CBSE Sample Papers for Class 11 Maths Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 1 with Solutions

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions:

• All questions are compulsory.
• The question paper consists of 36 questions divided into 4 Sections A, B, C and D.
• Section A comprises of 20 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 6 questions of 4 marks each and Section D comprises of 4 questions of 6 marks each.
• There is no overall choice. However internal choice has been provided in 6 questions of 1 mark, 2 questions of 2 marks, 2 questions of 4 marks and 2 questions of 6 marks. You have to attempt only one of the alternatives in all such questions.
• Write the serial number of questions before attempting.
• Use of a calculator is not permitted.

Section – A
Question numbers 1 to 10 carries 1 mark each. For each of these questions, four alternative choices have been provided of which only one is correct. Select the correct choice :

Question 1.
The value of \(\frac{1-\tan ^2 15^{\circ}}{1+\tan ^2 15^{\circ}}\) is.
(A) 1
(B) √3
(C) \(\frac{\sqrt{3}}{2}\)
(D) 2
Answer:
Option (C) is correct

Explanation: We know that,
\(\frac{1-\tan ^2 x}{1+\tan ^2 x}\) = cos 2x
= cos 2x [Here, x = 15°]
= cos 30°
= √3/2

Question 2.
The probability that at least one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then P(A̅) + P(B̅) is equal to
(A) 0.4
(B) 0.8
(C) 1.2
(D) 1.6
Answer:
Option (C) is correct.

Explanation:
Given that,
P (A ∪ B) = 0.6
and P (A ∩ B) = 0.2
P(A̅) + P(B̅) =2 – P(A ∩B) – P(A ∪ B)
= 2 – 0.2 – 0.6
= 1.2

Question 3.
The point represented by the complex number 2 is rotated about origin through an angle in the clockwise direction, the new position of the point is :
(A) 1 – 2i
(B) – 1 -2i
(C) 2 + i-
(D) -1 + 2i
OR
If \(\left(\frac{1+i}{1-i}\right)^x\) = 1, then
(A) x = 2n + 1
(B) x = 4n
(C) x = 2 n
(D) x = 4 n + 1, where n ∈ N
Answer:
Option (B) is correct

Explanation:
On rotating, about origin through an angle n/2 in the clockwise direction, the new position of the point is -1 – 2i.
z = 2 – i
z₁ = (2 – i) e^-π/2 [∵ e^x = cos x + i sin x]
z₁ = (2 – i)[cos \(\left(-\frac{\pi}{2}\right)\) + i sin \(\left(-\frac{\pi}{2}\right)\)]
z₁ = 2 – i(0 – i)
= -2i + i²
= -2i – 1

OR

Option (B) is correct.

Explanation:
We have, \(\left[\frac{1+i}{1-i}\right]\) = 1
so, on rationalizing,
\(\frac{1+i}{1-i}=\frac{2 i}{2}\) = i
Hence (i) x = 4n where n belongs to N so x = 4n

Question 4.
If a parallelepiped is formed by planes drawn through the points (5,8,10) and (3,6,8) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is
(A) 2√3
(B) 3√2
(C) √2
(D) √3
Answer:
Option (A) is correct

Explanation:
We have parallelepiped formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is distance between the two points which is 2√3 units.

Question 5.
The slope of a line which cuts off intercepts of equal lengths on the axes is
(A) -1
(B) 0
(C) 2
(D) √3
OR
The equations of the lines passing through the point (1,0) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
(A) √3x + y – √3=0, √3x – y – √3 = 0
(B) √3x + y + √3 = 0, √3x – y + √3=0
(C) x + √3y – √3 = 0, x — √3y — √3 = 0
(D) None of these
Answer:
Option (A) is correct

Explanation:
The slope of a line which cuts off intercepts of equal lengths on the axes is -1.

