WBJEE 2021 Paper 1 (Mathematics) Question Paper with Solutions

According to the question, we have


CC₁ = r + r₁ and

CC₂ = r + r₂

$\therefore$ CC₂ $-$ CC₁ = (r + r₂) $-$ (r + r₁) = r₂ $-$ r₁

$\Rightarrow$ Locus of centre is Hyperbola.

Given,

$\int\limits_{\log _e^2}^x {{{({e^x} - 1)}^{ - 1}}dx = \log _e^{{3 \over 2}}}$

$\Rightarrow \int\limits_{\log _e^2}^x {{1 \over {{e^x} - 1}}dx = \log _e^{{3 \over 2}}}$

$\Rightarrow \int\limits_{\log _e^2}^x {{{{e^{ - x}}} \over {{e^x}.{e^{ - x}} - {e^{ - x}}}}dx = \log _e^{{3 \over 2}}}$

$\Rightarrow \int\limits_{\log _e^2}^x {{{{e^{ - x}}} \over {1 - {e^{ - x}}}}dx = \log _e^{{3 \over 2}}}$

$\Rightarrow \left[ {{{\log }_e}|1 - {e^{ - x}}|} \right]_{\log _e^2}^x = \log _e^{{3 \over 2}}$

$\Rightarrow {\log _e}|1 - {e^{ - x}}| - {\log _e}|1 - {e^{ - \log _e^2}}| = \log _e^{{3 \over 2}}$

$\Rightarrow {\log _e}|1 - {e^{ - x}}| - {\log _e}\left| {1 - {1 \over 2}} \right| = \log _e^{{3 \over 2}}$

$\Rightarrow {\log _e}|1 - {e^{ - x}}| + \log _e^2 = \log _e^{{3 \over 2}}$

$\Rightarrow \log _e^2(1 - {e^{ - x}}) = \log _e^{{2 \over 3}}$

$\therefore$ $2(1 - {e^{ - x}}) = {3 \over 2}$

$\Rightarrow 1 - {e^{ - x}} = {3 \over 4}$

$\Rightarrow {e^{ - x}} = {1 \over 4}$

$\Rightarrow {e^x} = 4$

$\Rightarrow x = \log _e^4$

We have,

a($\alpha$ $\times$ $\beta$) + b($\beta$ $\times$ $\gamma$) + c($\gamma$ + $\alpha$) = 0

Clearly a($\alpha$ $\times$ $\beta$), ($\beta$ $\times$ $\gamma$) and ($\gamma$ + $\alpha$) are coplanar vector $\therefore$ $\alpha$, $\beta$ and $\gamma$ are also coplanar.

We have, $I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx}$

Let $f(x) = {{\sin x} \over x}$

f(x) is decreasing functions

$\Rightarrow f(\pi /3) < f(x) < f(\pi /4)$

$\Rightarrow {{3\sqrt 3 } \over {2\pi }} < f(x) < {{2\sqrt 2 } \over \pi }$

$\Rightarrow {{3\sqrt 3 } \over {2\pi }}\int_{\pi /4}^{\pi /3} {dx} < I < {{2\sqrt 2 } \over \pi }\int_{\pi /4}^{\pi /3} {dx}$

$\Rightarrow {{3\sqrt 3 } \over {2\pi }}\left( {{\pi \over 3} - {\pi \over 4}} \right) < I < {{2\sqrt 2 } \over \pi }(\pi /3 - \pi /4)$

$\Rightarrow {{\sqrt 3 } \over 8} < I < {{\sqrt 2 } \over 6}$

Given plane,

P₁ = lx + my = 0 and P₂ = z = 0

Plane through common line of P₁ and P₂

$\therefore$ P₃ = lx + my + nz = 0

angle between P₁ and P₃ = $\alpha$

$\therefore$ $\cos \alpha = {{{l^2} + {m^2}} \over {\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{l^2} + {m^2}} }}$

