JEE Main 2021 July 20 shift 1 Mathematics question paper and solutions are given here. JEE aspirants and students likewise can quickly access the question paper as well as the PDF from this page. The solutions to each of the questions have been prepared by our specialized team of experts. The solutions are structured in a step by step format which will ultimately help each student preparing for the exam to easily understand the problems. They will find only the correct answers to the questions asked in the JEE Main 2021 Mathematics paper.

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**Question 1: The mean of 6 numbers is 6.5 and its variance is 10.25. If 4 numbers are 2,4,5 and 7, then find the other two.**

**Answer: **a = 10 and b = 11 or a = 11 and b = 10

Let a and b be the other 2 numbers.

So (a + b + 2 + 4 + 5 + 7)/6 = 6.5

⇒ (a + b + 18)/6 = 6.5

⇒ a + b = 21 …(i)

σ^{2} = 10.25

⇒ (a^{2} + b^{2} + 2^{2} + 4^{2} + 5^{2} + 7^{2})/6 – 6.5^{2} = 10.25

⇒ (a^{2} + b^{2} + 4 + 16 + 25 + 49)/6 – 42.25 = 10.25

⇒ a^{2} + b^{2} = 221…(ii)

Solving (i) and (ii)

We get a = 10 and b = 11 or a = 11 and b = 10.

**Question 2: Find the coefficient of x ^{256} in (1- x)^{101}(x^{2} + x + 1)^{100}**

**Answer: **^{100}C_{85}

(1- x)^{101}(x^{2} + x + 1)^{100} = (1 – x)(1- x)^{100}(x^{2} + x + 1)^{100}

= (1 – x)(1 – x)^{100}(x^{2} + x + 1)^{100}

= (1 – x)(1 – x^{3})^{100}

= (1 – x^{3})^{100} – x(1-x^{3})^{100}

First term will not have x^{256}.

Coefficient of x^{256} is -1×coefficient of x^{255} in (1-x^{3})^{100}

= -1(-^{ 100}C_{85})

= ^{100}C_{85}

**Question 3: If **

**\(\begin{array}{l}\vec{a}\end{array} \)**

**,\(\begin{array}{l}\vec{b}\end{array} \), \(\begin{array}{l}\vec{c}\end{array} \) are three mutually perpendicular vectors equally inclined to \(\begin{array}{l}\vec{a}+\vec{b}+\vec{c}\end{array} \) at angle θ, then find the value of 36 cos**

^{2}2θ.

**Answer: **4

Similarly

⇒

From (i) we get

⇒ cos^{2}θ = 1/3

1 + cos 2θ = 2cos^{2}θ

36 cos^{2} 2θ = 36(2×1/3 – 1)^{2}

= 36/9

= 4

**Question 4: If f(x) = **

**\(\begin{array}{l}\left\{\begin{matrix} \sin x-e^{x} ,&x<0 \\ a+[-x], &0\leq x< 1 \\ 2x-b, & x\geq 1 \end{matrix}\right.\end{array} \)**

**is continuous in (-∞,1], find the value of a+b.**

**Answer: **a + b = 3

f(x) =

= -2

f(1^{–}) = a + (-1) = a – 1

f(1^{+}) = 2 – b

f(1^{–}) = f(1^{+})

⇒ a – 1 = 2 – b

⇒ a + b = 3

**Question 5: The number of solutions of tan ^{-1}√(x(x+1)) + sin^{-1}√(x^{2}+x+1) = π/2 is**

**Answer:**

Given tan^{-1}√(x(x+1)) + sin^{-1}√(x^{2}+x+1) = π/2

⇒ tan^{-1} √(x(x+1)) = π/2 – sin^{-1}√(x^{2}+x+1)

⇒ tan^{-1} √(x(x+1)) = cos^{-1}√(x^{2}+x+1)

⇒ cos^{-1} 1/√(x^{2}+x+1) = cos^{-1}√(x^{2}+x+1)

⇒ 1/√(x^{2}+x+1) = √(x^{2}+x+1)

⇒ (x^{2}+x+1) = 1

⇒ x^{2} + x = 0

⇒ x(x+1) =0

⇒ x = 0 or x = -1

Therefore, the number of solutions is 2.