OR

Option (A) is correct

Explanation:
The equations of the lines passing through the point (1, 0) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
√3x + y – √3 = 0 and √3x – y – √3 = 0

Question 6.
If – 3x + 17 < – 13, then
(A) x ∈ (10, ∞)
(B) x ∈ [-10, ∞)
(C) x ∈ (-∞, 10)
(D) x ∈ [-10, 10)
Answer:
Option (A) is correct.

Explanation:
We have,
– 3x + 17 < – 13
⇒ – 3x < – 30 ⇒ 3x > 30
⇒ x > 10
So x belongs to (10, infinity)

Question 7.
In a non-leap year, the probability of having 53 Tuesday or 53 Wednesday is
(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) None of these
Answer:
Option (B) is correct

Explanation:
In a non-leap year, the probability of having 53 Tuesday or 53 Wednesday is 2/7 Since there are only two ways in which Tuesday and Wednesday can occur in seven days of a week.

Question 8.
\(\lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1}\) is equal to:
(A) 1
(B) \(\frac{m}{n}\)
(C) –\(\frac{m}{n}\)
(D) \(\frac{m^2}{n^2}\)

OR

\(\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}\) is equal to:
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{-1}{2}\)
(D) -1
Answer:
Option (B) is correct

Explanation:
On applying the limits on given function we get m/n.

OR

Option (A) is correct

Explanation:
We know that,
\(\lim _{x \rightarrow 0} \frac{6 x(1-\cos 4 x)}{4 x(1-\cos 6 x)}=\frac{6}{4}\)
= \(\frac{3}{2}\)

Question 9.
If f(x) = 1 – x + x² – x³ + … – xⁿ + x¹⁰⁰, then f'(1) is equal to :
(A) 150
(B) -50
(C) -150
(D) 50
Answer:
Option (D) is correct

Explanation:
We have
f(x) = 1 – x + x² – x³ + … – xⁿ + x¹⁰⁰
f(x) = \(\frac{1\left(x^{100}-1\right)}{x-1}\)
f'(x) = 50

Question 10.
Mean deviation of nth observations x₁, x₂,…. x_n from their mean x is
(A) \(\sum_{i=1}^n\left(x_i-\bar{x}\right)\)
(B) \(\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right|\)
(C) \(\sum_{i=1}^n\left(x_i-\bar{x}\right)^2\)
(D) \(\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2\)
Answer:
Option (B) is correct

Explanation:
Mean deviation of nth observations x₁, x₂,…. x_n from their mean x is
\(\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right|\)

Question numbers 11 to 15 carry 1 mark each. Write whether the statement is true or false.

Question 11.
The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0.
Answer:
True

Explanation:
We have,
The point of intersection of the straight lines x + 2y -10= 0 and 2x + y + 5 = 0 is (-20/3,25/3). This point satisfies the equation 5x + 4y = 0.
So this statement is true.

Question 12.
The locus of the point of intersection of lines V3x – y – 4>/3fc = 0 and V3fcx + ky- 4^3 = 0 for different value of k is a hyperbola whose eccentricity is 2.
OR
The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.
Answer:
True

Explanation:
It is true that we get the locus of the point of intersection of lines for different value of k is a hyperbola whose eccentricity is 2.

OR

False

Explanation:
We have
Equation x + 3y = 0 does not satisfy the equation of circle so it is not the diameter of circle.
Hence this statement is False.

Question 13.
To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest is open to all, the number of ways in which the selection can be made is ⁵C₃ x ⁵C₉.
Answer:
True

Explanation:
The number of ways in which selection can be made is C(5, 3) × C(20, 9)
Hence given statement is True.

Question 14.
The sets {1, 2, 3,4} and {3, 4, 5,6} are equal.
Answer:
False

Explanation:
The two sets {1,2, 3,4} and {3,4,5,6} are not equal.

Question 15.
If xy > 0, then x > 0 and y < 0.
Answer:
False

Explanation:
Since xy > 0 so Either x > 0 and y > 0 or x < 0 and y < 0

Question numbers 16 to 20 carry 1 mark each.

