IIT JEE 1997 Mathematics Question Paper with Answer and Solution

(A) Given equation: $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$.
Squaring both sides: $(x + 1) + (x - 1) - 2\sqrt{(x + 1)(x - 1)} = 4x - 1$.
$2x - 2\sqrt{x^2 - 1} = 4x - 1$.
$-2\sqrt{x^2 - 1} = 2x - 1$.
Squaring both sides again: $4(x^2 - 1) = (2x - 1)^2$.
$4x^2 - 4 = 4x^2 - 4x + 1$.
$-4 = -4x + 1$.
$4x = 5 \implies x = 5/4$.
Checking $x = 5/4$ in the original equation: $\sqrt{5/4 + 1} - \sqrt{5/4 - 1} = \sqrt{9/4} - \sqrt{1/4} = 3/2 - 1/2 = 1$.
However,the $RHS$ is $\sqrt{4(5/4) - 1} = \sqrt{5 - 1} = \sqrt{4} = 2$.
Since $1 \neq 2$,the value $x = 5/4$ is an extraneous solution.
Therefore,the equation has no solution.

(A) Given that $\cos (\theta - \alpha ), \cos \theta, \cos (\theta + \alpha )$ are in $H.P.$
Therefore,their reciprocals $\frac{1}{\cos (\theta - \alpha )}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta + \alpha )}$ are in $A.P.$
This implies $\frac{2}{\cos \theta} = \frac{1}{\cos (\theta - \alpha )} + \frac{1}{\cos (\theta + \alpha )}$.
Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$,we get $\frac{2}{\cos \theta} = \frac{\cos (\theta + \alpha ) + \cos (\theta - \alpha )}{\cos (\theta - \alpha ) \cos (\theta + \alpha )} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
Simplifying,$\frac{1}{\cos \theta} = \frac{\cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
$\cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$.
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin^2 \alpha = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$,we have $\cos^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$.
$\cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$.
$\cos^2 \theta \sec^2 \frac{\alpha}{2} = 2$.
Taking the square root,$\cos \theta \sec \frac{\alpha}{2} = \pm \sqrt{2}$.

(D) Let $y = \cos x \cos(x + 2) - \cos^2(x + 1)$.
Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$,we have:
$y = \frac{1}{2}[\cos(x - (x + 2)) + \cos(x + x + 2)] - \cos^2(x + 1)$
$y = \frac{1}{2}[\cos(-2) + \cos(2x + 2)] - \cos^2(x + 1)$
Since $\cos(-2) = \cos 2$,we get:
$y = \frac{1}{2}\cos 2 + \frac{1}{2}\cos(2(x + 1)) - \cos^2(x + 1)$
Using $\cos(2\theta) = 2\cos^2 \theta - 1$:
$y = \frac{1}{2}\cos 2 + \frac{1}{2}(2\cos^2(x + 1) - 1) - \cos^2(x + 1)$
$y = \frac{1}{2}\cos 2 + \cos^2(x + 1) - \frac{1}{2} - \cos^2(x + 1)$
$y = \frac{1}{2}(\cos 2 - 1)$
Using $1 - \cos 2 = 2\sin^2 1$,we get $y = -\sin^2 1$.
This represents a straight line parallel to the $x$-axis,passing through $(\frac{\pi}{2}, -\sin^2 1)$.

(B) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
Let the foci be $F(ae, 0)$ and $F'(-ae, 0)$ and the end point of the minor axis be $B(0, b)$.
The center of the ellipse is $C(0, 0)$.
Given that $\angle FBF' = \frac{\pi}{2}$.
Since $\triangle FBF'$ is an isosceles triangle with $BF = BF'$,the altitude $BC$ bisects $\angle FBF'$.
Therefore,$\angle FBC = \frac{1}{2} \angle FBF' = \frac{\pi}{4}$.
In the right-angled triangle $\triangle BCF$,$\tan(\angle FBC) = \frac{CF}{BC}$.
$\tan(\frac{\pi}{4}) = \frac{ae}{b}$.
$1 = \frac{ae}{b} \Rightarrow b = ae$.
Squaring both sides,$b^2 = a^2 e^2$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^2$.
$1 - e^2 = e^2$.
$2e^2 = 1$.
$e^2 = 1/2$.
$e = 1/\sqrt{2}$.

