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1

JEE Advanced 2021 Paper 1 Online

Numerical
For x $$\in$$ R, the number of real roots of the equation $$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$$ is ________.
Your Input ________

Answer

Correct Answer is 4

Explanation

Given,

$$3{x^2} - 4\left| {{x^2} - 1} \right| + x - 1 = 0$$ .... (i)


For $$-$$1 $$\le$$ x $$\le$$ 1 i.e., x$$\in$$[$$-$$1, 1]

From Eq. (i), we get

$$3{x^2} - 4( - {x^2} + 1) + x - 1 = 0$$

$$ \Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$$

$$ \Rightarrow 7{x^2} + x - 5 = 0$$

$$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$$

Here, both values of x are acceptable.

For | x | > | i.e. x $$\in$$($$-$$ $$\infty$$, $$-$$1) $$\cup$$ (1, $$\infty$$)

From Eq. (i), we get

$$3{x^2} - 4({x^2} - 1) + x - 1 = 0$$

$$ \Rightarrow {x^2} - x - 3 = 0$$

$$ \Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$$

Again here, both values of x are acceptable.

Hence, total number of solutions is 4.
2

JEE Advanced 2018 Paper 1 Offline

Numerical
Let a, b, c three non-zero real numbers such that the equation $$\sqrt 3 a\cos x + 2b\sin x = c,x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$, has two distinct real roots $$\alpha $$ and $$\beta $$ with $$\alpha + \beta = {\pi \over 3}$$. Then, the value of $${b \over a}$$ is ............
Your Input ________

Answer

Correct Answer is 0.5

Explanation

We have, $$\alpha $$, $$\beta $$ are the roots of

$$\sqrt 3 a\cos x + 2b\sin x = c$$

$$ \therefore $$ $$\sqrt 3 a\cos \alpha + 2b\sin \alpha = c$$ ... (i)

and $$\sqrt 3 a\cos \beta + 2b\sin \beta = c$$ ... (ii)

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

$$\sqrt 3 a(\cos \alpha - \cos \beta ) + 2b(\sin \alpha - \sin \beta ) = 0$$

$$ \Rightarrow \sqrt 3 a\left( { - 2\sin \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) + 2b\left( {2\cos \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) = 0$$

$$ \Rightarrow \sqrt 3 a\sin \left( {{{\alpha + \beta } \over 2}} \right) = 2b\cos \left( {{{\alpha + \beta } \over 2}} \right)$$

$$ \Rightarrow \tan \left( {{{\alpha + \beta } \over 2}} \right) = {{2b} \over {\sqrt 3 a}}$$

$$ \Rightarrow \tan \left( {{\pi \over 6}} \right) = {{2b} \over {\sqrt 3 a}}$$ [$$ \because $$ $$\alpha $$ + $$\beta $$ = $${\pi \over 3}$$, given]

$$ \Rightarrow {1 \over {\sqrt 3 }} = {{2b} \over {\sqrt 3 a}} \Rightarrow {b \over a} = {1 \over 2}$$

$$ \Rightarrow {b \over a} = 0.5$$
3

IIT-JEE 2012 Paper 1 Offline

Numerical

The value of $$6 + {\log _{3/2}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$$ is __________.

Your Input ________

Answer

Correct Answer is 4

Explanation

$$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$$

Let $$\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}} \sqrt {...} } = y$$

$$\therefore$$ $$y = \sqrt {4 - {1 \over {3\sqrt 2 }}y} $$

$$ \Rightarrow {y^2} + {1 \over {3\sqrt 2 }}y - 4 = 0$$

$$ \Rightarrow 3\sqrt 2 {y^2} + y - 12\sqrt 2 = 0$$

$$\therefore$$ $$y = {{ - 1 \pm 17} \over {6\sqrt 2 }}$$ or $$y = {8 \over {3\sqrt 2 }}$$

Now,

$$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.y} \right) = 6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.{8 \over {3\sqrt 2 }}} \right)$$

$$ = 6 + {\log _{{3 \over 2}}}\left( {{4 \over 9}} \right) = 6 + {\log _{{3 \over 2}}}{\left( {{3 \over 2}} \right)^{ - 2}}$$

$$ = 6 - 2.{\log _{{3 \over 2}}}\left( {{3 \over 2}} \right) = 4$$

4

IIT-JEE 2011 Paper 2 Offline

Numerical
The number of distinct real roots of $${x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$$
Your Input ________

Answer

Correct Answer is 2

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