$\Rightarrow \cos \alpha = \sqrt {{{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}}$

$\Rightarrow {\cos ^2}\alpha = {{{l^2} + {m^2}} \over {{l^2} + {m^2} + {n^2}}}$

$\Rightarrow {\sec ^2}\alpha = {{{l^2} + {m^2} + {n^2}} \over {{l^2} + {m^2}}} = 1 + {{{n^2}} \over {{l^2} + {m^2}}}$

$\Rightarrow ({\sec ^2}\alpha - 1)({l^2} + {m^2}) = {n^2}$

$\Rightarrow {n^2} = {\tan ^2}\alpha ({l^2} + {m^2}) = n = \pm \sqrt {({l^2} + {m^2})} \tan \alpha$

$\therefore$ Equation becomes

$lx + my \pm z\tan \alpha \sqrt {{l^2} + {m^2}} = 0$

Given,

(0 1 2) M = (1 0 0) .... (i)

and (3 4 5) M = (0 1 0)

$\Rightarrow$ (6 8 10) M = (0 2 0) .... (ii)

On substracting Eq. (i) from Eq. (ii), we get

(6 8 10) M $-$ (0 1 2) M = (0 2 0) $-$ (1 0 0)

$\Rightarrow$ (6 7 8) M = ($-$1 2 0)

For x$\in$R, x $\ne$ $-$1

${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ..... + {x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i}\,.\,{x^i}}$

Here, coefficient of x¹⁷

$= {}^{2016}{C_{17}} + {}^{2015}{C_{16}} + {}^{2014}{C_{15}} + .... + {}^{1999}{C_0}$

$= {}^{2016}{C_{1999}} + {}^{2015}{C_{1999}} + {}^{2014}{C_{1999}} + ... + {}^{1999}{C_{1999}}$

($\because$ ${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$)

$= {}^{2017}{C_{2000}} = {{(2017)!} \over {(2000)!\,(2017 - 2000)!}}$

$= {{2017!} \over {(2000)!\,(17)!}}$

$\int_1^3 {{{|x - 1|} \over {|x - 2| + |x - 3|}}dx}$

| x $-$ 1 | critical point = 1

| x $-$ 2 | critical point = 2

| x $-$ 3 | critical point = 3

So, integration divided into two parts

(1) 1 to 2

and (2) 2 to 3

$\therefore$ $\int_1^2 {{{x - 1} \over { - (x - 2) - (x - 3)}}dx + \int_2^3 {{{x - 1} \over {(x - 2) - (x - 3)}}dx} }$

$= \int_1^2 {{{x - 1} \over { - 2x + 5}}dx + \int_2^3 {{{x - 1} \over 1}dx} }$

$= - {1 \over 2}\int_1^2 {{{ - 2x + 2} \over { - 2x + 5}}dx + \left[ {{{{x^2}} \over 2} - x} \right]_2^3}$

$= - {1 \over 2}\int_1^2 {{{ - 2x + 5 - 3} \over { - 2x + 5}}dx + \left( {{9 \over 2} - 3} \right) - (2 - 2)}$

$= - {1 \over 2}\left[ {\int_1^2 {{{ - 2x + 5} \over { - 2x + 5}}dx - \int_1^2 {{3 \over { - 2x + 5}}dx} } } \right] + {3 \over 2}$

$= - {1 \over 2}\left[ {\int_1^2 {dx + {3 \over 2}\left[ {\log | - 2x + 5|} \right]_1^2} } \right] + {3 \over 2}$

$= - {1 \over 2}\left[ {1 + {3 \over 2}\left[ {\log |1| - \log |3|} \right]} \right] + {3 \over 2}$

$= - {1 \over 2} + {3 \over 4}\log |3| + {3 \over 2}$

$= 1 + {3 \over 4}\log |3|$

Given, equation, 6^x + 8^x = 10^x

$\Rightarrow {\left( {{6 \over {10}}} \right)^x} + {\left( {{8 \over {10}}} \right)^x} = 1$

$\Rightarrow {\left( {{3 \over 5}} \right)^x} + {\left( {{4 \over 5}} \right)^x} = 1$

Let ${3 \over 5} = \sin \theta \Rightarrow {4 \over 5} = \cos \theta$

$\therefore$ ${(\sin \theta )^x} + {(\cos \theta )^x} = 1$

It is possible only x = 2

$\therefore$ given equation has exactly one real root.

