JEE Main 2020 Question Paper with Solution (Jan 6th second shift)

 

National test agency (NTA) conducted paper 2 of JEE Main (joint entrance exam main) for B. Arch. and B. Planning on 6th January 2020 in two shifts. As per our analysis of 2nd shift paper, it was lengthy and of moderate to hard level. Most of the questions asked in JEE Main 6th January 2020 second shift were from Calculus. We have curated memory based questions of today’s paper and their solutions for you. These questions asked in the second shift of the 6th January JEE Main 2020 paper. 

The following are the questions and solutions of 6th January second JEE Main January 2020 

Question 1. Let a, b દ R and a>). If the tangent at the point(2,2) to the circle x^2+y^2=8 touches the parabola y^2-4a(x-b), then b-a is equal to:

Solution

Slope of tangent to the equation x^2+y^2=8 at point (2,2) is 

2x+2yy'=0, put x = 2, y = 2

y’ = -1 (slope=m) 

Equation of slope to the parabola y^2-4a(x-b) in slope form is y=mx-ma+\frac{a}{m}

Put m = -1 and x = 2, y = 2

We get b-a = 4

 

Question 2. If the total number of ways in which 8-digit number can be formed by using all the digits 0,1,2,3,4,5,7,9 such that no two even digits appear together in (5!)k, then k is equal to:

Solution:

Digit  0,1,2,3,4,5,7,9.

Odd = 1, 3, 5, 7, 9 

_O_O_O_O_O_

Five odd (O) places can be filled by digits  1, 3, 5, 7, 9 in 5! Ways

Now, case 1 when first place is with digit  2

5  place can be filled with the digit 0 and 4 (_) in ^5C_2 ways

Now, case 2 when first place is with digit  4

5  place can be filled with the digit 0 and 2 (_) in ^5C_2 ways

Total ways (5!)(2 x^5C_2)

k = 20 

 

Question 3. 

Let g(f(x))=\left\{\begin{array}{ll}{(|x-3|)^{2}-b(|x-3|)-2,} & {\text { if } x<-1} \\ {(3 x+4)^{2}-b(3 x+4)-2,} & {\text { if } x \geq-1}\end{array}\right.  and b is real constant. If gof is continuous at x = -1, then b is equal to 

 

Solution:

g(f(x))=\left\{\begin{array}{ll}{(|x-3|)^{2}-b(|x-3|)-2,} & {\text { if } x<-1} \\ {(3 x+4)^{2}-b(3 x+4)-2,} & {\text { if } x \geq-1}\end{array}\right.

For continuity at x = -1,

LHL of g(f(x)) = RHL of g(f(x))

(|x-3|)^{2}-b(|x-3|)-2=(3 x+4)^{2}-b(3 x+4)-2 \text { at } x=-1

b = 5

 

Question 4. If the line joining the points (-1,2,5) and (3,4,-10) intersects the xy-plane at the point(x,y,z), the \frac{y}{x} is equal to?

Solution:

\begin{array}{l}{\text { The equation of a line passing through two points with position }} \\ {\text { vectors } \vec{a} \otimes \vec{b} \text { is }} \\ {\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})} \\ {\text { Given the line passes through the points }} \\ {\Rightarrow \vec{a}=-1 \hat{\imath}+2 \hat{\jmath}+5 \hat{k} \quad | \Rightarrow \vec{b}=3 \hat{\imath}+4 \hat{\jmath}-10 \hat{k}}\end{array}

\begin{array}{l}{\vec{b}-\vec{a}=(3-(-1)) \hat{i}+(4-2) \hat{j}+(-10-5) \hat{k}} \\ {\vec{b}-\vec{a}=4 \hat{i}+2 \hat{j}-15 \hat{k}} \\ {| \vec{r}=-\hat{i}+2 \hat{j}+5 \hat{k}+\lambda(4 \hat{i}+2 \hat{j}-15 \hat{k})} \\ {\text { Let the coordinates of the point where the line crosses the }} \\ {\text { XY plane be }(x, y, 0)} \\ {\quad \quad \text { So, } \vec{r}=x \hat{\imath}+y \hat{\jmath}+0 \hat{k} \quad \ldots(2)}\end{array}

\begin{array}{l}{| x \hat{\imath}+y \hat{\jmath}+0 \hat{k}=-\hat{i}+2 \hat{j}+5 \hat{k}+\lambda(4 \hat{i}+2 \hat{j}-15 \hat{k})} \\ {x \hat{\imath}+y \hat{\jmath}+0 \hat{k}=-\hat{i}+2 \hat{j}+5 \hat{k}+\lambda(\hat{4} \hat{i}+2 \hat{j}-15 \hat{k})} \\ {\lambda=1 / 3} \\ {y / x=8}\end{array}

