WBJEE 2022 Paper 1 (Mathematics) Question Paper with Solutions

Length of perpendicular from the centre of circle on tangent = radius of circle.

$r = \left| {{{3{x_1} + {y_1}} \over {\sqrt {{{(3)}^2} + {{(1)}^2}} }}} \right| \Rightarrow r = {5 \over {\sqrt {10} }} = \sqrt {{5 \over 2}}$

Equation of circle ${(x - b)^2} + {(y - k)^2} = {r^2}(h,b)$

Centre points

${(x - 2)^2} + {(y + 1)^2} = {\left( {\sqrt {{5 \over 2}} } \right)^2}$

${x^2} + 4 - 4x + {y^2} + 1 + 2y = {5 \over 2}$

Let the equation of other tangent which is passing through origin is $y = mx$

${x^2} + 4 - 4x + {m^2}{x^2} + 1 + 2mx = {5 \over 2}$ [$\because$ $y = mx$]

$\Rightarrow {x^2}(1 + {m^2}) + (2m - 4)x + {{10 - 5} \over 2} = 0$

$\Rightarrow (1 + {m^2}){x^2} + (2m - 4)x + {5 \over 2} = 0$

Compare with $a{x^2} + bx + c = 0$, where

$a = (1 + {m^2}),b = 2m - 4$ and $c = {5 \over 2}$

$\Rightarrow {(2m - 4)^2} - 4(1 + {m^2}) \times {5 \over 2} = 0$

$\Rightarrow 4{m^2} + 16 - 16m - 4 \times {5 \over 2} - 4{m^2} \times {5 \over 2} = 0$

$\Rightarrow 4{m^2} - 16m + 16 - 10 - 10{m^2} = 0$

$\Rightarrow - 6{m^2} - 16m + 6 = 0$

$\Rightarrow - 2(3{m^2} + 8m - 3) = 0$

$\Rightarrow (3m - 1)(m + 3) = 0$

$\Rightarrow 3m - 1 = 0,\,m + 3 = 0$

$m = {1 \over 3};m = - 3$

(i) $m = - 3,y = - 3x,3x + y = 0$

(ii) $m = {1 \over 3},y = {1 \over 3}x,3y - x = 0$

Other tangent $3y - x = 0$

$x - 3y = 0$

For continuous at x = 0,

$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$

$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{(x + b{x^2})}^{{1 \over 2}}} - {x^{{1 \over 2}}}} \over {b{x^{{1 \over 2}}}}}$

$= \mathop {\lim }\limits_{x \to {0^ + }} {{{x^{{1 \over 2}}}\left\{ {{{(1 + bx)}^{{1 \over 2}}} - 1} \right\}} \over {b{x^{{1 \over 2}}}}}$

$= \mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 + bx)}^{{1 \over 2}}} - 1} \over b}$

= 0 (when b $\ne$ 0)

$LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{\sin (a + 1)x + \sin x} \over x}\left( {{0 \over 0}} \right)$

$= \mathop {\lim }\limits_{x \to {0^ - }} {{(\cos (a + 1)x)(a + 1) + \cos x} \over 1}$

$= a + 1 + 1$

$= a + 2$

And $f(0) = c$ (given)

We know, for continuous function

$LHL = RHL = f(0)$

$\therefore$ $a + 2 = 0(b \ne 0) = c$

$\therefore$ c = 0, a = $-$2

and b = R $-$ {0}

Since $\overrightarrow c$ is a unit vector coplanar with $\overrightarrow a$ and $\overrightarrow b$

Let $\overrightarrow c$ = $x\overrightarrow a$ + $y\overrightarrow b$

$x(\widehat i + \widehat j - \widehat k) + y(\widehat i - \widehat j + \widehat k)$

$(x + y)\widehat i + (x - y)\widehat j - (x - y)\widehat k$

Since, $\overrightarrow c$ is perpendicular to $\overrightarrow a$.

