Table of Contents

**Important Questions for Class 12 Maths Term 2**

The students appearing in the CBSE Class 12 Term 2 must solve the CBSE Class 12 Term 2 Maths Important Questions given on this page. Mathematics requires practice, the more you practice maths questions your chances of getting good marks in Maths will increase. CBSE Class 12 Term 2 Maths Important Questions are best to practice. We have provided the solutions to CBSE Class 12 Term 2 Maths Important Questions. Bookmark this page to get all the updates from the Central Board of Secondary Education.

**Maths Class 12 Term 2 Important Questions: 2 Marks**

**Q.1: Determine the principal value of cos ^{-1}( -1/2).**

**Solution:**

Let us assume that, y = cos^{-1}( -1/2)

We can write this as:

cos y =- 1/2

cos y = cos (2π/3).

Thus, the Range of the principal value of cos^{-1} is [0, π ].

Therefore, the principal value of cos^{-1}( -1/2) is 2π /3.

**Q.2: Find the value of cot (tan ^{-1} α + cot^{-1} α).**

**Solution: **

Given that: cot (tan^{-1} 𝛂 + cot^{-1} 𝛂)

= cot (𝝅/𝟐) (since, tan^{-1} x + cot^{-1} x = 𝜋/2)

= cot (180°/2) ( we know that cot 90° = 0 )

= cot (90°)

= 0

Therefore, the value of cot (tan^{-1} α + cot^{-1} α) is 0.

**Q.3: The value of tan ^{-1} √3 – sec^{-1}(–2) is equal to:**

**(A) π (B) – π/3 (C) π/3 (D) 2π/3**

**Solution:**

Now, solve the first part of the expression: tan^{-1} √3

Let us take y = tan^{-1}√3

This can be written as:

tan y = √3

Now, use the trigonometry table to find the radian value

tan y = tan (π/3)

Thus, the range of principal value of tan^{-1} is (−π/2, π/2)

Therefore, the principal value of tan^{-1}√3 is π/3.

Now, solve the second part of the expression: sec^{-1}(–2)

Now, assume that y = sec^{-1} (–2)

sec y = -2

sec y = sec (2π/3)

We know that the principal value range of sec^{-1} is [0,π] – {π/2}

Therefore, the principal value of sec^{-1} (–2) = 2π/3

Now we have:

tan^{-1}(√3) = π/3

sec^{-1} (–2) = 2π/3

Now, substitute the values in the given expression:

= tan^{-1} √3 – sec^{-1} (−2)

= π/3 − (2π/3)

= π/3 − 2π/3

= (π − 2π)/3

= – π/3

Hence, the correct answer is an option (B)

**Q.4: Prove that sin ^{-1} (3/5) – sin^{-1} (8/17) = cos^{-1} (84/85).**

**Solution:**

Let sin^{-1} (3/5) = a and sin^{-1} (8/17) = b

Thus, we can write sin a = 3/5 and sin b = 8/17

Now, find the value of cos a and cos b

**To find cos a:**

Cos a = √[1 – sin^{2} a]

= √[1 – (3/5)^{2} ]

= √[1 – (9/25)]

= √[(25-9)/25]

= 4/5

Thus, the value of cos a = 4/5

**To find cos b:**

Cos b= √[1 – sin^{2} b]

= √[1 – (8/17)^{2} ]

= √[1 – (64/289)]

= √[(289-64)/289]

= 15/17

Thus, the value of cos b = 15/17

We know that cos (a- b) = cos a cos b + sin a sin b

Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:

cos (a – b) = (4/5)x (15/17) + (3/5)x(8/17)

cos (a – b) = (60 + 24)/(17x 5)

cos (a – b) = 84/85

(a – b) = cos^{-1} (84/85)

Substituting the values of a and b sin^{-1} (3/5)- sin^{-1} (8/7) = cos^{-1} (84/85)

Hence proved.

**Q. 5: Find the value of cos ^{-1} (1/2) + 2 sin^{-1} (1/2).**

**Solution:**

**First, solve for cos ^{-1} (1/2):**

Let us take, y = cos^{-1} (1/2)

This can be written as:

cos y = (1/2)

cos y = cos (π /3).