Given, radius of circular track is 10 m.


and speed of man = 10 m/sec. i.e. $v = r{{d\theta } \over {dt}} = 10$

Now, $\tan \theta = {y \over r}$

$\Rightarrow y = r\tan \theta$

$\Rightarrow {{dy} \over {dt}} = r{\sec ^2}\theta \,.\,{{d\theta } \over {dt}}$

$= {\sec ^2}\theta \,.\,r{{d\theta } \over {dt}}$

$= {\sec ^2}60^\circ \,.\,(10)$

$= {(2)^2} \times 10$

$= 40$ m/sec.

We have,

Mean, np = 4

and variance, npq = 2

$\therefore$ ${{npq} \over {np}} = {2 \over 4} = {1 \over 2}$

$\Rightarrow q = {1 \over 2}$ $\therefore$ $\Rightarrow p = {1 \over 2}$ and n = 8

$\therefore$ $P(x = 2) = {}^8{C_2}{\left( {{1 \over 2}} \right)^2}{\left( {{1 \over 2}} \right)^6}$

$={}^8{C_2}{\left( {{1 \over 2}} \right)^8}$

$= {{8 \times 7} \over {2 \times 1}} \times {1 \over {256}} = {7 \over {64}}$

Given equation

${y^2} = 2{x^3}$

$\Rightarrow 2y{{dy} \over {dx}} = 6{x^2}$

$\Rightarrow {{dy} \over {dx}} = {{3{x^2}} \over 4}$

$\Rightarrow {\left( {{{dy} \over {dx}}} \right)_{(h,k)}} = {{3{h^2}} \over k}$

Slope of tangent is perpendicular to line 4x = 3y

$\therefore$ $\left( {{{3{h^2}} \over k}} \right)\left( {{4 \over 3}} \right) = - 1$

$\Rightarrow 4{h^2} = - k$ .... (i)

and ${k^2} = 2{h^3}$ ..... (ii)

From Eqs. (i) and (ii)

${(4{h^2})^2} = 2{h^3}$

$\Rightarrow 16{h^4} = 2{h^3}$

$\Rightarrow h = {1 \over 8}$

$\therefore$ $k = - 4{\left( {{1 \over 8}} \right)^2}$

$\Rightarrow k = - 4 \times {1 \over {64}} = {{ - 1} \over {64}}$

$\therefore$ $(h,k) = \left( {{1 \over 8},{{ - 1} \over {64}}} \right)$ only

Given,

$g(x) = \int_x^{2x} {{{f(t)} \over t}dt}$

and f is a continuous function and $f(2x) = f(x)$

As, f is a continuous function so we can apply Leibnitz Rule for differentiation.

$\therefore$ $g'(x) = {d \over {dx}}(2x)\,.\,{{f(2x)} \over {2x}} - {d \over {dx}}(x)\,.\,{{f(x)} \over x}$

$= 2\,.\,{{f(2x)} \over {2x}} - 1\,.\,{{f(x)} \over x}$

$= {{f(x)} \over x} - {{f(x)} \over x}$ [as f(2x) = f(x)]

