfrac{xe^x}{\sqrt{1+e^x}}dx\)

Take 1 + e^x = t²            .... (1)

Differentiating with respect to x, we get

⇒ e^x dx = 2tdt

From equation (1), we get

e^x = t² - 1

So, x = log (t2 - 1)

Now,

I = \(\rm \displaystyle∫\dfrac{\log (t^2 - 1)}{\sqrt{t^2}}2tdt\)

= \(\rm 2 × \displaystyle∫\dfrac{\log (t^2 - 1)}{t} × tdt\)

= 2 ∫ log (t2 - 1) dt

Using integration by parts rule, we get

= 2 [log (t2 - 1) × t - 2 \(\rm \int \frac {t^2}{t^2-1}dt\) ]

= 2t log (t2 - 1) - 4 \(\rm \int \left[1+ \frac{1}{t^2-1} \right ]dt\)

= 2t log (t2 - 1) - 4t - 4 × \(\rm \frac{1}{2} \log \left(\frac{t-1}{t+1} \right) +c\)

= 2t log (t2 - 1) - 4t -  \(\rm2\log \left(\frac{t-1}{t+1} \right) +c\)

= 2t(log (t2 - 1) - 2) -  \(\rm2\log \left(\frac{t-1}{t+1} \right) +c\)

Resubstitute the value of t, we get

= 2 (x - 2) \(\rm \sqrt{1+e^x}\) - \( \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)

= \(\rm (2x-4)\sqrt{1+e^x}- \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)

Concept:

If function f(x) has an extreme at x = a then f'(a) = 0

Calculations:

Consider, the function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) 

Taking derivative w.r.to x , we get

⇒ \(\rm f'(x)= - \;p \sin x + \dfrac{3\sin 3x}{3}\)

⇒ \(\rm f'(x)= - \;p \sin x + \sin 3x\)

⇒ \(\rm f'(​​\dfrac {\pi}{6})= - \;p \sin ​​\dfrac {\pi}{6} + \sin 3​​\dfrac {\pi}{6}\)

⇒\(\rm f'(​​\dfrac {\pi}{6})= - ​​\dfrac p2 + 1\)

The function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) has an extremum at  \(\rm x=\dfrac{\pi}{6} \)

⇒f'(\(\dfrac{\pi}{6} \)) = 0

\(⇒ \rm - ​​\dfrac p2 + 1 = 0\)

⇒ p = 2

Concept:

The equation of tangent at point (x1 , y1) with slope m is given by 

\(\rm (y - y_1) = m (x - x_1)\)

Calculations:

Given curve is y = 2x sin x 

Taking derivative on both side, we get

\(\rm \dfrac {dy}{dx}= 2x \;cos\;x + 2 \;sin\;x\)

Put x = \(\rm \dfrac{\pi}{2}\) to find the equation of tangent at the point \(\left(\frac \pi 2, \pi\right)\).

\(\rm \dfrac {dy}{dx}= 2 \dfrac {\pi}{2} \;cos\; \dfrac {\pi}{2} + 2 \;sin\; \dfrac {\pi}{2}\)

\(\rm \dfrac {dy}{dx}= 2\)

The equation of tangent at point (x₁ , y₁) with slope m is given by 

\(\rm (y - y_1) = m (x - x_1)\)

\(\rm (y - {\pi}) = 2 (x - \dfrac{\pi}{2})\)

y = 2x

Hence, the equation of the tangent line to the curve y = 2x sin x at the point \(\left(\frac \pi 2, \pi\right)\) is 2x.

Concept:

If the function f(x) has an extremum at x = a then f'(a) = 0

Calculations:

Given, the function is \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) 

⇒ f'(x) = \(\rm p\;cos \; x + \dfrac {3\;cos \;3x}{3}\)

⇒ f'(x) = \(\rm p\;cos \; x + cos \;3x\)

⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \;3(\dfrac {\pi}{3})\)

⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\)

The function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\)

Therefore,  \(\rm f'(\dfrac {\pi}{3}) = 0\)

⇒ \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\) = 0

⇒ \(\rm \dfrac p 2 -1 = 0\)

⇒ \(\rm \dfrac p 2 =1\)

⇒ \(\rm p = 2\)

 Hence, the value of p for which the function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\) is 2.