Thus, the range of principal value of cos^{-1} is [0, π ]

Therefore, the principal value of cos^{-1} (1/2) is π/3.

**Now, solve for sin ^{-1} (1/2):**

Let y = sin^{-1} (1/2)

sin y = 1/2

sin y = sin ( π/6)

Thus, the range of principal value of sin^{-1} is [(-π)/2, π/2 ]

Hence, the principal value of sin^{-1} (1/2) is π/6.

Now we have cos^{-1} (1/2) = π/3 & sin^{-1} (1/2) = π/6

Now, substitute the obtained values in the given formula, we get:

= cos^{-1} (1/2) + 2sin^{-1} (1/2)

= π /3 + 2( π/6)

= π/3 + π/3

= ( π+π )/3

= 2π /3

Thus, the value of cos^{-1} (1/2) + 2 sin^{-1} (1/2) is 2π /3.

**Q. 6: Find the direction cosines of the line passing through the two points (– 2, 4, – 5) and (1, 2, 3).**

**Solution:**

We know that the direction cosines of the line passing through two points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) are given by

Using the distance formula,

From the given,

P(x_{1}, y_{1}, z_{1}) = (-2, 4, -5) and Q(x_{2}, y_{2}, z_{2}) = (1, 2, 3)

Hence, the direction cosines of the line joining the given two points are.

**Q. 7: Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear.**

**Solution:**

We know that the direction ratios of the line passing through two points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) are given by:

x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2}

Given points are A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11).

Direction ratios of the line joining A and B are:

1 – 2, – 2 – 3, 3 + 4

i.e. – 1, – 5, 7.

The direction ratios of the line joining B and C are:

3 –1, 8 + 2, – 11 – 3

i.e., 2, 10, – 14.

From the above, it is clear that direction ratios of AB and BC are proportional.

That means AB is parallel to BC. But point B is common to both AB and BC.

Hence, A, B, C are collinear points.

**Q. 8: If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.**

**Solution:**

Let the direction cosines of the line be l, m, and n.

l = cos 90° = 0

m = cos 135° = -1/√2

n = cos 45° = 1/√2

Hence, the direction cosines of the line are 0, -1/√2, and 1/√2.

**Q. 9: Find the angle between the pair of lines given by**

**Solution:**

From the given,

Let θ be the angle between the given pair of lines.

**Q. 10: Find the angle between the pair of lines given below.**

**(x + 3)/3 = (y -1)/5 = (z + 3)/4**

**(x + 1)/1 = (y – 4)/1 = (z – 5)/2**

**Solution:**

Given,

(x + 3)/3 = (y -1)/5 = (z + 3)/4

(x + 1)/1 = (y – 4)/1 = (z – 5)/2

The direction ratios of the first line are:

a_{1} = 3, b_{1} = 5, c_{1} = 4

The direction ratios of the second line are:

a_{2} = 1, b_{2} = 1, c_{2} = 2

Hence, the required angle is

**Q. 11: Find the distance between the lines l _{1} and l_{2} given by:**

**Solution:**

Given two lines are parallel.

The distance between the two given lines is

**Q. 12: Show that the lines (x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.**

**Solution:**

Given lines are:

(x – 5)/7 = (y + 2)/-5 = z/1 and x/1 = y/2 = z/3

The direction ratios of the given lines are 7, -5, 1 and 1, 2, 3, respectively.

We know that,

Two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular to each other if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Therefore, 7(1) + (-5) (2) + 1 (3)

= 7 – 10 + 3

=0

Hence, the given lines are perpendicular to each other.

**Q. 13: Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector**

**Solution:**

Given, the normal vector is:

We know that the equation of the plane with position vector is given by

Hence, the vector equation of the required plane is

**Q. 14: Find the intercepts cut off by the plane 2x + y – z = 5.**

**Solution:**

Given plane is 2x + y – z = 5 ……(i)

Dividing both sides of the equation (i) by 5,

(⅖)x + (y/5) – (z/5) = 1

We know that,

The equation of a plane in intercept form is (x/a) + (y/b) + (z/c) = 1, where a, b, c are intercepts cut off by the plane at x, y, z-axes respectively.

