WBJEE 2018 Paper 1 (Mathematics) Question Paper with Solutions

We have, {S_r} = \left| {\matrix{ {2r} & x & {n(n + 1)} \cr {6{r^2} - 1} & y & {{n^2}(2n + 3)} \cr {4{r^3} - 2nr} & z & {{n^3}(n + 1)} \cr } } \right|

\Rightarrow \sum\limits_{r = 1}^n {{S_r}} \left| {\matrix{ {2\sum\limits_{r = 1}^n r } & x & {n(n + 1)} \cr {\sum\limits_{r = 1}^n {(6{r^2} - 1)} } & y & {{n^2}(2n + 3)} \cr {\sum\limits_{r = 1}^n {(4{r^3} - 2nr)} } & z & {{n^3}(n + 1)} \cr } } \right|

= \left| {\matrix{ {n(n + 1)} & x & {{n^3}(n + 1)} \cr {{n^2}(2n + 3)} & y & {{n^2}(2n + 3)} \cr {{n^3}(n + 1)} & z & {{n^3}(n + 1)} \cr } } \right|

$= 0$ ($\because$ C₁ and C₃ are identical)

Hence, $\sum\limits_{r = 1}^n {{S_r}}$ is independent of x, y, z and n.

We have, f : [a, b] $\to$ R be differentiable on [a, b] and k $\in$ R, also f(a) = 0 = f(b) and J(x) = f'(x) + kf(x)

Let g(x) = kxf(x) which is continuous in [a, b] and differentiable in (a, b) such that

g(a) = 0 = g(b)

Then, for every c $\in$(a, b), g' (c) = 0 (by Rolle's theorem)

Now, g'(x) = kf(x) + kxf' (x)

$\Rightarrow$ g'(c) = kf(c) + kcf'(c)

$\Rightarrow$ kf(c) + kcf'(c) = 0

$\Rightarrow$ f(x) = 0, for every x = c $\in$ (a, b)

$\therefore$ J(x) = 0 has atleast one root in (a, b).

Let g(x) = e^$-$x f(x)

such that g(a) = 0, g(b) = 0 and g(x) is continuous and differentiable.

Then, for at least one value of c $\in$ (a, b) such that g'(c) = 0

Now, g'(x) = e^$-$x f'(x) + ($-$ e^$-$x) f(x)

$\Rightarrow$ g'(c) = e^$-$c f'(c) + (e^$-$c) f(c) = 0

$\Rightarrow$ e^$-$c f'(c) = e^$-$e f(c) $\Rightarrow$ f'(c) = f(c)

We have,

$(1 + {x^2}){{dy} \over {dx}} + 2xy - 4{x^2} = 0$

$\Rightarrow {{dy} \over {dx}} + \left( {{{2x} \over {1 + {x^2}}}} \right)y = {{4{x^2}} \over {1 + {x^2}}}$

Here, $IF = {e^{\int {{{2x} \over {1 + {x^2}}}} }}{e^{\log (1 + {x^2})}} = 1 + {x^2}$

$\therefore$ $y(1 + {x^2}) = \int {(1 + {x^2}) \times {{4{x^2}} \over {(1 + {x^2})}}dx + C}$

$\Rightarrow y(1 + {x^2}) = \int {4{x^2}dx + C}$

$\Rightarrow y(1 + {x^2}) = {{4{x^3}} \over 3} + C$

$\Rightarrow y(1 + {x^2}) = {{4{x^3}} \over 3} - 1$ [y(0) = $-$ 1]

$\Rightarrow y = {{4{x^3}} \over {3(1 + {x^2})}} - {1 \over {1 + {x^2}}}$

$\therefore$ $y(1) = {4 \over 6} - {1 \over 2} = {1 \over 6}$

We have, $f(x) = \sqrt {{{1 - |x|} \over {2 - |x|}}}$

f(x) is defined for all x satisfying

${{1 - |x|} \over {2 - |x|}}\, \ge \,0 \Rightarrow {{|x| - 1} \over {|x| - 2}}\, \ge 0$

$\Rightarrow \,|x|\, \le 1$ or $|x|\, > 2$

$\Rightarrow x \in [ - 1,1]$ or $x \in ( - \infty , - 2) \cup (2,\infty )$

$\Rightarrow x \in ( - \infty , - 2) \cup [ - 1,1] \cup (2,\infty )$

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

Since, ${{\sin x} \over x}$ is a decreasing function.