**
**

**Question 6: There are 6 batsmen, 7 bowlers and 2 wickets keepers. Find the number of ways of forming a team of 11 players having at least 4 batsmen, at least 5 bowlers and at least 1 wicket keeper.**

a. 567

b. 525

c. 462

d. 777

**Answer: **(d)

Team Size =11 players

Case 1: 4 batsmen and 5 bowlers and 2 wicket-keepers

⇒ ^{6}C_{4 }× ^{7}C_{5} × ^{2}C_{2}

= (6×5/2) × (7×6/2) × 1

= 315

Case 2: 4 batsmen and 6 bowlers and 1 wickets keeper

⇒ ^{6}C_{4 }× ^{7}C_{6} × ^{2}C_{1}

= (6×5/2) ×7×2

= 210

Case 3: 5 batsmen and 5 bowlers and 1 wickets keeper

⇒ ^{6}C_{5 }× ^{7}C_{5} × ^{2}C_{1}

= 6×(7×6/2) ×2

= 252

So total number of ways = 315 + 210 + 252

= 777

Hence option d is the answer.

**Question 7: All the letters of the word EXAMINATION are arranged. Find the probability that ‘M’ is at the 4 ^{th} position.**

a. 2/11

b. 1/11

c. 4/11

d. 8/11

**Answer: **(b)

EXAMINATION

Letters: AAIINNE T O X M

Total possible words

n(S) = 11!/2!^{3}

Now when M is at 4^{th} position

n(E)=10!/(2!)^{3}

P (M is at 4^{th} position) = (n(E))/(n(S))

= 1/11

Hence option b is the answer.

**Question 8: Find the number of integral terms in the expansion of **

**\(\begin{array}{l}(4^{\frac{1}{4}} + 5^{\frac{1}{6}})^{120}\end{array} \)**

a. 1

b. 21

c. 20

d. 11

**Answer: **(b)

T_{(r+1) }= ^{120}C_{r} 4^{(120 – r)/4}5^{r/6}

= ^{120}C_{r }2^{(120 – r)/2}5^{r/6}

For integral terms, r should be even and multiple of 6

i.e., r should be a multiple of 6.

r can be 0,6, 12,…,120

Therefore, the total number of integral terms = 21

Hence option b is the answer.

**
\(\begin{array}{l}\lim_{x\to0}\left \{ 2-\cos x\sqrt{\cos 2x} \right \}^{\frac{x+2}{x^{2}}}=e^{\alpha }\end{array} \) (1
\(\begin{array}{l}f(x)=\begin{vmatrix} 1 & -x &2x+1 \\ -x& 1 & -x\\ 2x+1& -x &1 \end{vmatrix}\end{array} \)
\(\begin{array}{l}\sum_{p+q+r=6}^{}\frac{6!}{p!q!r!}(ab)^{p}(bc)^{q}(ca)^{r}\end{array} \)
\(\begin{array}{l}\sum_{p+q+r=6}^{}\frac{6!}{p!q!r!}a^{p+r}b^{p+q}c^{q+r}\end{array} \)
****Question 9: If **

**\(\begin{array}{l}\lim_{x\to0}\left \{ 2-\cos x\sqrt{\cos 2x} \right \}^{\frac{x+2}{x^{2}}}=e^{\alpha }\end{array} \)**

**, then α = ?**

**Answer: 3**

^{∞}form)

⇒ e^{(2+1)2/(1+1)}

= e^{α}

⇒ e^{3} = e^{α}

= α = 3

**Question 10: Evaluate ∫ _{-1}^{1} log (√(1 – x) + √(1 + x))dx**

**Answer: **2 ln(2) – (1 – π/2)

f(-x) = f(x)

⇒ f is even

∴ I = 2∫_{0}^{1} log ((1 – x) + √(1 + x))dx…(i)

We will solve it by integration by parts.

Let u = ln(√(1-x)+√(1+x)

I = 2 ln(2) – (1 – π/2)

**Question 11: The logical equivalence of the Boolean expression ~(p∧~q)∨(q∨~p) is**

**Answer: ≡ p → q**

~(p∧~q) ∨ (q∨~p)

≡ (~p∨q) ∨ (q∨~p)

≡ ~p∨q

≡ p → q

**Question 12: Consider the curve y ^{2} = 2x and point A(2, 2). If the normal at A intersects the curve again at point B and the tangent at A intersects the x-axis at C, then area of ∆ABC is**

**Answer: **25/2

Equation of tangent at A

yy_{1} = x+x_{1}

⇒2y = x+2

Tangent at A intersects the x-axis at C

C(-2, 0)

Equation of normal at A

y+ 2x = 6

Solving normal equation with y^{2 }= 2x, we get B(9/2,-3)

So, area will be ½ × AC×AB

= ½ ×2√5×√[(25/4) + 25]

= 25/2

**Question 13: If for vectors A and B, A.B = |A×B|, then |A-B| is**

a. √(A^{2} + B^{2}√(2AB))

b. √(A^{2} + B^{2 }– √(2AB))

c. √(A^{2} + B^{2} – √2 AB)

d. √(A^{2} + B^{2} + √2 AB)

**Answer: **(c)

A.B = |A×B|

⇒ cos θ = sin θ

⇒ tan θ = 1

⇒ θ = π/4

|A-B|^{2} = A^{2} + B^{2} – 2A.B

= A^{2} + B^{2} – 2AB cos π/4

= A^{2} + B^{2} – √2AB

⇒ |A – B| = √(A^{2}+ B^{2} – √2 AB)

Hence option c is the answer.