$\therefore$ $x + y + x - y + x - y = 0$

$3x = y$

Since $|\overrightarrow c | = 1$

$\therefore$ ${(x + y)^2} + {(x - y)^2} + {(x - y)^2} = 1$

$\Rightarrow {(4x)^2} + {( - 2x)^2} + {( - 2x)^2} = 1$

$\Rightarrow 24{x^2} = 1 \Rightarrow x = {1 \over {2\sqrt 6 }}$

$\overrightarrow c = 4x\widehat i - 2x\widehat j + 2x\widehat k$

$= {2 \over {\sqrt 6 }}\widehat i - {1 \over {\sqrt 6 }}\widehat j + {1 \over {\sqrt 6 }}\widehat k$

i.e. $\overrightarrow c = {1 \over {\sqrt 6 }}(2\widehat i - \widehat j + \widehat k)$

$\overrightarrow a = {d_1}\widehat i + {d_2}\widehat j + {d_3}\widehat k$

Since, $\overrightarrow d$ is perpendicular to $\overrightarrow a$ and $\overrightarrow c$.

$\therefore$ $\overrightarrow d \,.\,\overrightarrow a = 0$

${d_1} + {d_2} - {d_3} = 0$

and $\overrightarrow d \,.\,\overrightarrow c = 0$

$2{d_1} - {d_2} + {d_3} = 0 \Rightarrow 3{d_1} = 0$

${d_1} = 0$

$\therefore$ ${d_2} = {d_3}$

Since, $|\overrightarrow d | = 1$

$\therefore$ $d_1^2 + d_2^2 + d_3^2 = 1$

${d_2} = {d_3} = {1 \over {\sqrt 2 }}$

Hence, $d = {1 \over {\sqrt 2 }}(\widehat j + \widehat k)$

The possible arrangements of n white and n black balls alternately in the following ways.

(i) WBWB ...... (ii) BWBW .......

Thus, the required number of ways

$n! \times n!+n! \times n!=2(n!)^2$

$gof:S \to U$

$gof$ is injective or (one-one) function.

So, $g(f{x_1}) = g(f{x_2})$

$f({x_1}) = f({x_2})$

$\therefore$ g is obviously injective.

Let $f(x) = x$ which is derivable in [0, 1].

Option A :

$\int_0^c {f(x)dx = (1 - c)f(c)}$

$\Rightarrow \int_0^c {x\,dx = (1 - c)\,.\,c}$

$\Rightarrow \left[ {{{{x^2}} \over 2}} \right]_0^c = (1 - c)\,c$

$\Rightarrow {{{c^2}} \over 2} = (1 - c)\,c$

$\therefore$ $c = 0$

or ${c \over 2} = 1 - c$

$\Rightarrow c = 2 - 2c$

$\Rightarrow 3c = 2$

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

$\therefore$ c = 0 does not belongs to (0, 1) but $c = {2 \over 3}$ belongs to (0, 1)

$\therefore$ Option A is correct.

Option B :

$\int_0^d {f(x)dx = (1 - d)f(d)}$

$\Rightarrow \int_0^d {x\,dx = (1 - d)\,.\,d}$

$\Rightarrow {{{d^2}} \over 2} = (1 - d)d$

$\therefore$ $d = 0$

or

${d \over 2} = 1 - d$

$\Rightarrow d = {2 \over 3}$ (which belongs to in between (0, 1))

$\therefore$ Option B is incorrect.

Option C :

$\int_0^c {f(x)dx}$

$= \int_0^c {x\,dx}$

$= \left[ {{{{x^2}} \over 2}} \right]_0^c$

$= {{{c^2}} \over 2}$

${{{c^2}} \over 2}$ exist all values of c between 0 and 1.

$\therefore$ Option C is incorrect.

Option D :

$\int_0^c {f(x) = dx}$

$= \int_0^c {x\,dx}$

$= \left[ {{{{x^2}} \over 2}} \right]_0^c$

$= {{{c^2}} \over 2}$

$\therefore$ $\int_0^c {f(x)\,dx}$ is not independent of c.

$\therefore$ Option D is incorrect.