$\therefore$ ${\pi \over {12}} \times {{\sin \pi /3} \over {\pi /3}} \le I \le {\pi \over {12}} \times {{\sin \pi /4} \over {\pi /4}}$

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

We know that, $\sin 30^\circ = {1 \over 2} = 0.5$

In 1st quadrant sin x is increasing function.

$\therefore$ $\sin 31^\circ > \sin 30^\circ$

$\Rightarrow \sin 31^\circ > 0.5$

We have, ${f_1}(x) = {e^x}$

${f_2}(x) = {e^{{f_1}(x)}}$

${f_{n + 1}}(x) = {e^{{f_n}(x)}}$

Now, ${f_n}(x) = {e^{{f_{n - 1}}(x)}}$

On taking log both sides, we get

$\log \left\{ {{f_n}(x)} \right\} = {f_{n - 1}}(x)\log e$

$\Rightarrow {d \over {dx}}\log \left( {{f_n}(x)} \right) = {d \over {dx}}{f_{n - 1}}(x)$ ($\because$ log e = 1)

$\Rightarrow {1 \over {{f_n}(x)}}{d \over {dx}}{f_n}(x) = {f_{n - 1}}'(x)$

$\Rightarrow {d \over {dx}}{f_n}(x) = {f_n}(x){f_{n - 1}}'(x)$ .... (i)

Now, ${f_1}'(x) = {e^x}$

and ${f_2}(x) = {e^{{f_1}(x)}}$

$\Rightarrow \log {f_2}(x) = {f_1}(x)\log e = {f_1}(x)$

$\Rightarrow {1 \over {{f_2}(x)}}.{f_2}'(x) = {f_1}'(x)$

$\Rightarrow {f_2}'(x) = {f_2}(x).{f_1}'(x)$

$= {f_2}(x).{e^x}$ ($\because$ ${f_1}'(x) = {e^x}$)

$= {f_2}(x).{f_1}(x)$ [$\because$ ${e^x} = {f_1}(x)$] ....(ii)

From Eq. (i),

${d \over {dx}}{f_n}(x) = {f_n}(x).{f_{n - 1}}(x)....{f_1}(x)$ [using Eq. (ii)]

We have,

$\int {f(x)} \sin x\cos xdx = {1 \over {2({b^2} - {a^2})}}\log (f(x)) + c$

$\Rightarrow f(x)\sin x\cos x = {1 \over {2({b^2} - {a^2})}}.{1 \over {f(x)}}.f'(x)$

$\Rightarrow f(x)\sin 2x = {1 \over {{b^2} - {a^2}}}.{{f'(x)} \over {f(x)}}$

$\Rightarrow \sin 2x = {1 \over {{b^2} - {a^2}}}{{f'(x)} \over {{{(f(x))}^2}}}$

$\Rightarrow \int {\sin 2xdx = {1 \over {{b^2} - {a^2}}}\int {{{f'(x)} \over {{{(f(x))}^2}}}} dx}$

$\Rightarrow {{ - \cos 2x} \over 2} = {1 \over {{b^2} - {a^2}}}.\left( {{{ - 1} \over {f(x)}}} \right)$

$\Rightarrow {{\cos 2x({b^2} - {a^2})} \over 2} = {1 \over {f(x)}}$

$\Rightarrow f(x) = {2 \over {({b^2} - {a^2})\cos 2x}}$

We have,

$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left\{ {{{\sec }^2}{\pi \over {4n}} + {{\sec }^2}{{2\pi } \over {4n}} + ... + {{\sec }^2}{{n\pi } \over {4n}}} \right\}$

$= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{{\sec }^2}\left( {{{r\pi } \over {4n}}} \right) = \int_0^1 {{{\sec }^2}\left( {{{\pi x} \over 4}} \right)} }$

$= {4 \over \pi }\left[ {\tan \left( {{{\pi x} \over 4}} \right)} \right]_0^1$

$= {4 \over \pi } \times 1 = {4 \over \pi }$

We have,

A = \left| {\matrix{ { - 1} & 7 & 0 \cr 2 & 1 & { - 3} \cr 3 & 4 & 1 \cr } } \right|

$= - 1(1 + 12) - 7(2 + 9)$

$= - 13 - 77 = - 90$ ... (i)

And let B = \left| {\matrix{ {13} & { - 11} & 5 \cr { - 7} & { - 1} & {25} \cr { - 21} & { - 3} & { - 15} \cr } } \right|

= 3 \times 5\left| {\matrix{ {13} & { - 11} & 1 \cr { - 7} & { - 1} & 5 \cr { - 7} & { - 1} & { - 1} \cr } } \right|