**Question 14: If the roots of the quadratic equation x ^{2 }+ 3^{1/4}x + 3^{1/2 }= 0 are α and β, then the value of α^{96}(α^{12} – 1) + β^{96}(β^{12} −1)**

a. 50⋅3^{24}

b. 51⋅3^{24}

c. 52⋅3^{24}

d. 104⋅3^{24}

**Answer: **(c)

x^{2} + 3^{1/2} = – 3^{1/4}x

⇒ x^{4} + 3 = -3^{1/2}x^{2}

⇒ x^{8} + 9 + 6x^{4} = 3x^{4}

⇒ x^{8} + 9 + 3x^{4} = 0

α^{8} = -9 – 3α^{4}

And α^{12} = -9α^{4} – 3α^{8}

= -9α^{4} -3(-9-3α^{4})

= 27

Similarly, β^{12} = 27

α^{96}(α^{12} – 1) + β^{96}(β^{12} – 1) = 27^{8}.26 + 27^{8}.26

= 52.3^{24}

Hence option c is the answer.

**Question 15: In ∆ABC, if AB = 5, ∠B=cos ^{-1} (3/5) and the radius of circumcircle of triangle is 5. Then the area of ∆ABC is**

a. 6+8√3

b. 3+4√3

c. 3+8√3

d. 6+4√3

**Answer: **(a)

cos B = ⅗

⇒ sinB = 4/5, R = 5

b = 2R sinB = 8

AB = c = 5

By cosine rule,

cos B = (a^{2}+c^{2}-b^{2})/2ac = 3/5

⇒ (a^{2}+25-64)/(2a(5))

= 3/5

⇒ a^{2}-39 = 6a

⇒ a^{2}-6a-39 = 0

⇒ a = ((6+8√3))/2

⇒ a = 3+4√3

∆ = abc/4R

= 6+8√3

Hence option a is the answer.

**Question 16: If the shortest distance between the lines **

**\(\begin{array}{l}r_{1}=\alpha\hat{i}+2\hat{j}+2\hat{k}+\lambda (\hat{i}-2\hat{j}+2\hat{k})\end{array} \)**

**, λ∈R, α>0 and\(\begin{array}{l}r_{2}=-4\hat{i}-\hat{k}+\mu (3\hat{i}-2\hat{j}-2\hat{k})\end{array} \) , µ∈R is 9, then the value of α is**

a. 2

b. 4

c. 6

d. √6

**Answer: **(c)

⇒ |(α + 4) ×8 + 16 + 12| = 108

⇒ (α + 4) ×8 = 80 (since α >0)

⇒ α = 6

Hence option c is the answer.

**Question 17: **Let y = mx+c, m>0 be the focal chord of y^{2} = −64x which is tangent to (x+10)^{2}+ y^{2 }= 4. Then the value of 4√2(m+c) is equal to

**Answer: **34

Focus of parabola is (−16,0)

So, −16m + c=0

⇒c = 16m …(i)

Now slope form of tangent to the circle is given by

y = m(x+10) ± 2√(1+m^{2})

So c = 10m ± 2√(1+m^{2}) …(ii)

So, from (i) and (ii)

16m = 10m ± 2√(1+m^{2})

⇒9m^{2} = 1 + m^{2}

⇒ m = 1/2√2

⇒ c = 16m

= 8/2√2

4√2 (m+c) = 34

**Question 18: A continuous differentiable function f(x) is increasing in (-∞, 3/2) and decreasing in (3/2, ∞). Then x = 3/2 is**

a. point of local maxima

b. point of local minima

c. point of inflection

d. None of these

**Answer: **(a)

A rough graph of f(x) can be drawn as

Thus x = 3/2 is a point of local maxima.

Hence, option a is the answer.

**Question 19: If z and ω are complex numbers such that |zω| = 1, arg(z) – arg(ω) = (3π )/2 then find arg **

**\(\begin{array}{l}\left ( \frac{1-2\bar{z}\omega }{1+3\bar{z}\omega } \right )\end{array} \)**

a. π/4

b. -π/4

c. 3π/4

d. -3π/4

**Answer:**

= (1-2i)/(1+3i)

= -½ – ½ i

arg( -½ – ½ i) = -3π/4

Hence option d is the answer.