For the given equation,

a = 5/2, b = 5, c = -5

Hence, the intercepts cut off by the plane are 5/2, 5 and -5.

**Q. 15: Find the equations of the planes that passes through three points (1, 1, 0), (1, 2, 1), and (– 2, 2, – 1).**

**Solution:**

Given points are (1, 1, 0), (1, 2, 1), and (– 2, 2, – 1).

Therefore, the plane will pass through the given three points.

We know that,

The equation of the plane through the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

(x – 1)(-2) -(y – 1) (3) + z (3) = 0

-2x + 2 – 3y + 3 + 3z = 0

-2x – 3y + 3z + 5 = 0

-2x – 3y + 3z = -5

Therefore, 2x + 3y – 3z = 5 is the required Cartesian equation of the plane.

**Q. No.16: Determine order and degree (if defined) of differential equation (y′′′) ^{2} + (y″)^{3} + (y′)^{4} + y^{5} = 0**

**Solution:**

Given differential equation is (y′′′)^{2} + (y″)^{3} + (y′)^{4} + y^{5} = 0

The highest order derivative present in the differential equation is y′′′.

Therefore, its order is 3.

The given differential equation is a polynomial equation in y′′′, y′′, and y′.

The highest power raised to y′′′ is 2.

Hence, its degree is 2.

**Q. No. 17: Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation d ^{2}y/dx^{2 }+ y=0.**

**Solution:**

The given function is y = a cos x + b sin x … (1)

Differentiating both sides of equation (1) with respect to x,

dy/dx = – a sinx + b cos x

d^{2}y/dx^{2} = – a cos x – b sinx

LHS = d^{2}y/dx^{2 }+ y

= – a cos x – b sinx + a cos x + b sin x

= 0

= RHS

Hence, the given function is a solution to the given differential equation.

**Q. No. 18: The number of arbitrary constants in the general solution of a differential equation of fourth order is:**

**(A) 0 (B) 2 (C) 3 (D) 4**

**Solution:**

We know that the number of constants in the general solution of a differential equation of order n is equal to its order.

Therefore, the number of constants in the general equation of the fourth-order differential equation is four.

Hence, the correct answer is D.

Note: The number of constants in the general solution of a differential equation of order n is equal to zero.

**Q. No. 19: Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.**

**Solution:**

Given,

y = a sin (x + b) … (1)

Differentiating both sides of equation (1) with respect to x,

dy/dx = a cos (x + b) … (2)

Differentiating again on both sides with respect to x,

d^{2}y/dx^{2} = – a sin (x + b) … (3)

Eliminating a and b from equations (1), (2) and (3),

d^{2}y/ dx^{2 }+ y = 0 … (4)

The above equation is free from the arbitrary constants a and b.

This the required differential equation.

**Class 12th Maths Term 2 Important Questions: 3 Marks**

**Q. No. 1 Find the differential equation of the family of lines through the origin.**

**Solution:**

Let y = mx be the family of lines through the origin.

Therefore, dy/dx = m

Eliminating m, (substituting m = y/x)

y = (dy/dx) . x

or

x. dy/dx – y = 0

**Q. No. 2: Form the differential equation of the family of circles having a centre on y-axis and radius 3 units.**

**Solution:**

The general equation of the family of circles having a centre on the y-axis is x^{2} + (y – b)^{2} = r^{2}

Given the radius of the circle is 3 units.

The differential; equation of the family of circles having a centre on the y-axis and radius 3 units is as below:

x^{2} + (y – b)^{2} = 3^{2}

x^{2} + (y – b)^{2} = 9 ……(i)

Differentiating (i) with respect to x,

2x + 2(y – b).y′ = 0

⇒ (y – b). y′ = -x

⇒ (y – b) = -x/y′ …….(ii)

Substituting (ii) in (i),

x^{2} + (-x/y′)^{2} = 9

⇒ x^{2}[1 + 1/(y′)^{2}] = 9

⇒ x^{2 }[(y′)^{2} + 1) = 9 (y′)^{2}

⇒ (x^{2} – 9) (y′)^{2} + x^{2} = 0

Hence, this is the required differential equation.