Question 16.
If R₁ = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and range of R₁.
Answer:
Given, R₁ = {(x, y) \ y = 2x + 7, where x ∈ R and -5 ≤ x ≤ 5}
Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R}
= [-5,5]

Also, y = 2x +7 (given)
⇒ When x = -5, then y = 2(-5) + 7 = -3 and when x = 5, then y = 2(5) + 7 = 17
Hence, range of R: = { -3 ≤ y ≤ 17, y ∈ R}
= [-3,17]

Question 17.
Which term is in independent of x In the expansion of (3x³ – \(\frac{1}{2 x^3}\))¹⁰
OR
Write number of terms in the expansion of {(2x + y³)⁴}⁷.
Answer:
Let (r+1)th term be independent of x in the given expression
Now T_r+1 = ⁿC_r (3x³)^10-r \(\left(\frac{-1}{2 x^3}\right)^r\)
= ⁿC_r 3^10-r \(\frac{-1}{2}\)x^30-6r
This term is independent of x, if 30 – 6r =0
⇒ r = 5
So, (5 + 1)th i.e 6th term is independent of x.

OR

Given,
{(2x + y³)⁴}⁷ = (2x + y³)²⁸
Hence, there are 29 terms in the expansion.

Commonly Made Error
Some students write the number of terms equal to n -1 which leads to incorrect answer.

Answering Tip
Total number of terms in expansion (a + b)ⁿ is n + 1.

Question 18.
Find the equation of a circle with centre (-1,2) and radius 4.
Answer:
The equation of a circle with centre (h, k) and radius V is given by
(x – h)² + (y – k)² = r² …(1)
Here, (h,k) = (-1, 2) and r = 4
⇒ (1) ⇒ {x-(-1)}² + (y – 2)² = 4²
⇒ (x + 1)² + (y – 2)² = 4²
⇒ x² + 2x + 1 + y² – 4y + 4 = 16
⇒ x² + y² + 2x – 4y – 11 = 0 is the required equation of circle.

Question 19.
Find the distance from the origin to (6, 6, 7).
Answer:
The origin is 0(0, 0, 0). So, the distance of P(6, 6, 7) from 0(0,0,0) is

Question 20.
Which term of the G.E 2,1, \(\frac{1}{2}\), ………….. is \(\frac{1}{1024}\)?
OR
The first term of a G.P is 2 and sum to infinity is 6 find common ratio.
Answer:
Given G.E is 2,1,\(\frac{1}{2}\), \(\frac{1}{4}\), …………..
First term a = 2
and common ratio = \(\frac{1}{2}\)
Let nth term of G.E be \(\frac{1}{1024}\)

Commonly Made Error
Some students use the incorrect formula for nth term of GP that leads to wrong result.

Answering Tip
It is necessary to revise the formula of the nth \ term of G.P because it is used in almost every ! question. !
OR
Given, a =2 and S_∞ = 6
S_∞ = \(\frac{a}{1-r}\)
6 = \(\frac{2}{1-r}\)
⇒ 6 – 6r = 2
⇒ Common Ratio r = \(\frac{4}{6}=\frac{2}{3}\)

Section – B
Question numbers 21 to 26 carries 2 marks each.

Question 21.
Differentiate with respect to x : e^xsin x + xⁿ cos x.
OR
Find the derivative of \(\frac{2}{x+1}-\frac{x^2}{3 x-1}\).
Answer:
Let, f(x) = e^xsin x + xⁿ cos x.
∴ f'(x) = \(\frac{d}{d x}\){e^xsinx + xⁿ cos x}
= \(\frac{d}{d x}\)(e^xsinx) + \(\frac{d}{d x}\)(xⁿcos x)
= sin x\(\frac{d}{d x}\)e^x + e^x\(\frac{d}{d x}\)sinx + cos x \(\frac{d}{d x}\)xⁿ + xⁿ\(\frac{d}{d x}\)cos x