$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x}\ln \sqrt {{{1 + x} \over {1 - x}}} } \right)$

Put $x = \cos 2\theta$

$\Rightarrow 2\theta = {\cos ^{ - 1}}(x)$

$\Rightarrow \theta = {1 \over 2}{\cos ^{ - 1}}(x)$

when $x \to 0$ then $\theta \to {\pi \over 4}$

$\therefore$ $\mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln \sqrt {{{1 + \cos 2\theta } \over {1 - \cos 2\theta }}} } \over {\cos 2\theta }}} \right)$

$= \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln \sqrt {{{2{{\cos }^2}\theta } \over {2{{\sin }^2}\theta }}} } \over {\cos 2\theta }}} \right)$

$= \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln {{({{\cot }^2}\theta )}^{{1 \over 2}}}} \over {\cos 2\theta }}} \right)$

$= \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln |\cot \theta |} \over {\cos 2\theta }}} \right)$

when $\theta \to {\pi \over 4}$ then $\cot \theta > 0$

$\therefore$ $|\cot \theta | = \cot \theta$

$= \mathop {\lim }\limits_{\theta \to {\pi \over 4}} {{\ln (\cot \theta )} \over {\cos 2\theta }}$ (${0 \over 0}$ form)

Applying L' Hospital Rule,

$= \mathop {\lim }\limits_{\theta \to {\pi \over 4}} {{{1 \over {\cot \theta }} \times ( - \cos e{c^2}\theta )} \over { - \sin 2\theta \times 2}}$

$= {{{1 \over {\cot {\pi \over 4}}} \times - \cos e{c^2}{\pi \over 4}} \over { - \sin {\pi \over 2} \times 2}}$

$= {{{1 \over 1} \times - {{(\sqrt 2 )}^2}} \over { - 1 \times 2}} = 0$

$= {2 \over 2} = 1$

$\cos y{{dy} \over {dx}} = {e^{x + \sin y}} + {x^2}{e^{\sin y}}$

On separating the variables, we get

$\cos y{e^{ - \sin y}}dy = ({e^x} + {x^2})dx$

On integrating both side,

$\int {\cos y{e^{ - \sin y}}dy = \int {({e^x} - {x^2})dx} }$

Let ${e^{ - \sin y}} = t$

$\therefore$ ${e^{ - \sin y}}( - \cos y)dy = dt$

$- \int {dt = {e^x} + {{{x^3}} \over 3}}$

$\Rightarrow - t = {e^x} + {{{x^3}} \over 3} + c$

$\Rightarrow - {e^{ - \sin y}} = {e^x} + {{{x^3}} \over 3} + c$

$\Rightarrow - {e^x} - {{{x^3}} \over 3} - {e^{ - \sin y}} = c$

${e^x} + {{{x^3}} \over 3} + {e^{ - \sin y}} = c$

Hence, $f(x) = \left( {{e^x} + {{{x^3}} \over 3}} \right)$

\left| {\matrix{ {(x - 2)} & {{{(x - 1)}^2}} & {{x^3}} \cr {(x - 1)} & {{x^2}} & {{{(x + 1)}^3}} \cr x & {{{(x + 1)}^2}} & {{{(x + 2)}^3}} \cr } } \right|

Applying ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1}$, then

\Delta x = \left| {\matrix{ {(x - 2)} & {{{(x - 1)}^2}} & {{x^3}} \cr 1 & {(2x - 1)} & {(3{x^2} + 3x + 1)} \cr 2 & {4x} & {6{x^2} + 12x + 8} \cr } } \right|

Applying ${R_3} \to {R_3} - 2{R_2}$, then

\left| {\matrix{ {(x - 2)} & {{{(x - 1)}^2}} & {{x^3}} \cr 1 & {(2x - 1)} & {3{x^2} + 3x + 1} \cr 0 & 2 & {6x + 6} \cr } } \right|

Expand w.r.t. C, then

$\Delta (x) = (x - 2)\{ (2x - 1)(6x - 6) - 2(3{x^2} + 3x + 1)\} - 1\{ {(x - 1)^2}(6x + 6) - 2{x^3}\}$

$\therefore$ Coefficient of x in $\Delta (x) = - 8 + 6 = - 2$

Given,

$f(x) = {a_0} + {a_1}|x| + {a_2}|x{|^2} + {a_3}|x{|^3}$

$\therefore$ f(x) = \left\{ {\matrix{ {{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3},} & {x \ge 0} \cr {{a_0} - {a_1}x + {a_2}{x^2} - {a_3}{x^3},} & {x < 0} \cr } } \right.