$= 0$

$\therefore$ $g'(x) = 0 \Rightarrow g(x) =$ constant function

We know that the equation of an ellipse whose center is at origin is

${{{{(x - 0)}^2}} \over {{a^2}}} + {{{{(y - 0)}^2}} \over {{b^2}}} = 1$

Differentiating both side w.r.t. x, we get

${{2x} \over {{a^2}}} + {{2y.{{dy} \over {dx}}} \over {{b^2}}} = 0$

$\Rightarrow {{y{{dy} \over {dx}}} \over {{b^2}}} = - {x \over {{a^2}}}$

Now separating variable and constant term we get,

${{y.{{dy} \over {dx}}} \over x} = - {{{b^2}} \over {{a^2}}}$

Differentiating both side w.r.t. x, we get

${{x.{d \over {dx}}\left( {y.{{dy} \over {dx}}} \right) - \left( {y.{{dy} \over {dx}}} \right).{d \over {dx}}(x)} \over {{x^2}}} = 0$

$\Rightarrow {{x.\left( {y.{{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}}.{{dy} \over {dx}}} \right) - y.{{dy} \over {dx}}.1} \over {{x^2}}} = 0$

$\Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0$

$\Rightarrow xy.y'' + x.{(y')^2} - yy' = 0$

[${{{d^2}y} \over {d{x^2}}} = y''$ and ${{{dy} \over {dx}} = y'}$]

Given, equation of ray of light $x + \sqrt 3 y = \sqrt 3$


On putting y = 0, we get $x = \sqrt 3$

$\therefore$ Coordinates of A are $\left( {\sqrt 3 ,0} \right)$

Now, let B(0, 1) then B' = (0, $-$1)

$\therefore$ Equation of AB' is ${{y - 0} \over {0 + 1}} = {{x - \sqrt 3 } \over {\sqrt 3 - 0}}$

$\Rightarrow \sqrt 3 y = x - \sqrt 3$

Given, f₁(x) = x, f₂(x) = 2 + log_e x

Now, let h(x) = f₂(x) $-$ f₁(x)

= (2 + log_e x) $-$ x

= 2 + log_e x $-$ x

Here, h(0⁺) < 0, h(1) > 0, h(e) > 0 and h(e²) < 0 and value of h(x) for all x $\ge$ e² is negative.

$\therefore$ h(x) = 0 has two roots is (0, 1) and (e, e²)

Required area

$= \int\limits_0^1 {(2x - {x^2})dx}$

$= \left[ {2.{{{x^2}} \over 2} - {{{x^3}} \over 3}} \right]_0^1$

$= \left[ {{x^2} - {{{x^3}} \over 3}} \right]_0^1$

$= \left( {1 - {1 \over 3}} \right)$

$= {2 \over 3}$

Let, the equation of line passes through the origin is y = mx.

This line and x = 1 line cut at point B.

y = m . 1 = m

$\therefore$ Point B = (1, m)

Area of $\Delta OAB = {1 \over 2} \times OA \times AB$

$= {1 \over 2} \times 1 \times m$

Given that straight line cut the area two equal parts.

$\therefore$ 2 $\times$ Area of $\Delta OAB = {2 \over 3}$

$\Rightarrow 2 \times {1 \over 2} \times 1 \times m = {2 \over 3}$

$\Rightarrow m = {2 \over 3}$

$\therefore$ Equation of straight line,

$y = {2 \over 3}x$

Given,

${2 \over 3}{\log _b}a + {3 \over 5}{\log _c}b + {5 \over 2}{\log _a}c = 3$

$\Rightarrow {{2\log a} \over {3\log b}} + {{3\log b} \over {5\log c}} + {{5\log c} \over {2\log a}} = 3$

Since, $A.M = {{{2 \over 3}{{\log }_b}a + {3 \over 5}{{\log }_c}b + {5 \over 2}{{\log }_a}c} \over 3}$

$= {3 \over 3} = 1$

and $G.M = {\left( {{2 \over 3}{{\log }_b}a\,.\,{3 \over 5}{{\log }_c}b\,.\,{5 \over 2}{{\log }_a}c} \right)^{{1 \over 3}}}$

$= {2 \over 3}{{\log a} \over {\log b}} \times {3 \over 5}{{\log b} \over {\log c}} \times {5 \over 2}{{\log c} \over {\log a}} = 1$

$\therefore$ $A.M. = G.M.$

$\therefore$ ${{2\log a} \over {3\log b}} = 1 \Rightarrow {a^2} = {b^3}$

$\Rightarrow a = {({3^6})^{{1 \over 2}}} = {3^3} = 27$

Given A and B be two non-singular skew symmetric matrices.