Using Product rule of differentiation
= sin x e^x + e^x. cos x + cos x .nx^n-1 + xⁿ.(-sin x)
f'(x) = e^x (sin x + cos x) + x^n-1 [n cos x – xsin x]

OR

Question 22.
Evaluate \(\lim _{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{x-a}\)
Answer:

Question 23.
An experiment involves tossing two coins and recording them in the following events
A : no tail
B : exactly one tail
C : at least one tail.
Write the sets representing events (i) A and C (ii) A but not B.
Answer:
When we toss two coins, the sample space is
S = {HH, HT, TH, TT}.
A = {HH}, B = {HT, TH},
C = {HT, TH, TT}
(i) A and C = A ∩ C = Φ
(ii) A but not B = A – B = {HH}

Question 24.
If cos (α + β)= \(\frac{4}{5}\) and sin (α – β) = \(\frac{5}{13}\) , where α, β lie between 0 and \(\frac{\pi}{4}\) then find the value of tan 2α.
OR
If A + B = \(\frac{\pi}{8}\), then prove that (1 + tanA) (1 + tan B) = 2
Answer:

OR
Given, A + B = \(\frac{\pi}{4}\)
Taking tangent both sides, we get
tan(A + B) = tan \(\frac{\pi}{4}\)
or \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
or tanA + tanB = 1 – tanAtanB
or tanA + tanB + tanAtanB = 1

Now, adding 1 both sides, we get
tan A + tan B + tan A tan B +1 = 2
(tan A +1) + tan B(1 + tan A) = 2
or (tanA + 1) (1 + tanB) = 2
or (1 + tan’A) (1 + tanB) = 2

Commonly Made Error
Generally students commit simplification errors in solving such type of questions.

Answering Tip
Learn concepts of trigonometric functions of i multiple and sub multiple angles.

Question 25.
Let A = {All prime numbers less than 10} and B = {all odd number less than 10}. Find (A – (A ∩ B)).
Answer:
Here, A = {2, 3, 5, 7} and B = {1, 3,5, 7,9}
A ∩ B — {2, 3, 5, 7} ∩ (1, 3, 5, 7, 9}
= {3, 5, 7}
A – (A ∩ B) = {2, 3,5, 7} – {3, 5, 7}
= {2}

Question 26.
Find the values of x for which the functions f(x) = 3x² – 1 and g(x) = 3 + x are equal.
Answer:
Given, f(x) = g(x)
∵ 3x² – 1 = 3 + x
⇒ 3x² – x – 4 = 0
or, (3x – 4)(x + 1) = 0
either x = -1 or x = \(\frac{4}{3}\)
x = -1, \(\frac{4}{3}\)
Hence, the values of x are -1 and \(\frac{4}{3}\) for which given function are equal.

Section – C
Question numbers 27 to 32 carry 4 marks each.

Question 27.
If sec x = √2 and \(\frac{3 \pi}{2}\) < x < 2π, find the values of \(\frac{1+\tan x+{cosec} x}{1+\cot x-{cosec} x}\)
OR
If tan (π cos θ) = cot (π sin θ), then prove that cos(θ – π) = \(\frac{1}{2 \sqrt{2}}\).
Answer:

OR
Given, tan (π cos θ) = cot (π sin θ)
∴ \(\frac{\sin (\pi \cos \theta)}{\cos (\pi \cos \theta)}=\frac{\cos (\pi \sin \theta)}{\sin (\pi \sin \theta)}\)
⇒ sin(π cos θ) sin(π sinθ)
= cos (π cos θ)cos (π ccosθ)
⇒ cos (π sinθ)cos (π cosθ) – sin(π cosθ) sin(π sinθ) = 0
⇒ cos (π sin θ + π cos θ) = 0
[∵ cos(A-B) = cos A cos B – sin A sin B]
⇒ cos(π sinθ + π cosθ) = cos\(\frac{\pi}{2}\)
⇒ π sinθ + π cosθ = \(\frac{\pi}{2}\)
⇒ sin θ + cos θ = \(\frac{1}{2}\)

Dividing and multiplying by √2, we get

Commonly Made Error
Some students make errors while simplifying and solving trigonometric equations which leads to prove the wrong result.