\Rightarrow f(x) = \left\{ {\matrix{ {{a_1} + 2{a_2}x + 3{a_3}{x^2},} & {x \ge 0} \cr { - {a_1} + 2{a_2}x - 3{a_3}{x^2},} & {x < 0} \cr } } \right.

$f(x)$ is differentiable at $x = 0$

$\therefore$ $\mathrm{L.H.D = R.H.D}$

$\Rightarrow - {a_1} + 0 + 0 = {a_1} + 0 + 0$

$\Rightarrow 2{a_1} = 0$

$\Rightarrow {a_1} = 0$

$\mathop {\lim }\limits_{ \in \to {0^ + }} \int_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1}$

$\Rightarrow \int_{{0^ + }}^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1}$

Differentiating both sides we get,

${{bx\cos 4x - a\sin 4x} \over {{x^2}}} = {{xa\cos 4x \times 4 - a\sin 4x} \over {{x^2}}}$

$\Rightarrow bx\cos 4x - a\sin 4x = 4ax\cos 4x - a\sin 4x$

Comparing coefficient of cos 4x and sin 4x both side we get,

b = 4a an $-$a = $-$a

$\therefore$ ${b \over a} = 4$

By checking options, when $a = {1 \over 4}$ and b = 1, then,

${b \over a} = 4$

$f(x) = {(x - 2)^{17}}{(x + 5)^{24}}$

$f'(x) = 17{(x - 2)^{16}}{(x + 5)^{24}} + {(x - 2)^{17}}(24){(x + 5)^{23}}$

$= {(x - 2)^{16}}{(x + 5)^{23}}\{ 17x + 85 + 24x - 48\}$

$=(x-2)^{16}(x+5)^{23}(41 x+37)=0$

$\therefore$ f(x) has neither a maximum nor a minimum at x = 2.

$V = a - k{t^2}$

${{dV} \over {dt}} = a - k{t^2}$

$\int_0^V {dV = \int_0^t {(a - k{t^2})dt} }$

$V = at - {{k{t^3}} \over 3}$ ..... (i)

When V is max $a = 0$

$\therefore$ $a - K{t^2} = 0$

$\Rightarrow t = \sqrt {{a \over k}}$

Using this is Eq. (i)

${V_{\max }} = a\sqrt {{a \over k}} - {k \over 3}\,.\,{a \over k}\sqrt {{a \over k}} = {2 \over 3}a\sqrt {{a \over k}}$

$z = x - iy$ ..... (i)

${(z)^{1/3}} = p + iq$ ..... (ii)

Value of $\left( {{x \over p} + {y \over q}} \right)/({p^2} + {q^2})$

Taking cube on both sides on equation,

${[{(z)^{1/3}}]^3} = {(p + iq)^3}$

$= {p^3} + {(iq)^3} + 3pqi(p + iq)$

$= {p^3} - i{q^3} + 3{p^2}qi - 3p{q^2}$

$z = {p^3} - 3p{q^2} - i({q^3} - 3{p^2}q)$

$z = p({p^2} - 3{q^2}) - iq({q^2} - 3{p^2})$ ...... (iii)

On comparing Eqs. (ii) and (iii), we get

$x = p({p^2} - 3{q^2}),y = q({q^2} - 3{p^2})$

Now,

${{\left( {{x \over p} + {y \over q}} \right)} \over {{p^2} + {q^2}}} = {{{{p({p^2} - 3{q^2})} \over p} + {{q({q^2} - 3{p^2})} \over q}} \over {{p^2} + {q^2}}}$