**Q. No. 3: Find the general solution of the differential equation dy/dx =1+y ^{2}/1+x^{2}.**

**Solution:**

Given differential equation is dy/dx =1+y^{2}/1+x^{2}

Since 1 + y^{2} ≠ 0, therefore by separating the variables, the given differential equation can be written as:

dy/1+y^{2} = dx/1+x^{2} …….(i)

Integrating equation (i) on both sides,

tan^{-1}y = tan^{-1}x + C

This is the general solution of the given differential equation.

**Q. No. 4: For each of the given differential equation, find a particular solution satisfying the given condition:**

**dy/dx = y tan x ; y = 1 when x = 0**

**Solution:**

dy/dx = y tan x

dy/y = tan x dx

Integrating on both sides,

log y = log (sec x) +C

log y = log (C sec x)

⇒ y = C sec x ……..(i)

Now consider y = 1 when x= 0.

1 = C sec 0

1 = C (1)

C = 1

Substituting C = 1 in (i)

y = sec x

Hence, this is the required particular solution of the given differential equation.

**Q. No. 5: Find the equation of a curve passing through (1, π/4) if the slope of the tangent to the curve at any point P (x, y) is y/x – cos ^{2}(y/x).**

**Solution:**

According to the given condition,

dy/dx = y/x – cos^{2}(y/x) ………….(i)

This is a homogeneous differential equation.

Substituting y = vx in (i),

v + (x) dv/dx = v – cos^{2}v

⇒ (x)dv/dx = – cos^{2}v

⇒ sec^{2}v dv = – dx/x

By integrating on both the sides,

⇒ ∫sec^{2}v dv = – ∫dx/x

⇒ tan v = – log x + c

⇒ tan (y/x) + log x = c ……….(ii)

Substituting x = 1 and y = π/4,

⇒ tan (π/4) + log 1 = c

⇒ 1 + 0 = c

⇒ c = 1

Substituting c = 1 in (ii),

tan (y/x) + log x = 1

**Q. No. 6: Integrating factor of the differential equation (1 – x ^{2})dy/dx – xy = 1 is**

**(A) -x**

**(B) x/ (1 + x ^{2})**

**(C) √(1- x ^{2})**

**(D) ½ log (1 – x ^{2})**

**Solution:**

Given differential equation is (1 – x^{2})dy/dx – xy = 1

(1 – x^{2})dy/dx = 1 + xy

dy/dx = (1/1 – x^{2}) + (x/1 – x^{2})y

dy/dx – (x/1 – x^{2})y = 1/1-x^{2}

This is of the form dy/dx + Py = Q

We can get the integrating factor as below:

Let 1 – x^{2} = t

Differentiating with respect to x

-2x dx = dt

-x dx = dt/2

Now,

I.F = **√**t = √(1- x^{2})

Hence, option C is the correct answer.

**Question 7:**

Evaluate: ∫ 3ax/(b^{2 }+c^{2}x^{2}) dx

**Solution:**

To evaluate the integral, I = ∫ 3ax/(b^{2 }+c^{2}x^{2}) dx

Let us take v = b^{2 }+c^{2}x^{2}, then

dv = 2c^{2}x dx

Thus, ∫ 3ax/(b^{2 }+c^{2}x^{2}) dx

= (3ax/2c^{2}x)∫dv/v

Now, cancel x on both numerator and denominator, we get

= (3a/2c^{2})∫dv/v

= (3a/2c^{2}) log |b^{2 }+c^{2}x^{2}| + C

Where C is an arbitrary constant

**Question 8:**

Determine ∫tan^{8}x sec^{4} x dx

**Solution:**

Given: ∫tan^{8}x sec^{4} x dx

Let I = ∫tan^{8}x sec^{4} x dx — (1)

Now, split sec^{4}x = (sec^{2}x) (sec^{2}x)

Now, substitute in (1)

I = ∫tan^{8}x (sec^{2}x) (sec^{2}x) dx

= ∫tan^{8}x (tan^{2 }x +1) (sec^{2}x) dx

It can be written as:

= ∫tan^{10}x sec^{2} x dx + ∫tan^{8}x sec^{2} x dx

Now, integrate the terms with respect to x, we get:

I =( tan^{11} x /11) + ( tan^{9} x /9) + C

Hence, ∫tan^{8}x sec^{4} x dx = ( tan^{11} x /11) + ( tan^{9} x /9) + C

**Question 9:**

**Write the anti-derivative of the following function: 3x ^{2}+4x^{3}**

**Solution:**

Given: 3x^{2}+4x^{3}

The antiderivative of the given function is written as:

∫3x^{2}+4x^{3 }dx = 3(x^{3}/3) + 4(x^{4}/4)

= x^{3} + x^{4}

Thus, the antiderivative of 3x^{2}+4x^{3} = x^{3} + x^{4}

**Question 10:**

**Determine the antiderivative F of “f” , which is defined by f (x) = 4x ^{3} – 6, where F (0) = 3**

**Solution:**

Given function: f (x) = 4x^{3} – 6

Now, integrate the function:

∫4x^{3} – 6 dx = 4(x^{4}/4)-6x + C

∫4x^{3} – 6 dx = x^{4} – 6x + C

Thus, the antiderivative of the function, F is x^{4} – 6x + C, where C is a constant

Also, given that, F(0) = 3,

Now, substitute x = 0 in the obtained antiderivative function, we get:

(0)^{4} – 6(0) + C = 3

Therefore, C = 3.

Now, substitute C = 3 in antiderivative function

Hence, the required antiderivative function is x^{4} – 6x + 3.

**Question 11:**

**Integrate the given function using integration by substitution: 2x sin(x ^{2}+ 1) with respect to x:**

**Solution:**

Given function: 2x sin(x^{2}+ 1)

We know that, the derivative of x^{2} + 1 is 2x.

Now, use the substitution method, we get

x^{2} + 1 = t, so that 2x dx = dt.

Hence, we get ∫ 2x sin ( x^{2 }+1) dx = ∫ sint dt

= – cos t + C

= – cos (x^{2} + 1) + C

Where C is an arbitrary constant

Therefore, the antiderivative of 2x sin(x^{2}+ 1) using integration by substitution method is = – cos (x^{2} + 1) + C

**Question 12:**

**Integrate: ∫ sin ^{3} x cos^{2}x dx**

**Solution:**

Given that, ∫ sin^{3} x cos^{2}x dx

This can be written as:

∫ sin^{3} x cos^{2}x dx = ∫ sin^{2} x cos^{2}x (sin x) dx

=∫(1 – cos^{2}x ) cos^{2}x (sin x) dx —(1)

Now, substitute t = cos x,

Then dt = -sin x dx

Now, equation can be written as:

Thus, ∫ sin^{3} x cos^{2}x dx = – ∫ (1-t^{2})t^{2} dt

Now, multiply t^{2} inside the bracket, we get

= – ∫ (t^{2}-t^{4}) dt

Now, integrate the above function:

= – [(t^{3}/3) – (t^{5}/5)] + C —(2)

Where C is a constant

Now, substitute t = cos x in (2)

= -(⅓)cos^{3}x +(1/5)cos^{5}x + C

Hence, ∫ sin^{3} x cos^{2}x dx = -(⅓)cos^{3}x +(1/5)cos^{5}x + C

**Q. No.:13 Find the area enclosed by the ellipse x ^{2}/a^{2} + y^{2}/b^{2} =1.**

**Solution:**

Given,

We know that,

Ellipse is symmetrical about both x-axis and y-axis.

Area of ellipse = 4 × Area of AOB

Substituting the positive value of y in the above expression since OAB lies in the first quadrant.

= 2ab × sin^{-1}(1)

= 2ab × π/2

= πab

Hence, the required area is πab sq.units.

**Class 12 Term 2 Maths Important Questions: 4 Marks**

**Q. No. 1: Find the area of the region bounded by y ^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.**

**Solution:**

We can draw the figure of y^{2} = 9x; x = 2, x = 4 and the x-axis in the first curve as below.

y^{2} = 9x

y = ±√(9x)

y = ±3√x

We can consider the positive value of y since the required area is in the first quadrant.