Such that AB = BA

$\Rightarrow$ A^T = $-$A and B^T = $-$B

Since matrices are non-singular

$\therefore$ order of these matrices are even.

Now, A²B²(A^TB)^$-$1 (AB^$-$1)^T

= A²B²($-$ AB)^$-$1 (AB^$-$1)^T

= $-$ A²B²(B^$-$1 A^$-$1) (B^T)^$-$1 A^T

= $-$ A²(B²B^$-$1)A^$-$1 ($-$B)^$-$1 ($-$ A)

= $-$ A²B . A^$-$1 B^$-$1 A

= $-$ A²B(BA)^$-$1 A

= $-$A²B(AB)^$-$1 A ($\because$ AB = BA)

= $-$ A²BB^$-$1 A^$-$1 A

= $-$ A²(BB^$-$1) (A^$-$1 A)

= $-$ A²

Let $I = \int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx}$

$\Rightarrow I = \int {{{2\sin x\cos x} \over {{{(a + b\cos x)}^2}}}dx}$

Put $a + b\cos x = t$

$\Rightarrow \cos x = {{t - a} \over b} \Rightarrow - b\sin xdx = dt$

$I = - {2 \over b}\int {\left( {{{t - a} \over b}} \right){1 \over {{t^2}}}dt}$

$= - {2 \over {{b^2}}}\int {{{t - a} \over {{t^2}}}dt}$

$= - {2 \over {{b^2}}}\left[ {\int {{1 \over t}dt - \int {{a \over {{t^2}}}dt} } } \right]$

$= - {2 \over {{b^2}}}\int {\left[ {\log (t) - a{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right] + C}$

$= - {2 \over {{b^2}}}\left[ {\log \left| t \right| + {a \over t}} \right] + C$

$\therefore$ $I = - {2 \over {{b^2}}}\left[ {\log \left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + C$

$\Rightarrow \alpha = - {2 \over {{b^2}}}$

Let coordinates of A are ($\alpha$, $-$ $\alpha$) and coordinates of B are ($\beta$, $\beta$)

$\beta$ + $\alpha$ = 2h and $\beta$ $-$ $\alpha$ = 2k

Now, Area of $\Delta AOB = {1 \over 2} \times OA \times OB$

$= {1 \over 2}\sqrt {2{\alpha ^2}} \sqrt {2{\beta ^2}} = \left| {\alpha \beta } \right| = c$

$\Rightarrow {\alpha ^2}{\beta ^2} = {c^2}$

$\Rightarrow 16{\alpha ^2}{\beta ^2} = 16{c^2}$

$\Rightarrow {\{ {(\beta + \alpha )^2} - {(\beta - \alpha )^2}\} ^2} = 16{c^2}$

$\Rightarrow {(4{h^2} - 4{k^2})^2} = 16{c^2}$

$\Rightarrow {h^2} - {k^2} = {c^2}$

$\therefore$ Locus of the mid point P of MR is ${({x^2} - {y^2})^2} = {c^2}$

Given, word 'EQUATION'

Here, consonants - Q, T, N

and Vowel - E, Q, A, I, O

$\therefore$ Number of word = ${}^5{C_3}$ $\times$ 3! $\times$ ${}^3{C_2}$ $\times$ 3!