**Question 20: If an invertible function f(x) is defined as f(x) = 3x – 2, g(x) is also an invertible function such that f ^{-1} (g^{-1} (x) )= x – 2, then g(x) is**

a. (x-8)/3

b. (x+8)/3

c. (x-3)/8

d. (x+3)/8

**Answer: (b)**

f^{-1} (g^{-1}(x) ) = x-2

⇒ f(x-2) = g^{-1 }(x)

3(x – 2) – 2 = g^{-1}(x)

3x – 8 = g^{-1} (x)

g^{-1} (x) = 3x – 8

or x = 3g(x) – 8

So g(x)=(x + 8)/3

Hence option b is the answer.

**Question 21: The probability of selecting integers a ∈ [-5,30], such that x ^{2 }+ 2(a+4)x – 5a + 64 >0 for all x∈ R is:**

**Answer: **1/2

x^{2}+ 2(a+4)x – (5a – 64)>0

D<0

Therefore 4(a+4)^{2 }+ 4(5a – 64) < 0

⇒ a^{2 }+ 13a – 48<0

⇒ (a + 16)(a – 3)<0

So, a ∈ (-16,3)

Total integers are 18

Probability = 18/36 = 1/2

**Question 22: If ∫ _{0}^{a }e^{x-[x]} dx = 10e – 9, then the value of a is (where [.] is greatest integer function)**

a. 9 + ln 2

b. 10 + ln 2

c. 10

d. 9

**Answer: **(b)

Let a = K + λ, 0 ≤ λ < 1

∫_{0}^{a} e^{{x}} dx = ∫_{0}^{K} e^{{x}} dx + ∫_{K}^{K+λ} e^{{x}} dx

= K ∫_{0}^{1} e^{x} dx + ∫_{0}^{λ} e^{x} dx

= K(e – 1)+(e^{λ }– 1)

=(Ke + e^{λ}) – (K+1)

Given, ∫_{0}^{a} e^{x-[x]} dx = 10e – 9

⇒ (Ke+e^{λ} )- (K+1) = 10e – 9

⇒ K = 10 and e^{λ }= 11 – 9 = 2

⇒ λ = ln 2

a = 10 + ln 2

Hence option b is the answer.

**Question 23: **

**\(\begin{array}{l}a_{ij}=\left\{\begin{matrix} 1 ,& i=j\\ -x ,&\left | i-j \right |=1 \\ 2x+1, & otherwise \end{matrix}\right.\end{array} \)**

**, A = [a**

_{ij}]_{3×3}. f(x) = Det(A). Then find the sum of local maximum and minimum values of fx.

a. 20/27

b. −20/27

c. 88/27

d. -88/27

**Answer: **(d)

= 4x^{3}– 4x^{2 }– 4x

f’(x) = 4(3x^{2}– 2x – 1)

f’(x) = 4(x-1)(x + ⅓)

f(1) = 4 – 4 – 4 = -4

f(-1/3) = (-4/27) – (4/9) + (4/3)

= (-4-12+36)/27

= 20/27

(20/27) – 4 = (20-108)/27 = -88/27

Hence option d is the answer.

**Question 24:Find the coefficient of a ^{3}b^{4}c^{5 }in (ab+bc+ca)^{6}. **

a. 60

b. 45

c. 40

d. 90

**Answer: **(a)

(ab + bc + ca)^{6} =

=

= For a^{3}b^{4}c^{5}, we need

p + r = 3

p + q = 4

q + r = 5

Solving we get, p = 1,q = 3,r = 2

Coefficient of a^{3}b^{4}c^{5} in (ab+bc+ca)^{6} is 6!/1!3!2! = 60.

Hence option a is the answer.

**Question 25: x (dy/dx)⋅ tan (y/x) = y tan (y/x) + x, y(1/2) = π/6. Area bounded by x = 0,x = 1/√2, y = y(x).**

**Answer: **(⅛)(π-1)

x (dy/dx). tan (y/x) = y tan (y/x) + x

(dy/dx) tan(y/x) = (y/x) tan(y/x) + 1

Let y = vx

dy/dx = v+x (dv/dx)

(v+x dv/dx) tanv = v tanv + 1

⇒ v+x dv/dx = v + cot v

⇒ dv/cotv = dx/x

tanv dv = dx/x

-log|cosv| = log|cx|

-cos v = cx

So y = x cos^{-1 }x

∫_{0}^{1/√2}x cos^{-1} x = (⅛)(π-1)

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