The required area is the shaded region enclosed by ABCD.

= 2 [(2)^{3} – (√2)^{3}]

= 2[8 – 2√2]

= 16 – 4√2

Hence, the required area is 16 – 4√2 sq.units.

**Q. No. 2: Find the area of the curve y = sin x between 0 and π.**

**Solution:**

Given,

y = sin x

Area of OAB

= – [cos π – cos 0]

= -(-1 -1)

= 2 sq. units

**Q. No. 3: Find the area of the region bounded by the two parabolas y = x ^{2} and y^{2} = x.**

**Solution:**

Given two parabolas are y = x^{2} and y^{2} = x.

The point of intersection of these two parabolas is O (0, 0) and A (1, 1) as shown in the below figure.

Now,

y^{2} = x

y = √x = f(x)

y = x^{2} = g(x), where, f (x) ≥ g (x) in [0, 1].

Area of the shaded region

= (⅔) – (⅓)

= ⅓

Hence, the required area is ⅓ sq.units.

**Q. No. 4: Smaller area enclosed by the circle x ^{2}+ y^{2} = 4 and the line x + y = 2 is**

**(A) 2 (π – 2)**

**(B) π – 2**

**(C) 2π – 1**

**(D) 2 (π + 2)**

**Solution:**

Option (B) is the correct answer.

Explanation:

Given,

Equation of circle is x^{2}+ y^{2} = 4……….(i)

x^{2}+ y^{2} = 2^{2}

y = √(2^{2} – x^{2}) …………(ii)

Equation of a lines is x + y = 2 ………(iii)

y = 2 – x

x | 0 | 2 |

y | 2 | 0 |

Therefore, the graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).

From the graph of a circle (i) and straight-line (iii), it is clear that points of intersections of circle

(i) and the straight line (iii) is A (2, 0) and B (0.2).

Area of OACB, bounded by the circle and the coordinate axes is

= [ 1 × √0 + 2 sin^{-1}(1) – 0√4 – 2 × 0]

= 2 sin^{-1}(1)

= 2 × π/2

= π sq. units

Area of triangle OAB, bounded by the straight line and the coordinate axes is

= 4 – 2 – 0 + 0

= 2 sq.units

Hence, the required area = Area of OACB – Area of triangle OAB

= (π – 2) sq.units

**Question 5:**

**Explain the continuity of the function f(x) = sin x . cos x**

**Solution:**

We know that sin x and cos x are continuous functions. It is known that the product of two continuous functions is also a continuous function.

Hence, the function f(x) = sin x . cos x is a continuous function.

**Question 6:**

**Determine the points of discontinuity of the composite function y = f[f(x)], given that, f(x) = 1/x-1.**

**Solution:**

Given that, f(x) = 1/x-1

We know that the function f(x) = 1/x-1 is discontinuous at x = 1

Now, for x ≠1,

f[f(x)]= f(1/x-1)

= 1/[(1/x-1)-1]

= x-1/ 2-x, which is discontinuous at the point x = 2.

Therefore, the points of discontinuity are x = 1 and x=2.

**Question 7:**

**If f (x) = |cos x|, find f’(3π/4)**

**Solution:**

Given that, f(x) = |cos x|

When π/2 <x< π, cos x < 0,

Thus, |cos x| = -cos x

It means that, f(x) = -cos x

Hence, f’(x) = sin x

Therefore, f’(3π/4) = sin (3π/4) = 1/√2

f’(3π/4) = 1/√2

**Question 8:**

**Verify the mean value theorem for the following function f (x) = (x – 3) (x – 6) (x – 9) in [3, 5]**

**Solution:**

f(x)=(x−3)(x−6)(x−9)

=(x−3)(x^{2}−15x+54)

=x^{3}−18x^{2}+99x−162

fc∈(3,5)

f′(c)=f(5)−f(3)/5−3

f(5)=(5−3)(5−6)(5−9)