= 15\left| {\matrix{ {13} & { - 11} & 1 \cr { - 7} & { - 1} & 5 \cr 0 & 0 & { - 6} \cr } } \right|({R_3} \to {R_3} - {R_2})

$= 15\{ 0 - 0 - 6( - 13 - 77)\}$

$= 15\{ ( - 6)( - 90) = 90 \times 90$ ... (ii)

From Eqs. (i) and (ii),

$B = {A^2}$

We have equations

${x^2} + {b_1}x + {c_1} = 0$

${D_1} = b_1^2 - 4{c_1}$

and ${x^2} + {b_2}x + {c_2} = 0$

${D_2} = b_2^2 - 4{c_2}$

Now, ${D_1} + {D_2} = b_1^2 + b_2^2 - 4({c_1} + {c_2})$

$b_1^2 + b_2^2 - 2{b_1}{b_2}$ [$\because$ ${b_1}{b_2} = 2({c_1} + {c_2})$]

$= {({b_1} - {b_2})^2} \ge 0$

$\Rightarrow$ At least one of D₁ and D₂ are non-negative real roots.

${(100)^{100}} - 1 = {(1 + 100)^{100}} - 1$

$= (1 + {}^{100}{C_1}.100 + {}^{100}{C_2}{.100^2}........) - 1$

$= {}^{100}{C_1}.100 + {}^{100}{C_2}.{(100)^2} + {}^{100}{C_3}.{(100)^3} + ........ + {}^{100}{C_{100}}.{(100)^{100}}$

$= {10^4}(1 + {}^{100}{C_2} + {}^{100}{C_3}{10^2} + ..... + {}^{100}{C_{100}}{(100)^{98}}$

$= {10^4}$ (1 + an integer multiple of 10)

We know that, if z₁, z₂ and z₃ are the vertices of an equilateral triangle. Then,

$z_1^2 + z_2^2 + z_3^2 - {z_1}{z_2} - {z_2}{z_3} - {z_3}{z_1} = 0$ .... (i)

Now, but we have

${{{z_1}} \over {{z_2}}} + {{{z_2}} \over {{z_1}}} = 1$

$\Rightarrow z_1^2 + z_2^2 = {z_1}{z_2}$

$\Rightarrow z_1^2 + z_2^2 - {z_1}{z_2} = 0$

Here, ${z_3} = 0$

Hence, given points form an equilateral triangle.

We have,

$I = \int_{1/2014}^{2014} {{{{{\tan }^{ - 1}}x} \over x}} dx$ ....(i)

Let $x = {1 \over t}$

$\Rightarrow dx = {{ - 1} \over {{t^2}}}dt$

Now, $I = \int_{2014}^{1/2014} {{{{{\tan }^{ - 1}}(1/t)} \over {1/t}}} \left( {{{ - 1} \over {{t^2}}}dt} \right)$

$= \int_{1/2014}^{2014} {{{{{\cot }^{ - 1}}t} \over t}dt}$

$= \int_{1/2014}^{2014} {{{{{\cot }^{ - 1}}x} \over x}dx}$ .... (ii)

On adding Eqs. (i) and (ii), we get

$2I = \int_{1/2014}^{2014} {{{\pi /2} \over x}dx = {\pi \over 2}(\log x)_{1/2014}^{2014}}$

$= {\pi \over 2}(\log 2014 - \log 1/2014)$

$\Rightarrow I = {\pi \over 4}\left( {\log 2014 - \log {1 \over {2014}}} \right)$

$= {\pi \over 4}(\log 2014 + \log 2014)$

$= {\pi \over 4}(2\log 2014) = {\pi \over 2}\log 2014$

We have,

$f(x) = 3{x^{10}} - 7{x^8} + 5{x^6} - 21{x^3} + 3{x^2} - 7$

$\therefore$ $f(1 - h) = 3{(1 - h)^{10}} - 7{(1 - h)^8} + 5{(1 - h)^6} - 21{(1 - h)^3} + 3{(1 - h)^2} - 7$

$= 3(1 - 10h + 45{h^2} - 120{h^3} + ..... + {h^{10}}) - 7(1 - 8h + 28{h^2} - 56{h^3} + ..... + {h^8}) + 5(1 - 6h + 15{h^2} - 20{h^3} + ..... + {h^6}) - 21(1 - 3h + 3{h^2} - {h^3}) + 3(1 - 2h + {h^2}) - 7$