Question 5. If  \int e^{2 x}(\cos x+7 \sin x) d x=e^{2 x} g(x)+c, where c ia s constant of integration, then \mathrm{g}(0)+\mathrm{g}\left(\frac{\pi}{2}\right)  is equal to:

Solution:

Use the concept

\begin{array}{l}{\int e^{a x} \cos b x=\frac{e^{a x}}{a^{2}+b^{2}}(a \cos b x+b \cos b x)+c} \\ {\int e^{a x} \sin b x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+c} \\ {\int e^{2 x}(\cos x+7 \sin x) d x=e^{2 x}(3 \sin (x)-\cos (x))+C} \\ {g(x)=3 \sin x) d x=e^{2 x}(3 \sin (x)-\cos (x))+C} \\ {\int e^{2 x}(\cos x+7 \sin x) d x=e^{2 x}(3 \sin (x)-\cos (x))+C} \\ {g(x)=-1} \\ {g(\pi / 2)=-3} \\ {g(0)+g(\pi / 2)=2}\end{array}

Question 6. If for some c<0, the quadratic equation, 2 \mathrm{cx}^{2}-2(2 \mathrm{c}-1) \mathrm{x}+3 \mathrm{c}^{2}=0 has two distinct real roots, \frac{1}{\mathrm{a}} \text { and } \frac{1}{\mathrm{b}}, then the value of the determinant \left[\begin{array}{ccc}{1+a} & {1} & {1} \\ {1} & {1+b} & {1} \\ {1} & {1} & {1+c}\end{array}\right] is:

 

Solution:

 

For two distinct real roots 

 As Sum of the roots = - \frac{B}{A} 

So \frac{1}{a}+\frac{1}{b}=2-\frac{1}{c}

And the product of the roots = \frac{C}{A}

 So \frac{1}{a}*\frac{1}{c}=\frac{3}{2}c\Rightarrow \frac{2}{3}=abc

And

 =abc[1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]=abc[1+2-\frac{1}{c} +\frac{1}{c}]= 3abc= 3*\frac{2}{3}=2

 

Question 7.

If the volume of parallelepiped whose coterminous edges are 

\begin{array}{l}{\vec{a}=\hat{i}+\hat{j}+2 \hat{k}} \\ {\vec{b}=2 \hat{i}+\lambda \hat{j}+\hat{k}} \\ {\vec{c}=2 \hat{i}+2 \hat{j}+\lambda \hat{k}}\end{array}

is 35m^{3} \text { then the value of } \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}-\vec{c} \cdot \vec{a} \text { is }  

  1. -10

  2. 12

  3. 22

  4. -14

 

Solution:


 

Question 8. Which of the following is not equivalent to \sim \mathrm{p} \Lambda \mathrm{q}?

\begin{array}{ll}{(a) \sim(q \rightarrow p)} & {(b) \sim p \wedge(\sim p \rightarrow q)} \\ {(c) \sim(p \vee \sim q)} & {(d) \sim p \rightarrow \sim q}\end{array}

Solution: 

Draw truth table

      

 

Correct option (D)

 

Question 9. If x = x(y) is the solution of the differential equation, \mathrm{ydx}-\left(\mathrm{x}+2 \mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}=0, with \mathrm{x}(-\pi)=\pi^{2}, then x is equal to:

Solution:

\begin{aligned} \text { yd } x-\left(x+2 y^{2}\right) d y &=0 \\ \frac{y d x-x d y}{y^{2}} &=2 d y \\ \int d\left(\frac{x}{y}\right) &=\int 2 d y \\ \frac{x}{y}=2 & \\ x=& 2 \pi^{2}+C \\ C=&-\pi^{2}+C \\ C &=-\pi^{2} \\ x &=2 y^{2}-\pi^{2} \end{aligned}

 

Question 10. If A={1,2,3,4}, then the number of functions on the set A, which are not one-one, is:

  1. 240

  2. 248

  3. 232

  4. 256

Solution:

Total number of functions = n^m = 4^4= 256

Total number of one-one function ^np_m= 4!=24

 the number of functions on the set A, which are not one-one, is 256 - 24 = 232

 