Concept:

Integration by parts:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

Let, \(\rm I=\displaystyle\int_1^2 e^x \left(\dfrac{1}{x}- \dfrac{1}{x^2}\right)dx \)

\(\rm =\displaystyle\int_1^2 (\dfrac{e^x}{x})dx -\displaystyle\int_1^2 \left( \dfrac{1}{x^2}\right)dx \)

\(\rm =[\frac1 x\displaystyle\int_1^2 {e^x}dx]_1^2+\displaystyle\int_1^2 \left( \dfrac{e^x}{x^2}\right)dx -\displaystyle\int_1^2 \left( \dfrac{e^x}{x^2}\right)dx \)

\(\rm =[\frac1 x( e^x)]_1^2\)

\(\rm =\frac1 2( e^2)-e\)

\(\rm =e\left(\dfrac{e}{2}-1\right)\)

Hence, option (3) is correct.

Concept:

\(\rm \int x^ndx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

I = \(\rm \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx\)

\(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^3 \sqrt{x^4 \left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

\(=\rm \frac{1}{4}\int \frac{4x^2 - 4}{x^5 \sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

\(=\rm \frac{1}{4}\int \frac{\frac{4}{x^3} - \frac{4}{x^5 }}{\sqrt{\left(2 - \frac{2}{x^2} + \frac{1}{x^4} \right )}}dx\)

Let \(\rm 2 - \frac{2}{x^2} + \frac{1}{x^4} = t\)

Differentiating with respect to x, we get

\(\rm \left(\frac{4}{x^3} - \frac{4}{x^5 }\right)dx = dt\)

Now,

I \(=\rm \frac{1}{4}\int \frac{dt}{\sqrt t}\)

\(=\rm \frac{1}{4}\int t^{-1/2}\;dt\)

\(=\rm \frac{1}{4}\times \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)

\(=\rm \frac{1}{2}t^{\frac{1}{2}}+c\)

\(=\rm \frac{1}{2} \sqrt {\rm 2 - \frac{2}{x^2} + \frac{1}{x^4}}+c\)

\(=\rm \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2}+C\)

Concept:

\(\rm \frac{d(\log x)}{dx} = \frac 1 x\)

Calculation:

Given:  y = e^{log (log x)}

To Find: \(\rm \frac{dy}{dx}\)

As we know that, e^{log x} = x

∴ e^{log (log x)} = log x

Now, y = log x

Differentiating with respect to x, we get

\(\rm \frac{dy}{dx}=\rm \frac{d(\log x)}{dx} = \frac 1 x\)

Concept:

Integration by parts: Integration by parts is a method to find integrals of products

• The formula for integrating by parts is given by,
• ∫u v dx = u∫v dx −∫u' (∫v dx) dx

Where u is the function u(x) and v is the function v(x)

ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

Let I = \(\int^\pi_0 x^3 \sin xdx\)

Apply by parts rule, we get

\(\rm =x^3 \int^\pi_0sinxdx- \int^\pi_03x^2(-cosx)dx\)

\(\rm =[x^3(-cosx)]_0^\pi+3[x^2\int^\pi_0cosxdx- \int^\pi_02x(sinx)dx]_0^\pi\)

\(\rm =\pi^3+0-6\int^\pi_0x(sinx)dx\)

\(\rm=\pi^3-6[x\int^\pi_0sinxdx- \int^\pi_0(-cosx)dx]\)

\(\rm =\pi^3-6[\pi- 0]\)

= π3 - 6π

Concept:

• For a given curve y = f(x), the slope (m) of the tangent at a given point x = a is given by m = f'(a).

• Chain Rule of Derivatives: \(\rm \dfrac{d}{dx}f[g(x)]=\dfrac{d}{d\ g(x)}f[g(x)]\times \dfrac{d}{dx}g(x)\).

• \(\rm \dfrac{d}{dx}x^n=nx^{n-1}\).

Calculation:

Let us first find out f'(x) for the curve y = f(x) = \(\rm \sqrt{4x-3}-1\).

f'(x) = \(\rm \dfrac{d}{dx}(\sqrt{4x-3}-1)\)

= \(\rm \dfrac{d}{d(4x-3)}(\sqrt{4x-3})\times \dfrac{d}{dx}(4x-3)\)

= \(\rm \dfrac{1}{2}\left(\dfrac{1}{\sqrt{4x-3}}\right)\times4\)

= \(\rm \dfrac{2}{\sqrt{4x-3}}\).

Let's say that the tangent at a point (a, b) has slope m = \(\dfrac23\).