Answering Tip
Don’t forget to divide and multiply by √2 in calculation.

Question 28.
If α and β are different complex numbers with |β| = 1, then find \(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|\)
OR
If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that (a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².
Answer:
Given that |β| = 1

OR

Given that (a + ib) (c + id) (e + if) (g + ih) = A + iB
Taking modulus on both sides, we get
| (a + ib)(c + id)(e + if)(g + ih)\ = |A + iB|
|a + ib| |c + id| |e + if| |g + ih| = |A + iB| (∵|z₁z₂…z_n| = |z₁||z₂||z₃|…|z_n|)
\(\sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2}=\sqrt{A^2+B^2}\)
(∵ if z = a + ib, then |z| = \(\sqrt{a^2+b^2}\))

Squaring on both sides, we get
(a² + b²)(c² + d²)(e² + f²)(g² + h²) = (A² + B²)
Hence proved.

Question 29.
Find the equation of the set of all points whose distance from (0,4) are \(\frac{2}{3}\) of their distance from the line y = 9.
Answer:
Let the point be P(x, y)
∴ Distance from (0,4) = \(\sqrt{x^2+(y-4)^2}\)
So, the distance from the line y = 9 is \(\left|\frac{y-9}{\sqrt{1}}\right|\)
∴ \(\sqrt{x^2+(y-4)^2}=\frac{2}{3}\left|\frac{y-9}{\sqrt{0+1}}\right|\)
⇒ x² + y² – 8y + 16 = \(\frac{4}{9}\)(y² – 18y + 81)
⇒ 9x² + 9y² – 72y +144 = 4y² – 72y + 324
⇒ 9x² + 5y² = 180 is the required equation of set of all points.

Question 30.
The product of first three terms of G.P. is 1000. If 6 is added to its second term and 7 is added to its third term, the terms become in A.P Find G.P
Answer:
Let the numbers in G.P be \(\frac{a}{r}\), a, 1
Product \(\frac{a}{r}\) . a .ar = 1000
⇒ a³ = 1000
⇒ a = 10

According to question
A.P a₁ = \(\frac{a}{r}=\frac{10}{r}\)
a₂ = 0 + 6 = 10 + 6 = 16
a₃ = ar + 7 = 10r + 7
Also, a₃ = a₂ + 2(a₂ – a₁)
[a, b, c are in A.P]

10r + 7 = \(\frac{10}{r}\) + 2[16 – \(\frac{10}{r}\)]
⇒ 10r² + 7r = 10 + 32r – 20
⇒ 10r² – 25r + 10 = 0
⇒ (r – 2)(10r – 5) = 0
⇒ r = 2 or \(\frac{1}{2}\)
Substituting the value of a and r in eq (i), we get
G.P : 5,10,20,… when r = 2
and G.P : 20,10,5,… when r = \(\frac{1}{2}\)

Question 31.
Find the mean and variance of frequency distribution given below.