$\Rightarrow f(1 - h) = - 24 + 53h + {h^2}( - 46) + {h^3}( - 47) + ....$ and $f(1) = - 24$

$\therefore$ $\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over {{h^3} + 3h}}$

$= \mathop {\lim }\limits_{h \to 0} {{ - 24 + 53h + {h^2}( - 46) + {h^3}( - 47) + ... - ( - 24)} \over {h({h^2} + 3)}}$

$= \mathop {\lim }\limits_{h \to 0} {{53h + {h^2}( - 46) + {h^3}( - 47) + ....} \over {h({h^2} + 3)}}$

$= \mathop {\lim }\limits_{h \to 0} {{53 + h( - 46) + {h^2}( - 47) + ....} \over {{h^2} + 3}} = {{53} \over 3}$

We have,

${a_r} = {(\cos 2r\pi + i\sin 2r\pi )^{1/9}} = {e^{{{2\pi i} \over 9}}}$

Now, \left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right| = \left| {\matrix{ {{a^{{{2\pi i} \over 9}}}} & {{a^{{{4\pi i} \over 9}}}} & {{a^{{{6\pi i} \over 9}}}} \cr {{a^{{{8\pi i} \over 9}}}} & {{a^{{{10\pi i} \over 9}}}} & {{a^{{{12\pi i} \over 9}}}} \cr {{a^{{{14\pi i} \over 9}}}} & {{a^{{{16\pi i} \over 9}}}} & {{a^{{{18\pi i} \over 9}}}} \cr } } \right|

= {e^{{{2\pi i} \over 9}}} \times {e^{{{8\pi i} \over 9}}}\left| {\matrix{ 1 & {{e^{{{2\pi i} \over 9}}}} & {{e^{{{4\pi i} \over 9}}}} \cr 1 & {{e^{{{2\pi i} \over 9}}}} & {{e^{{{4\pi i} \over 9}}}} \cr {{e^{{{14\pi i} \over 9}}}} & {{e^{{{16\pi i} \over 9}}}} & {{e^{{{18\pi i} \over 9}}}} \cr } } \right|

$= {e^{{{2\pi i} \over 9}}} \times {e^{{{8\pi i} \over 9}}} \times 0$ ($\because$ R₁ and R₂ are identical)

$= 0$

We have,

$\int {{e^{\sin x}}} .\left( {{{x{{\cos }^3}x - \sin x} \over {{{\cos }^2}x}}} \right)dx = {e^{\sin x}}f(x) + c$

$\Rightarrow \int {{e^{\sin x}}} (x\cos x - \sec x\tan x)dx = {e^{\sin x}}f(x) + c$

$\Rightarrow \int {{e^{\sin x}}} (x\cos x - 1 + 1 - \sec x\tan x)dx = {e^{\sin x}}f(x) + c$

$\Rightarrow \int {[{e^{\sin x}}\cos x(x - \sec x) + {e^{\sin x}}(1 - \sec x\tan x)]dx = {e^{\sin x}}f(x) + c}$

$\Rightarrow \int {{d \over {dx}}\{ {e^{\sin x}}(x - \sec x)\} dx = {e^{\sin x}}f(x) + c}$

$\Rightarrow {e^{\sin x}}(x - \sec x) = {e^{\sin x}}f(x) + c$

$\Rightarrow f(x) = x - \sec x$

Number of ways of selection of n objects from 2n objects, where as n objects are identical in out of 2n objects.

n identical and no different object = 1 ways = ${}^n{C_0}$

n $-$ 1 identical and 1 different object = 1 $\times$ ${}^n{C_1}$

n $-$ 2 identical and 2 different object = 1 $\times$ ${}^n{C_2}$

0 identical and n different objects = 1 $\times$ ${}^n{C_n}$ = ${}^n{C_0}$ + ${}^n{C_1}$ + ${}^n{C_2}$ + ... + ${}^n{C_n}$ = 2ⁿ

We have,

f : R $\to$ R be a twice continuously differentiable function such that f(0) = f(1) = f'(0) = 0

Now, for at least one value of c₁$\in$(0, 1),

f'(c₁) = 0 (by Rolle's theorem)

Again, f'(0) = 0 = f'(c₁)

$\Rightarrow$ f'(c) = 0 for some c$\in$(0, c₁) (by Rolle's theorem)

We have,

equation of parabola y = x²

Let P($\alpha$ , $\alpha$²) is a point on the parabola.