∴ m = f'(a) = \(\rm \dfrac23\)

⇒ \(\rm \dfrac{2}{\sqrt{4a-3}}=\dfrac23\)

⇒ \(\rm \sqrt{4a-3}=3\)

⇒ 4a - 3 = 9

⇒ a = 3.

And b = f(a) = f(3) = \(\rm \sqrt{4(3)-3}-1=\sqrt9 -1 = 3-1 =2\).

∴ The required point is (a, b) = (3, 2).

Concept:

\(\rm \tan ^2x+1=sec^2x\)

tan x = sin x /cos x

sin 2x = 2 sin x cos x

The table below shows the sign of trigonometric ratios in different quadrants:

Calculation:

Here, \(\tan x = \dfrac{-3}{4}\)

Squaring and adding 1 to both the sides, we get 

\(\rm \tan ^2x+1=(-\frac 3 4)^2+1\\ sec^2x=\frac {25}{16}\\ sec x=\pm\frac54\)

x is in fourth quadrant so, sec x= 5/4 and cos x = 4/5                       ......(\(\dfrac{3\pi}{2}< x < 2\pi\))

Now

 \(\rm \frac {\sin x}{\cos x}=\frac{-3}{4}\\ sin x= \frac{-3}{4}\times \frac{4}{5}=\frac{-3}{5}\)

Now, sin 2x  = 2sin x cos x 

\(=2\times \frac{-3}{5}\times \frac{4}{5}\\ =\frac{-24}{25}\)

Hence, option (4) is correct.

Calculation:

I = \(\rm \int_1^4 \frac{x^2 + x}{\sqrt {2x+1}}\;dx\)

Let 2x + 1 = t²         ..... (1)

Differentiaiting with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

From equation (1), we get

x = \(\rm \frac{t^2-1}{2}\)

Now,

I = \(\rm \int_{\sqrt 3}^{3} \frac{(\rm \frac{t^2-1}{2})^2 + \rm \frac{t^2-1}{2}}{\sqrt{t^2}}\;tdt \)

= \(\rm \int_{\sqrt 3}^{3} \left(\frac{t^4-2t^2+1}{4}+ \rm \frac{t^2-1}{2} \right )dt\)

= \(\rm \int_{\sqrt 3}^{3} \left(\frac{t^4-2t^2+1+2t^2-2}{4} \right )dt\)

= \(\rm \frac{1}{4}\int_{\sqrt 3}^{3} (t^4-1)dt\)

= \(\rm \frac{1}{4} \left[\frac{t^5}{5}-t \right ]_{\sqrt 3}^{3}\)

= \(\frac{57-\sqrt 3}{5} \)

Concept:

\(\rm \frac{dx^n}{dx}=nx^{n-1}\)

Calculation:

Given: y2 = 4ax

Differentiating with respect to x, we get

\(\rm \Rightarrow {2y}\frac{dy}{dx} = 4a\)

\(\rm \Rightarrow \frac{dy}{dx} = \frac{4a}{{2y}} \)

\(\rm \therefore \frac{dy}{dx} = \frac{2a}{y} \)

CONCEPT:

If \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

The are 7 indeterminate forms which are as follows:

• \((\frac{0}{0})\)
• \(\left( {\frac{{ \pm ∞ }}{{ \pm ∞ }}} \right)\)
• (∞ - ∞)
• (0 × ∞)
• 00
• 1∞
• ∞0

CALCULATION:

Given: \(\mathop {\lim }\limits_{x \to 0} \frac{{-{3x^2} - 7x + 8}}{{{7x^2} + 2x + 2}} = k\)

As we know that, if \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\mathop {\lim }\limits_{x \to 0} \frac{{-{3x^2} - 7x + 8}}{{{7x^2} + 2x + 2}}\)does not result into any indeterminate form

So, we can substitute x = 0 in the expression \(\frac{{-{3x^2} - 7x + 8}}{{{7x^2} + 2x + 2}}\) in order to find the value of k

⇒ \(\mathop {\lim }\limits_{x \to 0} \frac{{-{3x^2} - 7x + 8}}{{{7x^2} + 2x + 2}} = 4 = k\)

Hence, Option D is the correct answer.