Question 11. The function f(x)=e^{x+1}(4x^2-16x+11):

  1. Decreasing in 

Solution

y=f(x)=e^{x+1}(4x^2-16x+11)

So

 \frac{dy}{dx}= e^{x+1}(8x-16)+ e^{x+1}(4x^2-16x+11)=0\\ \Rightarrow 4x^2-16x+11+8x-16=0\\ \Rightarrow 4x^2-8x-5=0\\ \Rightarrow 2x(2x-5)+1(2x-5)=0\\ x=-\frac{1}{2} , \frac{5}{2}

So 

\frac{d^2y}{dx^2}= e^{x+1}(8 )+e^{x+1}(8x-16)+e^{x+1}(8x-16)+ e^{x+1}(4x^2-16x+11)=0\\ \Rightarrow e^{x+1}(8+8x-16+8x-16+4x^2-16x+11 )=0\\ \Rightarrow 4x^2-13=0\\ \Rightarrow at \ x=-\frac{1}{2}\Rightarrow \frac{d^2y}{dx^2} <0 \ \ So \ it \ is \ maxima \\ \Rightarrow at \ x= \frac{5}{2}\Rightarrow \frac{d^2y}{dx^2} >0 \ \ So \ it \ is \ minima \\ \\

From the options, we can see that,

from  the f(x) must be increased to reach maxima.

and from ( \frac{5}{2},\infty ) it must begin increasing since at x= \frac{5}{2}  it is minima.

 

Question 12. A ray of light is projected from the origin at angle of 60o with the positive direction of x - axis towards the line, y = 2, which gets reflected from the point (∝, 2). Then the distance of the reflected ray of light from the point (2,2) is:

  1. 3-\sqrt{3}
     

  2. 1 -\frac{1}{\sqrt{3}}
     

  3. \sqrt{3}-1
     

  4. 2(1-\frac{1}{\sqrt{3}})
     

Solution

Distance PB2(1-\frac{1}{\sqrt{3}})

 

Question 13.
\begin{array}{l}{\text { Let } \mathrm{A}=\left[\begin{array}{cc}{x} & {2 y} \\ {-1} & {y}\end{array}\right] ; \mathrm{x}, \text { y } \varepsilon \mathrm{R}} \\ {\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc}{1} & {0} \\ {0} & {\propto}\end{array}\right](\propto \varepsilon \mathrm{R}), \text { then } \propto+\mathrm{y}^{2} \text { is equal to : }}\end{array}
 

Solution-

 A=\left[\begin{array}{cc}{x} & {2 y} \\ {-1} & {y}\end{array}\right]

and A^{T}=\left[\begin{array}{cc}{x} & {-1} \\ {2 y} & {y}\end{array}\right]

So , \begin{aligned} A A^{T} &=\left[\begin{array}{cc}{x} & {2 y} \\ {-1} & {y}\end{array}\right]\left[\begin{array}{cc}{x} & {-1} \\ {2 y} & {y}\end{array}\right] \\ \Rightarrow A A^{T}=\left[\begin{array}{cc}{x^{2}+4 y^{2}} & {-x+2 y^{2}} \\ {-x+2 y^{2}} & {1+y^{2}}\end{array}\right]=\left[\begin{array}{cc}{1} & {0} \\ {0} & {\alpha}\end{array}\right] \end{aligned}

This gives us

\begin{aligned} x &=2 y^{2} \ldots(1) \\ x^{2}+4 y^{2} &=1 \Rightarrow x^{2}+4 y^{2}-1=0 \ldots(2) \\ \text { from }(1) & \text { and }(2) 4 y^{4}+4 y^{2}-1=0 \\ \Rightarrow y^{2}=& \frac{1}{2} \Rightarrow y=\frac{1}{\sqrt{2}} \\ \text { And } 1+y^{2}=& \alpha \Rightarrow 1+\frac{1}{2}=\alpha=\frac{3}{2} \\ \Rightarrow \alpha+y^{2} &=\frac{3}{2}+\frac{1}{2}=2 \end{aligned}

 

Question 14. If 10sin4∝ + 15cos4∝ = 6, , then 27cosec6∝ + 8sec6∝ is equal to:

\text { Question } 14 . \text { lf } 10 \sin 4 \alpha+15 \cos 4 \alpha=6, \quad \text { a } \varepsilon\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \text { then } 27 \csc 6 \alpha+8 \sec 6 \alpha \text { is equal to: }

\begin{aligned} & 10 \sin ^{4} \alpha+15 \cos ^{4} \alpha=10 \sin ^{4} \alpha+15\left(1-\sin ^{2} \alpha\right) \\=& 10 \sin ^{4} \alpha+15\left(1+\sin ^{4} \alpha-2 \sin ^{2} \alpha\right) \\=& 25 \sin ^{4} \alpha+15-30 \sin ^{2} \alpha \end{aligned}

so given 25sin^4\alpha+15-30sin^2\alpha=6; 25 \sin ^{4} \alpha+15-30 \sin ^{2} \alpha=6

So let sin^2\alpha =t; \sin ^{2} \alpha=t

So

 \begin{aligned} & 25 t^{2}-30 t+9=0 \\ \Rightarrow & 25 t^{2}-15 t-15 t+9=0 \\ \Rightarrow & 5 t(5 t-3)-3(5 t-3) \\ \Rightarrow & t=\frac{3}{5}=\sin ^{2} \alpha \Rightarrow \cos ^{2} \alpha=1-\sin ^{2} \alpha=1-\frac{3}{5}=\frac{2}{5} \\ \Rightarrow & \csc ^{2} \alpha=\frac{5}{3} \text { and } \sec ^{2} \alpha=\frac{5}{2} \end{aligned}

So 

27 \csc ^{6} \alpha+8 \sec ^{6} \alpha=27 *\left(\frac{5}{3}\right)^{3}+8 *\left(\frac{5}{2}\right)^{3}=5^{3}+5^{3}=250

Question 15. The area (in sq.units) of the region is:

\text { Question 15. The area (in sq.units) of the region }\left\{(\mathrm{x}-\mathrm{y}): \frac{1}{2} \leq \mathrm{y} \leq \sin \mathrm{x}, 0 \leq \mathrm{x} \leq \pi\right\}_{\mathrm{is:}}

  1. ; 3-\frac{2 \pi}{3}

  2. ; \sqrt{3}-\frac{\pi}{6}

  3. ; 3-\frac{\pi}{3}

  4. ; \sqrt{3}-\frac{\pi}{3}

Solution:

=

 

Question 16. The sum of the values of x satisfying the equation, ,, is:

Total sum = 16 + 9 + 1 =26


 

Question 17. Let Sn denote the sum of the first n terms of an A.P, a1, a2, a3,......an. If a5 + a9 = 1 and S9 = 6, then which of the following is not true?

  1. a6 + a8 = 1

  2. S6 = 19/2

  3. S13 = 13/2

  4. a13 = 0

 

Solution:

 

 

Question 18. If , then is equal to:

  1. 9

  2. 3

  3. 6

  4. 2

 

Solution:

(1+x)n=nC0+nC1x+nCx2+..........................

 

 

We get 

From here


 

Question 19. is equal to:

  1. -2 loge 2

  2. 1

  3. 1 - loge 2

  4. -loge 2

 

Solution:

= -loge 2


 

Question 20. Let a function, f:(-1,3)⟶R be defined as f(x) = min{x[x], |x[x]-2| + 2}, where {x} denotes the greatest integer ≤ x, then f is:

 

Solution:

 

Question 21. Let the ellipse, pass through the point (2,3) and have eccentricity equal to . Then, the equation of the normal to the ellipse at (2,3) is:

 

  1. 2y - x = 4

  2. 2x - y = 1

  3. 3x - 2y = 0

  4. 3x - y =3


 

Solution:

Only B option has slope 2


 

Question 22. A bag contains 6 red balls and 10 green balls. 3 balls are drawn from it one by one randomly without replacement. If the third drawn ball is red, then the probability that the first 2 drawn balls are green is:

  1. 3/7

  2. 9/49

  3. 9/56

 

Solution: 

 

Question 23. If a variable plane in 3-D space moves in such a way that the sum of its reciprocal of intercepts on the x and y-axis exceeds, the reciprocal of its intercept on the z-axis by 2, then all such planes will pass through the point:
 

 

Solution: Correct answer is A option

 

Question 24. Let the data 4, 10, x, y, 27 be in increasing order. If the median of data is 18 and its mean deviation about mean is 7.6, then the mean of this data is:

  1. 17

  2. 16

  3. 16.5

  4. 15.5

 

Solution: 

X will be 18 and after solving y will be 18.8 and mean will be 15.5

Correct option D part

 

Question 25. If m be the latest value of  attained at z = zo, then the ordered pair (|zo|, m) is equal to:

Solution: 

Correct option is 1st option

 

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