= 10 $\times$ 6 $\times$ 3 $\times$ 6

= 1080

Given, a, b, c are in G.P. and $\log \left( {{{5c} \over {2a}}} \right),\log \left( {{{7b} \over {5c}}} \right),\log \left( {{{2a} \over {7b}}} \right)$ and A.P.

$\therefore$ b² = ac

and

$2\log \left( {{{7b} \over {5c}}} \right) = \log \left( {{{5c} \over {2a}}} \right) + \log \left( {{{2a} \over {7b}}} \right)$

$\Rightarrow {{49{b^2}} \over {25{c^2}}} = {{5c} \over {7b}}$

$\Rightarrow {(7b)^3} = {(5c)^3}$

$\Rightarrow 7b = 5c \Rightarrow c = {7 \over 5}b$

and ${b^2} = ac$

$\Rightarrow {b^2} = a\left( {{7 \over 5}b} \right)$

$\Rightarrow b = {7 \over 5}a$

Sides are ${{5b} \over 7},b,{7 \over 5}b$

$\therefore$ a, b, c are the length of the sides of scalene triangle.

We have,

\left| {\matrix{ {{a^2} + 10} & {ab} & {ac} \cr {ab} & {{b^2} + 10} & {bc} \cr {ac} & {bc} & {{c^2} + 10} \cr } } \right|

= {1 \over {abc}}\left| {\matrix{ {a({a^2} + 10)} & {a{b^2}} & {a{c^2}} \cr {{a^2}b} & {b({b^2} + 0)} & {b{c^2}} \cr {{a^2}c} & {{b^2}c} & {c({c^2} + 10)} \cr } } \right|

Taking common a, b, c from R₁, R₂ and R₃ respectively

= {{abc} \over {abc}}\left| {\matrix{ {{a^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2}} & {{b^2} + 10} & {{c^2}} \cr {{a^2}} & {{b^2}} & {{c^2} + 10} \cr } } \right|

Applying C₁ $\to$ C₁ + C₂ + C₃, we get

= \left| {\matrix{ {{a^2} + {b^2} + {c^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 10} & {{b^2} + 10} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 0} & {{b^2}} & {{c^2} + 10} \cr } } \right|

= ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 1 & {{b^2} + 10} & {{c^2}} \cr 1 & {{b^2}} & {{c^2} + 10} \cr } } \right|

R₂ $\to$ R₂ $-$ R₁ and R₃ $\to$ R₃ $-$ R₁

= ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 0 & {10} & 0 \cr 0 & 0 & {10} \cr } } \right|

$= ({a^2} + {b^2} + {c^2} + 10)\,(100)$

$\because$ If is divisible by 100.

Let direction cosines of line is (l, l, l).

$\therefore$ l² + l² + l² = 1

$\Rightarrow$ 3 l² = 1 $\Rightarrow$ l = ${1 \over {\sqrt 3 }}$

$\therefore$ Direction cosines of line is $\left( {{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }}} \right)$

Now, equation of a line passing through P(2, $-$1, 2)

${{x - 2} \over {1/\sqrt 3 }} = {{y + 1} \over {1/\sqrt 3 }} = {{z - 2} \over {1/\sqrt 3 }} = k$ .... (i)

$\Rightarrow$ x = k + 2, y = k $-$ 1 and z = k + 2

Since line (i) meets the plane

2x + y + z = 9 at Q

$\therefore$ 2(k + 2) + (k $-$ 1) + (k + 2) = 9

$\Rightarrow$ 2k + 4 + k $-$ 1 + k + 2 = 9

$\Rightarrow$ 4k + 5 = 9

$\Rightarrow$ 4k = 4 $\Rightarrow$ k = 1

$\therefore$ Coordinates of Q are (3, 0, 3)

$\therefore$ $PQ = \sqrt {{{(3 - 2)}^2} + {{(0 + 1)}^2} + {{(3 - 2)}^2}}$

$= \sqrt {1 + 1 + 1} = \sqrt 3$ units.