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1

JEE Advanced 2020 Paper 2 Offline

Numerical
Let the function $$f:(0,\pi ) \to R$$ be defined by $$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$$

Suppose the function f has a local minimum at $$\theta $$ precisely when $$\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \} $$, where $$0 < {\lambda _1} < ...{\lambda _r} < 1$$. Then the value of $${\lambda _1} + ... + {\lambda _r}$$ is .............
Your Input ________

Answer

Correct Answer is 0.5

Explanation

The given function f : R $$ \to $$ R be defined by

$$f(\theta ) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$$

$$ = 1 + \sin 2\theta + {(1 - \sin 2\theta )^2}$$

$$ = 1 + \sin 2\theta + 1 + {\sin ^2}2\theta - 2\sin 2\theta $$

$$ = {\sin ^2}2\theta - \sin 2\theta + 2$$

$$ = {\left( {\sin 2\theta - {1 \over 2}} \right)^2} + {7 \over 4}$$

The local minimum of function 'f' occurs when

$$\sin 2\theta = {1 \over 2}$$

$$ \Rightarrow 2\theta = {\pi \over 6},\,{{5\pi } \over 6},\,{{13\pi } \over 6},\,...$$

$$ \Rightarrow \theta = {\pi \over {12}},\,{{5\pi } \over {12}},\,{{13\pi } \over {12}},\,...$$

but $$\theta \in \{ {\lambda _1}\pi ,\,{\lambda _2}\pi ,\,...,\,{\lambda _r}\pi \} $$,

where $$0 < {\lambda _1} < .... < {\lambda _r} < 1$$.

$$ \therefore $$ $$\theta = {\pi \over {12}},\,{{5\pi } \over {12}}$$

So, $${\lambda _1} + ... + {\lambda _r} = {1 \over {12}} + {5 \over {12}} = 0.50$$
2

JEE Advanced 2020 Paper 2 Offline

Numerical
Let the function f : [0, 1] $$ \to $$ R be defined by

$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$

Then the value of $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$ is ..........
Your Input ________

Answer

Correct Answer is 19

Explanation

The given function f : [0, 1] $$ \to $$ R be define by

$$f(x) = {{{4^x}} \over {{4^x} + 2}}$$

$$ \because $$ $$f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {2 \over {2 + {4^x}}}$$

$$ \therefore $$ $$f(x) + f(1 - x) = {{{4^x}} \over {{4^x} + 2}} + {2 \over {2 + {4^x}}}$$

$$ = {{{4^x} + 2} \over {{4^x} + 2}}$$

So, f(x) + f(1 $$-$$ x) = 1 .....(i)

$$ \therefore $$ $$f\left( {{1 \over {40}}} \right) + f\left( {{2 \over {40}}} \right) + f\left( {{3 \over {40}}} \right) + ... + f\left( {{{39} \over {40}}} \right) - f\left( {{1 \over 2}} \right)$$

$$ = \left[ {f\left( {{1 \over {40}}} \right) + f\left( {{{39} \over {40}}} \right)} \right] + \left[ {f\left( {{2 \over {40}}} \right) + f\left( {{{38} \over {40}}} \right)} \right] + ... + \left[ {f\left( {{{18} \over {40}}} \right) + f\left( {{{22} \over {40}}} \right)} \right] + \left[ {f\left( {{{19} \over {40}}} \right) + f\left( {{{21} \over {40}}} \right)} \right] + \left[ {f\left( {{{20} \over {40}}} \right) - f\left( {{1 \over 2}} \right)} \right]$$

$$ = \{ 1 + 1 + ... + 1 + 1\} + f\left( {{1 \over 2}} \right) - f\left( {{1 \over 2}} \right)$$

$$ = \{ 1 + 1 + ... + 1 + 1\}$$(19 times) {from Eq. (i)}

= 19.
3

JEE Advanced 2020 Paper 1 Offline

Numerical
For a polynomial g(x) with real coefficients, let mg denote the number of distinct real roots of g(x). Suppose S is the set of polynomials with real coefficients defined by

$$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $$;

For a polynomial f, let f' and f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (mf' + mf''), where f $$ \in $$ S, is ..............
Your Input ________

Answer

Correct Answer is 5

Explanation

Given set S of polynomials with real coefficients

$$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $$

and for a polynomial $$f \in S$$, Let

$$f(x) = {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3})$$

it have $$-$$1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = $$-$$1 and x = 1, so f'(x) having at least three roots x = $$-$$1, 1 and $$\alpha $$. Where $$\alpha $$$$ \in $$($$-$$1, 1) and f''(x) having at least two roots in interval ($$-$$1, 1)

So, mf' = 3 and mf'' = 2

$$ \therefore $$ Minimum possible value of (mf' + mf'') = 5
4

JEE Advanced 2020 Paper 1 Offline

Numerical
Let f : [0, 2] $$ \to $$ R be the function defined by

$$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$$

If $$\alpha ,\,\beta \in [0,2]$$ are such that $$\{ x \in [0,2]:f(x) \ge 0\} = [\alpha ,\beta ]$$, then the value of $$\beta - \alpha $$ is ..........
Your Input ________

Answer

Correct Answer is 1

Explanation

The given function f : [0, 2] $$ \to $$ R defined by

$$f(x) = (3 - \sin (2\pi x))\sin \left( {\pi x - {\pi \over 4}} \right) - \sin \left( {3\pi x + {\pi \over 4}} \right)$$

$$ = (3 - \sin (2\pi x))\left[ {{{\sin \pi x} \over {\sqrt 2 }} - {{\cos \pi x} \over {\sqrt 2 }}} \right] - \left\{ {{{\sin 3\pi x} \over {\sqrt 2 }} + {{\cos (3\pi x)} \over {\sqrt 2 }}} \right\}$$

$$ = (3 - \sin (2\pi x)){{[\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}[3\sin (\pi x) - 4{\sin ^3}(\pi x) + 4{\cos ^3}(\pi x) - 3\cos (\pi x)]$$

$$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[3 - \sin (2\pi x) - 3 + 4\{ {\sin ^2}(\pi x) + {\cos ^2}(\pi x) + \sin (\pi x)\cos (\pi x)\} ]$$

$$ = {{\sin (\pi x) - \cos (\pi x)} \over {\sqrt 2 }}[4 + \sin (2\pi x)]$$

As, $$f(x) \ge 0\forall \in [\alpha ,\beta ]$$, where $$\alpha ,\beta \in [0,2]$$, so

$$\sin (\pi x) - \cos (\pi x) \ge 0$$

as $$4 + \sin (2\pi x) > 0\,\forall x \in R$$.

$$ \Rightarrow \pi x \in \left[ {{\pi \over 4},{{5\pi } \over 4}} \right] \Rightarrow x \in \left[ {{1 \over 4},{5 \over 4}} \right]$$

$$ \therefore $$ $$\alpha = {1 \over 4}$$ and $$\beta = {5 \over 4}$$

Therefore the value of $$(\beta - \alpha ) = 1$$

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