JEE Advanced 2015 · Paper 1 · Mathematics Q59

MathematicsMatrix match+8 / -4Question 59 of 60

Match the entries in Column I with the entries in Column II.

Column I

A
In $\mathbb R^2$ , if the magnitude of the projection vector of the vector $\alpha\hat i+\beta\hat j$ on $\sqrt3\hat i+\hat j$ is $\sqrt3$ and if $\alpha=2+\sqrt3\beta$ , then possible value(s) of $|\alpha|$ is (are)
B
Let $a$ and $b$ be real numbers such that the function $f(x)=\begin{cases}-3ax^2-2,&x<1\\bx+a^2,&x\ge1\end{cases}$ is differentiable for all $x\in\mathbb R$ . Then possible value(s) of $a$ is (are)
C
Let $\omega\ne1$ be a complex cube root of unity. If $(3-3\omega+2\omega^2)^{4n+3}+(2+3\omega-3\omega^2)^{4n+3}+(-3+2\omega+3\omega^2)^{4n+3}=0$ , then possible value(s) of $n$ is (are)
D
Let the harmonic mean of two positive real numbers $a$ and $b$ be 4. If $q$ is a positive real number such that $a,5,q,b$ is an arithmetic progression, then the value(s) of $|q-a|$ is (are)

Column II

Answer and worked solution

Official answer: A → P, Q; B → P, Q; C → P, Q, S, T; D → Q, T

(A) Projection. $\dfrac{|\sqrt3\alpha+\beta|}2=\sqrt3$ with $\alpha=2+\sqrt3\beta$ :
$|2\sqrt3+4\beta|=2\sqrt3\implies\beta=0\text{ or }-\sqrt3,$
so $\alpha=2$ or $-1$ . Then $|\alpha|=1,2$ : P, Q.

(B) Differentiability at 1.

Continuity: $-3a-2=b+a^2$ .
Equal derivatives: $-6a=b$ .

These give $a^2-3a+2=0$ , so $a=1,2$ : P, Q.

(C) Cube roots. The second and third bases are $\omega$ and $\omega^2$ times the first. The sum is $z^{4n+3}\left(1+\omega^{4n+3}+\omega^{2(4n+3)}\right)$ , which is zero exactly when $3\nmid n$ . So $n=1,2,4,5$ : P, Q, S, T.

(D) HM and AP. Write $a=5-d$ , $b=5+2d$ and $\frac{2ab}{a+b}=4$ . This gives $2d^2-3d-5=0$ , so $d=\frac52$ or $-1$ (both keep all terms positive). Then $|q-a|=2|d|=5$ or 2: Q, T.

Answer: A → P, Q; B → P, Q; C → P, Q, S, T; D → Q, T

Topics:Dot product Differentiability Roots of unity Harmonic progression

← Previous question Next question →

JEE Advanced 2015 · Paper 1, Mathematics question 59