=2(−1)(−4)=−8

f(3)=(3−3)(3−6)(3−9)=0

f′(c)=8−0/2=4

∴f′(c)=3c^{2}−36c+99

3c^{2}−36c+99=4

3c^{2}−36c+95=0

ax^{2}+bx+c=0

a=3

b=−36

c=95

c=36±√(36)^{2}−4(3)(95)/2(3)

=36±√1296−1140/6

=36±12.496

c=8.8&c=4.8

c∈(3,5)

f(x)=(x−3)(x−6)(x−9) on [3,5]

**Question 9:**

**Explain the continuity of the function f = |x| at x = 0.**

**Solution:**

From the given function, we define that,

f(x) = {-x, if x<0 and x, if x≥0

It is clearly mentioned that the function is defined at 0 and f(0) = 0. Then the left-hand limit of f at 0 is

Lim_{x→0- }f(x)= lim_{x→0- }(-x) = 0

Similarly for the right hand side,

Lim_{x→0+ }f(x)= lim_{x→0+} (x) = 0

Therefore, for the both left hand and the right hand limit, the value of the function coincide at the point x = 0.

Therefore, the function f is continuous at the point x =0.

**Question 10:**

**If y= tan x + sec x , then show that d ^{2.}y / dx^{2 }= cos x / (1-sin x)^{2}**

**Solution:**

Given that, y= tan x + sec x

Now, the differentiate wih respect to x, we get

dy/dx = sec^{2} x + sec x tan x

= (1/ cos^{2} x) + (sin x/ cos^{2} x)

= (1+sinx)/ (1+sinx)(1-sin x)

Thus, we get.

dy/dx = 1/(1-sin x)

Now, again differentiate with respect to x, we will get

d^{2}y / dx^{2 }= -(-cosx )/(1- sin x)^{2}

d^{2}y / dx^{2} = cos x / (1-sinx)^{2}.

**Q.11:** Show that the Signum Function f: R → R, given by

**Solution:**

**Check for one to one function:**

For example:

f(0) = 0

f(-1) = -1

f(1) = 1

f(2) = 1

f(3) = 1

Since, for the different elements say f(1), f(2) and f(3), it shows the same image, then the function is not one to one function.

**Check for Onto Function:**

For the function,f: R →R

In this case, the value of f(x) is defined only if x is 1, 0, -1

For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.

Thus, the function “f” is not onto function.

Hence, the given function “f” is neither one-one nor onto.

**Q.12:** **If f: R → R is defined by f(x) = x ^{2} − 3x + 2, find f(f(x)).**

**Solution:**

Given function:

f(x) = x^{2} − 3x + 2.

To find f(f(x))

f(f(x)) = f(x)^{2} − 3f(x) + 2.

= (x^{2} – 3x + 2)^{2} – 3(x^{2} – 3x + 2) + 2

By using the formula (a-b+c)^{2} = a^{2}+ b^{2}+ c^{2}-2ab +2ac-2ab, we get

= (x^{2})^{2} + (3x)^{2} + 2^{2}– 2x^{2} (3x) + 2x^{2}(2) – 2x^{2}(3x) – 3(x^{2} – 3x + 2) + 2

Now, substitute the values

= x^{4} + 9x^{2} + 4 – 6x^{3} – 12x + 4x^{2} – 3x^{2} + 9x – 6 + 2

= x^{4} – 6x^{3} + 9x^{2} + 4x^{2} – 3x^{2} – 12x + 9x – 6 + 2 + 4

Simplify the expression, we get,

f(f(x)) = x^{4} – 6x^{3} + 10x^{2} – 3x

**Q.13: Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.**

**Solution:**

**Check the binary operation * is commutative :**

We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

L.H.S =(a, b) * (c, d)

=(a + c, b + d)

R. H. S = (c, d) * (a, b)

=(a + c, b + d)

Hence, L.H.S = R. H. S

Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

* is commutative (a, b) * (c, d) = (a + c, b + d)

**Check the binary operation * is associative :**

We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

L.H.S = (a, b) * ( (c, d) * (x, y) ) = (a+c+x, b+d+y)

R.H.S = ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)

Thus, L.H.S = R.H.S

Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

Thus, the binary operation * is associative

**Checking for Identity Element:**

e is identity of * if (a, b) * e = e * (a, b) = (a, b)

where e = (x, y)

Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)

= (x + a , b + y) = (a, b)

Now, (a + x, b + y) = (a, b)

Now comparing these, we get:

a+x = a

x = a -a = 0

Next compare: b +y = b

y = b-b = 0

Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.