CONCEPT:

If \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

The are 7 indeterminate forms which are as follows:

• \((\frac{0}{0})\)
• \(\left( {\frac{{ \pm ∞ }}{{ \pm ∞ }}} \right)\)
• (∞ - ∞)
• (0 × ∞)
• 00
• 1∞
• ∞0

CALCULATION:

Given: \(\mathop {\lim }\limits_{x \to 7} g\left( x \right) = k\)  where \(g(x) = \sqrt {8x - 7}\)

As we know that, if \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\mathop {\lim }\limits_{x \to 7} g\left( x \right)\) does not result into any indeterminate form

So, we can substitute x = 7 in the expression \(g(x) = \sqrt {8x - 7}\) in order to find the value of k

⇒ \(\mathop {\lim }\limits_{x \to 7} \sqrt {8x -7} = 7 = k\)

Hence, option C is the correct answer.

\(\int\limits_0^{\pi/2}\sqrt{sin\theta}d\theta\int\limits_0^{\pi/2}\sqrt{cos\theta}d\theta\) 

\(\int\limits_0^{\pi/2}\sqrt{sin^{1/2}\theta}d\theta\int\limits_0^{\pi/2}\sqrt{cos^{1/2}\theta}d\theta\) 

\(=\cfrac{\Gamma\left(\frac{\frac12+1}{2}\right)\Gamma(\frac{0+1}2)}{2\Gamma\left(\frac{\frac12+0+2}2\right)}\) \(\times\cfrac{\Gamma\left(\frac{0+1}2\right)\Gamma\left(\frac{\frac12+1}{2}\right)}{2\Gamma\left(\frac{0+\frac12+2}2\right)}\)

\((\because\int\limits_0^{\pi/2}sin^m\theta cos^n \theta d\theta= \cfrac{\Gamma(\frac{m+1}2)\Gamma(\frac{n+1}2)}{2\Gamma}(\frac{m+n+2}2))\) 

\(= \cfrac{\Gamma(\frac34)\Gamma(\frac12)\Gamma(\frac12)\Gamma(\frac34)}{4\Gamma(\frac54)\Gamma(\frac54)}\) 

\(= \cfrac{\sqrt\pi\times\sqrt{\pi}\,\Gamma(\frac34)\Gamma(\frac34)}{4\Gamma(\frac54)\Gamma(\frac54)}\) 

\((\because\Gamma(1/2)=\sqrt{\pi})\)

\(= \cfrac{\pi}4\frac{[\Gamma(\frac34)]^2}{[\Gamma(\frac54)]^2}\) 

 \(=\frac{\pi}4\left(\cfrac{\Gamma(\frac34)}{\Gamma(\frac54)}\right)^2\)

\(\frac{d(u,v)}{d(x,y)}\) \(=\begin{vmatrix}\partial u/\partial x&\partial v/\partial x\\\partial u/\partial y&\partial v/\partial y\end{vmatrix}\)

\(=\begin{vmatrix}\frac\partial{\partial x}x^2&\frac{\partial y^2}{\partial x}\\\frac{\partial}{\partial y}x^2&\frac{\partial y^2}{\partial y}\end{vmatrix}\) \(=\begin{vmatrix}2x&0\\0&2y\end{vmatrix}\)

= 4xy - 0 = 4xy.

Let I = \(\int\frac{sin\,x(1-cos\,x)}{(1+cos\,x)(5-cos\,x)}dx\)

\(=\int\frac{sin\,x(1-cos\,x)}{(2-(1-cos\,x))(4+(1-cos\,x))}dx\)

Let 1 - cos x = t

sin x dx = dt

∴ I \(=\int\frac{t\,dt}{(2-t)(4+t)}\)

\(=\int\frac{t\,dt}{8-2t-t^2}\)

= -1/2 ∫\(\frac{-2t\,dt}{8-2t-t^2}\)

= -1/2 ∫\(\frac{(-2t-2+2)\,dt}{8-2t-t^2}\)

= -1/2 [∫\(\frac{-2t-2}{8-2t-t^2}dt\) + ∫\(\frac{2}{8-2t-t^2}dt\)]

= -1/2 log |8 - 2t - t²| - ∫\(\frac1{9-(t+1)^2}dt\)

= -1/2 log |8 - 2t - t²| - \(\frac1{2\times3}\) log \(|\frac{3+(t-1)}{3-(t-1)}|+c\)

= -1/2 log |8 - 2t - t²| - 1/6 log \(|\frac{2+t}{4-t}|+c\)

= -1/2 log |8 - 2(1 - cos x) - (1 - cos x)²| - 1/6 log \(|\frac{2+1-cos\,x}{4-(1-cos\,x)}|+c\)

= -1/2 log |5 - cos²x + 4 cos x| - 1/6 log \(|\frac{3-cos\,x}{3+cos\,x}|+c.\)