Answer:

Question 32.
Show that the coefficient of middle term in the expansion of (1 + x)²⁰ is equal to the sum of the coefficients of ‘ two middle of the terms in the expansion of (1 + x)¹⁹.
Answer:
The index 20 in (1 + x)²⁰ is even.
Middle term
= ²⁰C₁₀ (1)¹⁰x¹⁰ = ²⁰C₁₀ x¹⁰
Now coefficient of middle term in (1+ x)’9
The index 19 in (1 + x)¹⁹ is odd.
So, middle term as I]!j±J. and the next term
i.e., T₁₀ and T₁₁
T₁₀ = T₉₊₁ = ¹⁹C₉ × 1¹⁰ × x⁹ = ¹⁹C₉ x⁹
and T₁₁ = T₁₀₊₁ = ¹⁹C₁₀ × 1⁹ × x¹⁰ = ¹⁹C₁₀ x¹⁰
Now, sum of coefficient of middle terms in
(1 + x)¹⁹ = ¹⁹C₉ + ¹⁹C₁₀ = ²⁰C₁₀
= coefficient of middle term in (1 + x)²⁰
Hence proved

Section – D

Question 33.
If \(\frac{2 \sin a}{1+\cos a+\sin a}\) = y, then prove that \(\frac{1-\cos a+\sin a}{1+\sin a}\) is also equal to y.
Answer:

Question 34.
Solve the following system of inequalities graphically
2x + y ≤ 24, x + y ≥ 11, 2x + 5y ≤ 40 , x,y ≥ 0.
OR
Show that the solution set of the following system of linear inequalities is an unbounded region .
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.
Answer:
Inequality (x ≥ 0) represents the region on the right of y-axis and y-axis itself. Inequality (y ≥ 0) represent the region above x-axis and x-axis itself.
2x + y ≤ 24 ………….(1)
x + y ≥ 11 …………….(2)
2x + 5y ≤ 40 …………(3)
We first draw the graph of lines
2x + y = 24, x + y = 11 and 2x + 5y = 40
Now, 2x + y = 24, passes through A(12, 0) and B(0, 24)
Again, x + y = 11, passes through C(11, 0) and D(0, 11)

Further 2x + 5y = 40, passes through £(20, 0) and F(0,8) 1
Shaded area PQAC is the solution area.

Commonly Made Error
Some students do not shade the feasible region in the graph of the inequality that leads to deduction of marks.

Answering Tip
The points of the line are also included in the solution of the inequality and the graph of the inequality lies below or above the graph of the equation.

OR

We have 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Line 2x + y = 8 passes through the points (0, 8) and (4,0).
Line x + 2y = 10 passes through points (10, 0) and (0, 0) lie on the opposite side of the line 2x + y = 8.
For (0, 0), (0) + 2(0) – 10 < 0.
Therefore, the region satisfying the inequality x + 2y ≥ 10 and (0, 0) lie on the opposite side of the line x + 2y = 10.
Also, for x ≥ 0, y ≥ 0, region lies in the First quadrant. The common region is plotted as shown in the following figure.

It is clear from the graph that common shaded portion is unbounded.

Question 35.
If f(x) = \(\frac{x-1}{x+1}\) then show that
(i) f\(\left(\frac{1}{x}\right)\) = -f(x)
(ii) f\(\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}\)
Answer:

Question 36.
Evaluate \(\lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}\)
OR
If p, q, r are in G.P and the equations px² + 2qx + r =0 and dx² + 2ex + f = 0 have a common root, then show that \(\frac{d}{p}, \frac{e}{q}, \frac{f}{r}\)are in A.P.
Answer:

OR
It is given that p, q, r are in G.R
∴ q² = pr
Now, px² + 2qx + r = 0
⇒ px² + 2 yfprx + r = 0
⇒ \((\sqrt{p} x+\sqrt{r})^2\) =0
⇒ \(\sqrt{p} x+\sqrt{r}\) = 0
⇒ x = \(-\sqrt{\frac{r}{p}}\)

It is given that the equations px² + 2qx + r = 0 and dx² + lex + f = 0 have a common root and the equation px² + 2qx + r = 0 has equal roots equal to \(-\sqrt{\frac{r}{p}}\)

Commonly Made Error
Students have notation errors such as writing p² = qr instead of q²= pr. Incorrect signs are given for a term of roots of equations. Arithmetic errors are numerous.

Answering Tip
It is necessary to identify the given question, understand and then proceed to prove the result.