Therefore, the operation * does not have any identity element.

**Q.14: Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.**

**Solution:**

**Checking for Inverse:**

f(x) = 4x + 3

Let f(x) = y

y = 4x + 3

y – 3 = 4x

4x = y – 3

x = (𝑦 − 3)/4

Let g(y) = (𝑦 − 3)/4

where g: Y → N

**Now find gof:**

gof**=** g(f(x))

= g(4x + 3) = [(4𝑥 + 3) − 3]/4

= [4𝑥 + 3 − 3]/4

=4x/4

= x = I_{N}

**Now find fog:**

fog= f(g(y))

= f [(𝑦 − 3)/4]

=4[(𝑦 − 3)/4] +3

= y – 3 + 3

= y + 0

= y = I_{y}

Thus, gof = I_{N}and fog = I_{y},

Hence, f is invertible

Also, the Inverse of f = g(y) = [𝒚 – 3]/ 4

**Q. 15: Let A = R {3} and B = R – {1}. Consider the function f: A →B defined by f (x) = (x- 2)/(x -3). Is f one-one and onto? Justify your answer.**

**Solution:**

Given function:

f (x) = (x- 2)/(x -3)

**Checking for one-one function:**

f (x_{1}) = (x_{1}– 2)/ (x_{1}– 3)

f (x_{2}) = (x_{2}-2)/ (x_{2}-3)

Putting f (x_{1}) = f (x_{2})

(x_{1}-2)/(x_{1}-3)= (x_{2}-2 )/(x_{2} -3)

(x_{1}-2) (x_{2}– 3) = (x_{1}– 3) (x_{2}-2)

x_{1} (x_{2}– 3)- 2 (x_{2} -3) = x_{1} (x_{2}– 2) – 3 (x_{2}– 2)

x_{1} x_{2} -3x_{1} -2x_{2} + 6 = x_{1} x_{2} – 2x_{1} -3x_{2} + 6

-3x_{1}– 2x_{2} =- 2x_{1} -3x_{2}

3x_{2} -2x_{2} = – 2x_{1} + 3x_{1}

x_{1}= x_{2}

Hence, if f (x_{1}) = f (x_{2}), then x_{1} = x_{2}

Thus, the function f is one-one function.

Checking for onto function:

f (x) = (x-2)/(x-3)

Let f(x) = y such that y B i.e. y ∈ R – {1}

So, y = (x -2)/(x- 3)

y(x -3) = x- 2

xy -3y = x -2

xy – x = 3y-2

x (y -1) = 3y- 2

x = (3y -2) /(y-1)

For y = 1, x is not defined But it is given that. y ∈ R – {1}

Hence, x = (3y- 2)/(y- 1) ∈ R -{3} Hence, f is onto.

**Related Post:**

CBSE Class 12th Hindi Term 2 Important Questions

CBSE Class 12th Biology Term 2 Important Questions

CBSE Class 12th Chemistry Term 2 Important Questions

CBSE Class 12th Physics Term 2 Important Questions

**CBSE Class 12th Term 2 Maths Important Questions with Answer: FAQs**

**Q. Is NCERT enough for Class 12 Term 2 Mathematics?**

Yes, NCERT is more than enough for Class 12 Term 2 Mathematics. The students can practice the CBSE Class 12 Term 2 Mathematics Important Questions given on this page.

**Q. Where can I find CBSE Class 12 Term 2 Mathematics Important Questions?**

You can find the CBSE Class 12 Term 2 Mathematics Important Questions here. We have given the CBSE Class 12 Term 2 Mathematics Important Questions based on the latest CBSE Term 2 Exam pattern on this page.

**Q. When CBSE will conduct the Class 12 Term 2 Exam for Mathematics?**

The CBSE will conduct the Class 12 Term 2 Exam for Mathematics on the 7th of June 2022, Tuesday.