(B) Given that $p, q, r$ are mutually perpendicular vectors of the same magnitude,let $|p| = |q| = |r| = c$.
Thus,$p \cdot q = q \cdot r = r \cdot p = 0$ and $p \cdot p = q \cdot q = r \cdot r = c^2$.
Using the vector triple product identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$,we expand each term:
$p \times \{(x - q) \times p\} = (p \cdot p)(x - q) - (p \cdot (x - q))p = c^2(x - q) - (p \cdot x)p$.
Similarly,$q \times \{(x - r) \times q\} = c^2(x - r) - (q \cdot x)q$ and $r \times \{(x - p) \times r\} = c^2(x - p) - (r \cdot x)r$.
Summing these,we get:
$c^2(x - q + x - r + x - p) - [(p \cdot x)p + (q \cdot x)q + (r \cdot x)r] = 0$.
$c^2(3x - (p + q + r)) - [(p \cdot x)p + (q \cdot x)q + (r \cdot x)r] = 0$.
If we substitute $x = \frac{1}{2}(p + q + r)$,then $p \cdot x = \frac{1}{2}c^2$,$q \cdot x = \frac{1}{2}c^2$,and $r \cdot x = \frac{1}{2}c^2$.
Substituting these into the equation:
$c^2(3(\frac{1}{2}(p + q + r)) - (p + q + r)) - [\frac{1}{2}c^2 p + \frac{1}{2}c^2 q + \frac{1}{2}c^2 r] = 0$.
$c^2(\frac{3}{2}(p + q + r) - (p + q + r)) - \frac{1}{2}c^2(p + q + r) = 0$.
$c^2(\frac{1}{2}(p + q + r)) - \frac{1}{2}c^2(p + q + r) = 0$.
This confirms $x = \frac{1}{2}(p + q + r)$ is the solution.

(D) Given $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$.
Since the derivative of a determinant is the sum of determinants obtained by differentiating one row at a time,and the second and third rows are constants,the third derivative $f'''(x)$ is given by differentiating the first row three times:
$f'''(x) = \left| \begin{array}{ccc} \frac{d^3}{dx^3}(x^3) & \frac{d^3}{dx^3}(\sin x) & \frac{d^3}{dx^3}(\cos x) \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$
Calculating the derivatives:
$\frac{d^3}{dx^3}(x^3) = 6$
$\frac{d^3}{dx^3}(\sin x) = -\cos x$
$\frac{d^3}{dx^3}(\cos x) = \sin x$
Thus,$f'''(x) = \left| \begin{array}{ccc} 6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$.
At $x = 0$:
$f'''(0) = \left| \begin{array}{ccc} 6 & -\cos(0) & \sin(0) \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right| = \left| \begin{array}{ccc} 6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$.
Since the first two rows are identical,the value of the determinant is $0$.
Therefore,the result is $0$,which is independent of $p$.

(C) For $f(x) = \frac{x}{\sin x}$,we have $f'(x) = \frac{\sin x - x \cos x}{\sin^2 x} = \frac{\cos x(\tan x - x)}{\sin^2 x}$.
Since $0 < x \le 1$ (in radians),we know that $\tan x > x$ and $\cos x > 0$. Thus,$f'(x) > 0$,which implies $f(x)$ is an increasing function.
For $g(x) = \frac{x}{\tan x} = x \cot x$,we have $g'(x) = \cot x - x \csc^2 x = \frac{\sin x \cos x - x}{\sin^2 x} = \frac{\sin 2x - 2x}{2 \sin^2 x}$.
Let $h(x) = \sin 2x - 2x$. Then $h'(x) = 2 \cos 2x - 2 = 2(\cos 2x - 1)$. Since $\cos 2x < 1$ for $x \in (0, 1]$,$h'(x) < 0$.
Since $h(0) = 0$ and $h(x)$ is decreasing,$h(x) < 0$ for $x > 0$. Therefore,$g'(x) < 0$,which implies $g(x)$ is a decreasing function.
Thus,$f(x)$ is an increasing function.