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

$\because$ $\left[ {{{dy} \over {dx}} = 2x \Rightarrow {{dy} \over {d{x_{(\alpha ,{\alpha ^2})}}}} = 2\alpha } \right]$

$\Rightarrow y = 2\alpha x - {\alpha ^2}$

Also, given $y = - {x^2} + 4x - 4$

$\therefore$ $- {x^2} + 4x - 4 = 2\alpha x - {\alpha ^2}$

$\Rightarrow {x^2} + 2x(\alpha - 2) + (4 - {\alpha ^2}) = 0$

Discriminant = 0

$4{(\alpha - 2)^2} - 4(4 - {\alpha ^2}) = 0$

$\Rightarrow {(\alpha - 2)^2} - (4 - {\alpha ^2}) = 0$

$\Rightarrow {\alpha ^2} - 4\alpha + 4 - 4 + {\alpha ^2} = 0$

$\Rightarrow {\alpha ^2} - 2\alpha = 0$

$\Rightarrow \alpha = 0,\,\alpha = 2$

2b = (n + 2)th form

$= a + (n + 2 - 1)d$

$\Rightarrow 2b = a + (n + 1)d$

$\Rightarrow d = {{2b - a} \over {n + 1}}$$\therefore$ m th mean = $a + m\left( {{{2b - a} \over {n + 1}}} \right)$

and also, b = (n + 2)th form

$= 2a + (n + 2 - 1)d$

$= 2a + (n + 1)d$

$\Rightarrow d = {{b - 2a} \over {n + 1}}$

$\therefore$ m th mean = $2a + m\left( {{{b - 2a} \over {n + 1}}} \right)$

According to the question,

$a + m\left( {{{2b - a} \over {n + 1}}} \right) = 2a + m\left( {{{b - 2a} \over {n + 1}}} \right)$

$\Rightarrow {a \over b} = {m \over {n + 1 - m}}$

We have, $x = {1 \over 2}vt$

$\Rightarrow x = {1 \over 2}{{dx} \over {dt}}t$ $\because$ $\left[ {v = {{dx} \over {dt}}} \right]$

$\Rightarrow {{2dt} \over t} = {{dx} \over x}$

$\Rightarrow 2.\int {{{dt} \over t} = } \int {{{dx} \over x}}$

$\Rightarrow 2\log |t| + \log |c| = \log |x|$

$\Rightarrow \log ({t^2}.\,c) = \log x$

$\Rightarrow x = {t^2}c$

$\Rightarrow {{dx} \over {dt}} = 2tc$

$\Rightarrow {{{d^2}x} \over {d{t^2}}} = 2c$

$\Rightarrow$ acceleration f is constant.

Given, ${y^2} = 2d(x + \sqrt d )$ ... (i)

$\Rightarrow$ 2y y₁ = 2d $\Rightarrow$ d = y y₁

From Eq. (i),

${y^2} = 2y\,{y_1}(x + \sqrt {y\,{y_1}} )$

$\Rightarrow {y^2} - 2y\,{y_1}x = \sqrt {y\,{y_1}} .2y\,{y_1}$

$\Rightarrow {({y^2} - 2y\,{y_1}x)^2} = 4{(y\,{y_1})^3}$

So, degree of above equation is 3.

Given, $M = \int\limits_0^{\pi /2} {{{\cos x} \over {x + 2}}dx}$ and $N = \int\limits_0^{\pi /4} {{{\sin x\cos x} \over {{{(x + 1)}^2}}}dx}$

$= \int\limits_0^{\pi /4} {{1 \over 2}.{{\sin 2x} \over {{{(x + 1)}^2}}}dx}$

Put $2x = t \Rightarrow dx = {{dt} \over 2}$

$\therefore$ $N = \int_0^{\pi /2} {{{\sin t} \over {4{{(t/2 + 1)}^2}}}dt}$

$= \int_0^{\pi /2} {{{\sin t} \over {{{(t + 2)}^2}}}dt}$

$\therefore$ $M - N = \int_0^{\pi /2} {(\cos x).{1 \over {(x + 2)}}dx} - \int_0^{\pi /2} {{{\sin x} \over {{{(x + 2)}^2}}}dx}$

$= \left( {{{\sin x} \over {x + 2}}} \right)_0^{\pi /2} - \int_0^{\pi /2} { - {{\sin x} \over {{{(x + 2)}^2}}}dx} - \int_0^{\pi /2} {{{\sin x} \over {{{(x + 2)}^2}}}dx}$

$= {{\sin \pi /2} \over {\pi /2 + 2}} = {1 \over {{{\pi + 4} \over 2}}} = {2 \over {\pi + 4}}$