$= {{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$

$= {{ - 2({p^2} + {q^2})} \over {{p^2} + {q^2}}} = - 2$

We have

${a_n} = {({1^2} + {2^2} + {3^2}\, + \,...\, + \,{n^2})^n}$

and ${b_n} = {n^n}(n!)$

Also, written as

${a_n} = {\left[ {{{n(n + 1)(2n + 1)} \over 6}} \right]^n}$

and ${b_n} = {n^n}(n!)$

By putting the values of n = 1, 2, 3, 4, ...., we get

${a_n} > {b_n},\forall n$

Determinant of order 2 will be of the form

\left| {\matrix{ a & b \cr c & d \cr } } \right|

$\Delta = ad - bc$

The total number of ways of choosing a, b, c and d is

${2^4} = 16$

Now, $\Delta \ne 0$ if and only if either ad = 1

bc = 0 or ad = 0, bc = 1 for ad = 1 (a, d) = (1, 1)

For bc = 0 (b, c) = (0, 1), (1, 0), (0, 0)

So, three cases are there.

Similarly, for ad = 0 and bc = 1

There will be three cases.

$\therefore$ Number of favourable cases = 3 + 3 = 6

So, required probability $=\frac{6}{16}=\frac{3}{8}$

By given condition $2[\log 2b - \log 3c]$

$= \log a - \log 2b + \log 3c - \log a$

$\log 3c - \log 2b$

$\Rightarrow 3[\log 2b - \log 3c] = 0$

$2b = 3c$

Also, ${b^2} = ac$

$\therefore$ ${{9{c^2}} \over 4} = ac$

$\Rightarrow c = {{4a} \over 9}$

$b = {{3c} \over 2} = {3 \over 2}.{4 \over 9}a = {{2a} \over 3}$

Since, $a + b = a + {{2a} \over 3} = {{5a} \over 3} > c$

$b + c = {{2a} \over 3} + {{4a} \over 9} = {{10a} \over 9} > a$

$c + a = {{13a} \over 9} > b$

$\therefore$ a, b, c are the sides of triangle. Also, as a is the greatest side,

$\angle$A of $\Delta$ABC

$\cos A = {{{b^2} + {c^2} - {a^2}} \over {2bc}} = {{{{4{a^2}} \over 9} + {{16{a^2}} \over {81}} - {a^2}} \over {2\,.\,{2 \over 3}a\,.\,{4 \over 9}a}}$

$= {{ - 29} \over {48}} < 0$

$\therefore$ $\Delta$ABC is an obtuse angle.

$I = \int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{\sin x}}} \over {{{(\cos x)}^{\sin x}} + {{(\sin x)}^{\cos x}}}}dx}$ ..... (1)

Replace x with $\left( {{\pi \over 2} - x} \right)$,

$\therefore$ $I = \int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{\sin \left( {{\pi \over 2} - x} \right)}}} \over {{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{\sin \left( {{\pi \over 2} - x} \right)}} + {{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{\cos \left( {{\pi \over 2} - x} \right)}}}}dx}$

$I = \int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{\cos x}}} \over {{{(\sin x)}^{\cos x}} + {{(\cos x)}^{\sin x}}}}dx}$ ...... (2)

Adding 1 and 2 we get,

$2I = \int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{\cos x}} + {{(\cos x)}^{\sin x}}} \over {{{(\sin x)}^{\cos x}} + {{(\cos x)}^{\sin x}}}}dx}$

$= \int\limits_0^{{\pi \over 2}} {1\,dx}$

$= {\pi \over 2}$

$\Rightarrow I = {\pi \over 4}$

$y = {e^{{{\tan }^{ - 1}}x}}$

Differentiating both sides with respect to x, we get

$\Rightarrow {{dy} \over {dx}} = {e^{{{\tan }^{ - 1}}(x)}} \times {1 \over {1 + {x^2}}}$

$\Rightarrow {y_1}(1 + {x^2}) = {e^{{{\tan }^{ - 1}}x}}$

$\Rightarrow {y_1}(1 + {x^2}) = y$

Differentiating both side with respect to x, we get

$\Rightarrow {y_2}(1 + {x^2}) + {y_1}(2x) = {y_1}$

$\Rightarrow {y_2}(1 + {x^2}) + {y_1}(2x - 1) = 0$