(B) Let $L = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $.
We can rewrite the expression as $L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{2n} {\frac{1}{n} \cdot \frac{r/n}{{\sqrt {1 + {{(r/n)}^2}} }}} $.
Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{kn} {f(r/n)} = \int_0^k {f(x)dx}$.
Here,$f(x) = \frac{x}{{\sqrt {1 + {x^2}} }}$ and the upper limit is $k = 2$.
So,$L = \int_0^2 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$.
Let $u = 1 + x^2$,then $du = 2x dx$,or $x dx = \frac{1}{2} du$.
When $x = 0$,$u = 1$. When $x = 2$,$u = 5$.
$L = \int_1^5 {\frac{1}{{2\sqrt u }}} du = \left[ {\sqrt u } \right]_1^5 = \sqrt{5} - 1$.

(C) Given $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$.
Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx$,we have $a + b = 1 - k + k = 1$.
Thus,$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)(1 - (1 - x))\} dx$.
$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)x\} dx$.
$I_1 = \int_{1 - k}^k f\{x(1 - x)\} dx - \int_{1 - k}^k x f\{x(1 - x)\} dx$.
$I_1 = I_2 - I_1$.
$2I_1 = I_2$.
Therefore,$I_1/I_2 = 1/2$.

(B) Let the determinant be $\Delta$. Apply the column transformation $C_1 \to C_1 + C_3$.
Using the formula $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$ and $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$,we get:
$C_1 = \begin{bmatrix} 1+a^2 \\ 2\cos px \cos dx \\ 2\sin px \cos dx \end{bmatrix}$.
Now,apply $C_1 \to C_1 - 2\cos dx \cdot C_2$:
$C_1 = \begin{bmatrix} 1+a^2 - 2a\cos dx \\ 2\cos px \cos dx - 2\cos px \cos dx \\ 2\sin px \cos dx - 2\sin px \cos dx \end{bmatrix} = \begin{bmatrix} 1+a^2 - 2a\cos dx \\ 0 \\ 0 \end{bmatrix}$.
Expanding along the first column:
$\Delta = (1+a^2 - 2a\cos dx) \cdot [\cos px \sin(p+d)x - \sin px \cos(p+d)x]$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\Delta = (1+a^2 - 2a\cos dx) \cdot \sin((p+d)x - px) = (1+a^2 - 2a\cos dx) \sin dx$.
Since the final expression $(1+a^2 - 2a\cos dx) \sin dx$ does not contain $p$,the determinant is independent of $p$.

(A) Given $g(x) = \int_0^x \cos^4 t \,dt$.
We need to find $g(x+\pi) = \int_0^{x+\pi} \cos^4 t \,dt$.
Using the property of definite integrals, $\int_0^{x+\pi} f(t) \,dt = \int_0^x f(t) \,dt + \int_x^{x+\pi} f(t) \,dt$.
Thus, $g(x+\pi) = g(x) + \int_x^{x+\pi} \cos^4 t \,dt$.
Since $\cos^4 t$ is a periodic function with period $\pi$, the integral over any interval of length $\pi$ is equal to the integral over $[0, \pi]$.
Therefore, $\int_x^{x+\pi} \cos^4 t \,dt = \int_0^\pi \cos^4 t \,dt = g(\pi)$.
Hence, $g(x+\pi) = g(x) + g(\pi)$.